Calculus in three dimensions and applications

Calculus in Three Dimensions and Applications Eamonn Gaffney. Based extensively on Lecture Notes by Richard Earl. Trinity Term 2012

Syllabus 1. Div, grad and curl in Euclidean coordinates. 2. Volume, surface and line integrals. 3. Stokes’ Theorem and the Divergence Theorem (proofs excluded). Illustration by rederivation of models studied from Hilary Term, for example, the heat equation from the Divergence Theorem. 4. Gravity as a conservative force. Gauss’s Theorem. The equivalence of Poisson’s equation and the inverse-square law.

Recommended Texts • Erwin Kreyszig, Advanced Engineering Mathematics (Wiley, 8th Edition, 1999). • D. E. Bourne & P. C. Kendall, Vector Analysis and Cartesian Tensors (Stanley Thornes, 1992). • H. M. Schey, Div, Grad and Curl and all that (W. W. Norton, Third Edition, 1996).

Further Reading • Mary L. Boas, Mathematical Models in the Physical Sciences (Wiley 2nd Edition, 1983). • Gerald B. Folland Advanced Calculus (Prentice Hall, 2002)

Website https://www.maths.ox.ac.uk/courses/material

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1. DIV, GRAD AND CURL Definition 1 By a scalar field φ on R3 we shall mean a map φ : R3 → R. Definition 2 By a vector field F on R3 we shall mean a map F : R3 → R3 . • We will typically assume that scalar and vector fields are smooth – their partial derivatives exist with respect to x, y and z to all orders – though for brevity this will not always be stated. • Occasionally we may consider more general scalar fields φ : Rn → R and vector fields F : Rn → Rm . Example 3 The temperature in the lecture theatre at any given instant is a scalar field T (x, y, z) . At the same instant the velocity of the air flow would be a vector field v (x, y, z) . If we include time as a fourth co-ordinate then T and v would now be fields on a subset of R4 . 3 Example 4 The three co-ordinates x, y, z are each scalar p fields on R . The position vector 2 2 2 r = (x, y, z) is a vector field; its magnitude r = |r| = x + y + z is a scalar field, though note it is not smooth at (0, 0, 0) .

We shall also consider scalar and vector fields defined on proper subsets of R3 (or more generally Rn ). The domains of these fields will usually be open, so that we can define their partial derivatives. Definition 5 A set U ⊆ R3 is said to be open if for every x ∈ U there exists ε > 0 such that  B (x; ε) = y ∈ R3 : |y − x| < ε ⊆ U . We refer to B (x; ε) as the open ball of radius ε about x. We will move straight on to defining three differential operators div, grad and curl and introduce some examples and identities. In due course, we will look to make more sense of their definitions physically. Definition 6 (Gradient) Let φ : R3 → R be a scalar field. Then the gradient of φ, written grad φ, is  grad φ =

∂φ ∂φ ∂φ , , ∂x ∂y ∂z

 .

Note that grad takes scalar fields to vector fields. DIV, GRAD AND CURL

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Definition 7 (Divergence) Let F : R3 → R3 be a vector field where F (x, y, z) = (F1 (x, y, z) , F2 (x, y, z) , F3 (x, y, z)) , with respect to a Cartesian coordinate system. Then the divergence of F, written div F, is

div F =

∂F1 ∂F2 ∂F3 + + . ∂x ∂y ∂z

Note that div takes vector fields to scalar fields. Definition 8 (Curl) Let F : R3 → R3 be a vector field where F (x, y, z) = (F1 (x, y, z) , F2 (x, y, z) , F3 (x, y, z)) . with respect to a Cartesian coordinate system. Then the curl of F, written curl F, is i   j k ∂ ∂ ∂ ∂F ∂F ∂F ∂F ∂F ∂F 3 2 1 3 2 1 curl F = ∂x ∂y ∂z = − , − , − . ∂y ∂z ∂z ∂x ∂x ∂y F1 F2 F3 Note that curl takes vector fields to vector fields. p Example 9 Let r = (x, y, z) and r = |r| = x2 + y 2 + z 2 . Then ∂x + ∂x i ∂ curl r = ∂x x div r =

∂y ∂z + = 3; ∂y ∂z j k ∂ ∂ ∂y ∂z = 0, y z

and if r 6= 0 then grad r =

x y z p p ,p 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2

!

r = . r

Example 10 With r and r as above, and f a differentiable function on (0, ∞) note i j k ∂ ∂ ∂ curl (f (r) r) = ∂x ∂y ∂z f (r) x f (r) y f (r) z   ∂r ∂r ∂r ∂r ∂r ∂r 0 0 0 0 0 0 = zf (r) − yf (r) , xf (r) − zf (r) , yf (r) − xf (r) ∂y ∂z ∂z ∂x ∂x ∂y  zy yz xz zx yx xy  = f 0 (r) − , − , − = 0. r r r r r r 4

DIV, GRAD AND CURL

Example 11 Let c = (c1 , c2 , c3 ) be a constant vector. Then ∂ ∂ ∂ (c2 z − c3 y) + (c3 x − c1 z) + (c1 y − c2 x) = 0; ∂x ∂y ∂z grad (c · r) = grad (c1 x + c2 y + c3 z) = (c1 , c2 , c3 ) = c; div (c ∧ r) =

i j k ∂ ∂ ∂ curl (c ∧ r) = ∂x ∂y ∂z c2 z − c3 y c3 x − c1 z c1 y − c2 x

= (2c1 , 2c2 , 2c3 ) = 2c.

Definition 12 The differential operator  ∇=

∂ ∂ ∂ , , ∂x ∂y ∂z



is called del or nabla. Notation 13 By a slight abuse of notation we can write     ∂φ ∂φ ∂φ ∂ ∂ ∂ grad φ = , , = , , φ; ∂x ∂y ∂z ∂x ∂y ∂z   ∂ ∂ ∂ ∂F1 ∂F2 ∂F3 + + = , , div F = · (F1 , F2 , F3 ) ; ∂x ∂y ∂z ∂x ∂y ∂z   ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 − , − , − curl F = ∂y ∂z ∂z ∂x ∂x ∂y  i  ∂ ∂j k ∂ ∂ ∂ ∂ = ∂x ∂y ∂z = , , ∧ (F1 , F2 , F3 ) . ∂x ∂y ∂z F1 F2 F3 So we often write grad φ = ∇φ,

div F = ∇ · F,

curl F = ∇ ∧ F.

(1.1)

Remark 14 The notation grad, div, curl and that of (1.1) are equally common in mathematical texts, and so these lecture notes and the problem sheets will deliberately make use of both notations. Remark 15 For any scalar field φ we have       ∂ ∂φ ∂ ∂φ ∂ ∂φ ∂ 2φ ∂ 2φ ∂ 2φ div (grad φ) = + + = + + 2, ∂x ∂x ∂y ∂y ∂z ∂z ∂x2 ∂y 2 ∂z which is the Laplacian. As div grad = ∇ · ∇ DIV, GRAD AND CURL

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then we often write ∇2 φ =

∂ 2φ ∂ 2φ ∂ 2φ + + 2. ∂x2 ∂y 2 ∂z

Also, for vector fields F = (F1 , F2 , F3 ) with respect to a Cartesian coordinate system, it is common to write
∇2 F = ∇2 F1 , ∇2 F2 , ∇2 F3 . Thus the Laplacian can both take scalar fields to scalar fields and vector fields to vector fields. Remark 16 Whilst ∇ looks, and occasionally behaves, like a standard vector it is very unwise to assume this behaviour of ∇ without proof; most standard vector identities are not true when they involve ∇. For example, the commutativity of the dot product does not extend: ∇ · F 6= F · ∇. The LHS is div F. The RHS at first glance doesn’t mean anything, it certainly isn’t a scalar. Rather   ∂ ∂ ∂ ∂ ∂ ∂ , , + F2 + F3 = F1 F · ∇ = (F1 , F2 , F3 ) · ∂x ∂y ∂z ∂x ∂y ∂z which is a differential operator. In fact F · ∇ is standard notation for it. When F is of unit size, then F · ∇ equals the directional derivative in the direction of F. In summary, we have the following table: operator div = ∇· grad = ∇ curl = ∇∧ div grad = ∇2 F·∇

maps: vector fields to scalar fields scalar fields to vector fields vector fields to vector fields scalar fields to scalar fields vector fields to vector fields scalar fields to scalar fields vector fields to vector fields

The next proposition shows that the formulae for calculating each of div, grad and curl do not depend on our choice i, j, k of a Cartesian co-ordinate system. Recall that: Definition 17 The vectors e1 , e2 , e3 are said to be an orthonormal basis of R3 if  1 if i = j, ei · ej = δ ij := 0 if i = 6 j. So the vectors are all of unit length and mutually perpendicular. Further it is said to be a right-handed orthonormal basis if e1 ∧ e2 = e3 and left-handed if e1 ∧ e2 = −e3 . So e1 = i, e2 = j, e3 = k is an example of a right-handed orthonormal basis. 6

DIV, GRAD AND CURL

Proposition 18 Let e1 , e2 , e3 be an orthonormal basis of R3 with associated co-ordinates X, Y, Z defined by the identity Xe1 + Y e2 + Ze3 = xi + yj + zk. (1.2) Then

∂ ∂ ∂ ∂ ∂ ∂ + e2 + e3 =i +j +k . ∂X ∂Y ∂Z ∂x ∂y ∂z Proof. From (1.2) we have that e1

x = (e1 · i) X + (e2 · i) Y + (e3 · i) Z, X = (e1 · i) x + (e1 · j) y + (e1 · k) z, and four other similar equations for y, z, Y, Z. It follows from the chain rule that ∂ ∂x ∂ ∂y ∂ ∂z ∂ ∂ ∂ ∂ = + + = (e1 · i) + (e1 · j) + (e1 · k) ∂X ∂X ∂x ∂X ∂y ∂X ∂z ∂x ∂y ∂z and hence ∂ ∂ ∂ + e2 + e3 ∂X ∂Y ∂Z   ∂ ∂ ∂ = e1 (e1 · i) + (e1 · j) + (e1 · k) + e2 [. . .] + e3 [. . .] ∂x ∂y ∂z ∂ ∂ ∂ + [. . .] + [. . .] = [e1 (e1 · i) + e2 (e2 · i) + e3 (e3 · i)] ∂x ∂y ∂z ∂ ∂ ∂ = i +j +k ; ∂x ∂y ∂z this last line follows from the orthonormality of the bases as e1

i = αe1 + βe2 + γe3 leads to α = e1 · i, β = e2 · i, γ = e3 · i, by dotting the equation with e1 , e2 , e3 respectively.

DIV, GRAD AND CURL

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Corollary 19 Let f : R3 → R3 be a vector field given by f = (f1 , f2 , f3 ), let φ : R3 → R be a scalar field and let e1 , e2 , e3 be a right-handed orthonormal basis of R3 . Suppose that F1 e1 + F2 e2 + F3 e3 = f1 i + f2 j + f3 k; Φ (X, Y, Z) = φ (x, y, z) . That is, F and Φ represent f and φ in the new co-ordinates X, Y, Z. Then ∂Φ ∂Φ ∂Φ e1 + e2 + e3 ; ∂X ∂Y ∂Z ∂F1 ∂F2 ∂F3 div f = + + ; ∂Y ∂Z ∂X e1 e2 e3 ∂ ∂ ∂ . curl f = ∂X ∂Y ∂Z F1 F2 F3

grad φ =

That is, we calculate grad, div and curl using the same formulae, irrespective of what righthanded orthonormal co-ordinate system we are using. Proof. The previous proposition showed that the formula for grad does not change under an orthonormal change of co-ordinates. From Geometry I we know that the formula for the dot product is similarly invariant as is the formula for the cross product provided the change is to another right-handed system. Hence the formulae for div = ∇· and curl = ∇∧ do not change when we change to another right-handed orthonormal system.

Certain adjustments to the above formulae need to be made for co-ordinate systems that aren’t orthonormal – e.g. cylindrical polar co-ordinates or spherical polar co-ordinates. The relevant formulae below are for reference only; certainly you are not expected to memorize them, and if needed in an exam there would be no presumption of their knowledge so that they would be included in a hint or given and you would be asked to derive them. The main point to take away from this is that the grad, div, curl formulae are different for co-ordinate systems that aren’t Euclidean. Definition 20 Cylindrical polar co-ordinates r, θ, z are defined by the identity r = (x, y, z) = (r cos θ, r sin θ, z) . If we note dr = (cos θ, sin θ, 0) dr + (−r sin θ, r cos θ, 0) dθ + (0, 0, 1) dz 8

DIV, GRAD AND CURL

then we may write dr = er dr + reθ dθ + ez dz where er = (cos θ, sin θ, 0) ,

eθ = (− sin θ, cos θ, 0) ,

ez = (0, 0, 1) .

Note, for any values of r, θ, z, the vectors er , eθ , ez make a right-handed orthonormal basis of R3 . Proposition 21 The formulae for grad, div, curl in terms of cylindrical polar co-ordinates of fields ψ (r, θ, z) and F = Fr er + Fθ eθ + Fz ez are 1 ∂ψ ∂ψ ∂ψ er + eθ + ez ; ∂r r ∂θ ∂z 1 ∂Fθ ∂Fz 1 ∂ (rFr ) + + ; ∇·F = r ∂r r ∂θ ∂z e re e 1 ∂r ∂ θ ∂z . ∇∧F = r ∂r ∂θ ∂z Fr rFθ Fz ∇ψ =

Proof. By way of example we shall prove only the first identity here. Recall that Cartesian co-ordinates are given in terms of cylindrical polar ones by x = r cos θ,

y = r sin θ,

z=z

Say φ (x, y, z) = ψ (r, θ, z) . Then ∂ψ ∂φ ∂x ∂φ ∂y ∂φ ∂z ∂φ ∂φ = + + = cos θ + sin θ, ∂r ∂x ∂r ∂y ∂r ∂z ∂r ∂x ∂y ∂ψ ∂φ ∂x ∂φ ∂y ∂φ ∂z ∂φ ∂φ = + + = − r sin θ + r cos θ, ∂θ ∂x ∂θ ∂y ∂θ ∂z ∂θ ∂x ∂y ∂ψ ∂φ = . ∂z ∂z Hence ∂ψ 1 ∂ψ ∂ψ er + eθ + ez ∂r r ∂θ ∂z    ∂φ ∂φ 1 ∂φ ∂φ ∂φ = cos θ + sin θ (cos θ, sin θ, 0) + − r sin θ + r cos θ (− sin θ, cos θ, 0) + k ∂x ∂y r ∂x ∂y ∂z     ∂φ ∂φ 2 ∂φ 2 ∂φ ∂φ 2 2 = cos θ + sin θ i + sin θ + cos θ j + k ∂x ∂x ∂y ∂y ∂z = ∇φ. DIV, GRAD AND CURL

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Definition 22 Spherical polar co-ordinates r, θ, φ are defined by the identity r = (x, y, z) = (r sin θ cos φ, r sin θ sin φ, r cos θ) . If we note dr equals (sin θ cos φ, sin θ sin φ, cos θ) dr + r (cos θ cos φ, cos θ sin φ, − sin θ) dθ + r sin θ (− sin φ, cos φ, 0) dφ then we may write dr = er dr + reθ dθ + r sin θeφ dφ where er = (sin θ cos φ, sin θ sin φ, cos θ) ,

eθ = (cos θ cos φ, cos θ sin φ, − sin θ) ,

eφ = (− sin φ, cos φ, 0) .

Note, for any values of r, θ, φ the vectors er , eθ , eφ , make a right-handed orthonormal basis. Proposition 23 The formulae for grad, div, curl in terms of spherical polar co-ordinates of fields ψ (r, φ, θ) and F = Fr er + Fφ eφ + Fθ eθ are 1 ∂ψ 1 ∂ψ ∂ψ er + eθ + eφ ; ∂r r ∂θ r sin θ ∂φ  
∂ 2 ∂ ∂ 1 r sin θFr + (r sin θFθ ) + (rFφ ) ; ∇·F = 2 r sin θ ∂r ∂θ ∂φ er reθ r sin θeφ 1 ∂ ∂ ∂ . ∇∧F = 2 ∂φ r sin θ ∂r ∂θ Fr rFθ r sin θFφ ∇ψ =

Those interested in reading up on the more general theory of curvilinear co-ordinates (which explains the similarities in the previous two propositions) should see Boas (pp.427-435) or Kreyszig (pp.498-504).

1.1 Identities Proposition 24 Let F be a vector field on R3 and φ be a scalar field on R3 . Then curl grad φ = 0; 10

div curl F = 0. DIV, GRAD AND CURL


Proof. With the notation ∇φ = φx , φy , φz where subscripts now denote partial differentiation, we have i j k ∂ ∂ ∂
curl (∇φ) = ∂x ∂y ∂z = φzy − φyz , φxz − φzx , φyx − φxy = 0. φ φ φ x y z Again with subscripts denoting partial differentiation, we also have   ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 div (curl F) = div − , − , − ∂y ∂z ∂z ∂x ∂x ∂y h i h i h i = (F3 )yx − (F2 )zx + (F1 )zy − (F3 )xy + (F2 )xz − (F1 )yz = 0.

Proposition 25 (Product Rules) Let F be a vector field on R3 and φ, ψ be scalar fields on R3 . Then ∇ (φψ) = φ∇ψ + ψ∇φ; div (φF) = ∇φ · F + φ div F; curl (φF) = φ curl F + ∇φ ∧ F. Corollary 26 For any scalar field φ and integer n we have ∇ (φn ) = nφn−1 ∇φ. Proof. (i) With subscripts denoting partial differentiation we have   ∇ (φψ) = (φψ)x , (φψ)y , (φψ)z


= φψ x + ψφx , φψ y + ψφy , φψ z + ψφz = φ ψ x , ψ y , ψ z + ψ φx , φy , φz = φ∇ψ + ψ∇φ. (ii) div (φF) =

div (φF) = (φF1 )x + (φF2 )y + (φF3 )z = φ (F1 )x + φx F1 + φ (F2 )y + φy F2 + φ (F3 )z + φz F3 h i
= φ (F1 )x + (F2 )y + (F3 )z + φx , φy , φz · (F1 , F2 , F3 ) = φ div F + (∇φ) · F.

IDENTITIES

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(iii) curl (φF) equals   ∂ (φF3 ) ∂ (φF2 ) ∂ (φF1 ) ∂ (φF3 ) ∂ (φF2 ) ∂ (φF1 ) − , − , − ∂y ∂z ∂z ∂x ∂x ∂y  
∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 = φ − , − , − + φy F3 − φz F2 , φz F1 − φx F3 , φx F2 − φy F1 ∂y ∂z ∂z ∂x ∂x ∂y i j k = φ curl F + φx φy φz F1 F2 F3 = φ curl F + (∇φ) ∧ F.

Proposition 27 (Further Identities) For vector fields F, G and scalar fields φ, ψ: ∇ (F · G) ∇ · (F ∧ G) ∇ ∧ (F ∧ G) ∇ ∧ (∇ ∧ F) ∇2 (φψ)

= = = = =

(F · ∇) G + (G · ∇) F + F ∧ (∇ ∧ G) + G ∧ (∇ ∧ F) G · (∇ ∧ F) − F · (∇ ∧ G) F (∇ · G) − G (∇ · F) + (G · ∇) F − (F · ∇) G ∇ (∇ · F) − ∇2 F ψ∇2 φ + 2∇φ · ∇ψ + φ∇2 ψ

Remark 28 All the identities above can be proven by simply grinding out the calculations. But there are neater ways to write out the formula for div, grad and curl as given below: ∇φ =

X ∂φ ei , ∂xi i

∇·F =

X i

ei ·

∂F , ∂xi

∇∧F =

X i

ei ∧

∂F , ∂xi

F·∇ =

X

(F · ei )

i

∂ , ∂xi

where the dummy variable i ranges over 1, 2, 3 in each case and e1 , e2 , e3 is any right-handed orthonormal basis. For example X i

ei ∧

∂F ∂F = e1 ∧ + e2 ∧ ∂xi ∂x1 e1 e2 e3 0 0 = 1 ∂F1 ∂F2 ∂F3 ∂x1

∂x1

∂x1

∂F ∂F + e3 ∧ ∂x2 ∂x3 e1 e2 e3 + 0 1 0 ∂F ∂F ∂F3 2 1 ∂x2

∂x2

∂x2

e1 + 0 ∂F 1 ∂x3

e2 0

e3 1

∂F2 ∂x3

∂F3 ∂x3



∂F3 ∂F2 ∂F3 ∂F1 ∂F2 ∂F1 + e3 + e1 − e3 − e1 + e2 ∂x1 ∂x1 ∂x2 ∂x2 ∂x3 ∂x3 = ∇ ∧ F. = −e2

With these formulae the proofs of the identities above are much shorter; on the other hand the formulae need to be applied with much more care and attention to detail than was previously the case. 12

DIV, GRAD AND CURL

Proof. The third and fifth identities appear in Exercise Sheet 1, Question 1. To prove the first identity recall that u ∧ (v ∧ w) = (u · w) v − (u · v) w and hence

= = = = =

(F · ∇) G + (G · ∇) F + F ∧ (∇ ∧ G) + G ∧ (∇ ∧ F) ! ! X X X ∂F ∂G ∂F ∂G X + (G · ei ) +F∧ ei ∧ +G∧ ei ∧ (F · ei ) ∂xi ∂xi ∂xi ∂xi i i i i     X X ∂G X ∂F X ∂G ∂F (F · ei ) + (G · ei ) + F ∧ ei ∧ + G ∧ ei ∧ ∂xi ∂xi ∂xi ∂xi i i i i        X ∂F ∂G ∂G ∂F ∂F ∂G + (G · ei ) + F· ei − (F · ei ) + G· ei − (G · ei ) (F · ei ) ∂x ∂x ∂x ∂x ∂x ∂xi i i i i i i   X ∂G ∂F ei F · +G· ∂xi ∂xi i X ∂ (F · G) = ∇ (F · G) . ei ∂xi i

To prove the second identity, and recalling the symmetries of the scalar triple product, we note X ∂ (F ∧ G) ∇ · (F ∧ G) = ei · ∂xi i   X ∂F ∂G = ei · ∧G+F∧ ∂xi ∂xi i     X X ∂F ∂G = G · ei ∧ − F · ei ∧ ∂xi ∂xi i i ! ! X X ∂F ∂G = G· ei ∧ −F· ei ∧ = G · (∇ ∧ F) − F · (∇ ∧ G) . ∂xi ∂xi i i To prove the fourth identity recall again that u ∧ (v ∧ w) = (u · w) v − (u · v) w and then ! X X ∂ ∂F ∇ ∧ (∇ ∧ F) = ei ∧ ej ∧ ∂xi ∂xj i j   XX ∂ 2F = ei ∧ ej ∧ ∂xi ∂xj i j    XX  XX ∂ 2F ∂ 2F = ei · ej − (ei · ej ) ∂xi ∂xj ∂xi ∂xj i j i j " #  X XX ∂F ∂ 2F ∂ X ei · − δ ij = ej ∂x ∂x ∂xi ∂xj j i i i j j " # X X ∂ 2F ∂ X ∂F = ej ei · − = ∇ (∇ · F) − ∇2 F 2 ∂x ∂x ∂x j i i j i i

IDENTITIES

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1.2 Physical Interpretation Recall from the Hilary Term course on PDEs that Fourier’s Law states: Example 29 (Fourier’s Law) In an isotropic medium with constant thermal conductivity k, with temperature T (x, t) at position x and at time t, Fourier’s Law states that q = −k∇T where q is the heat flux – that is the amount of energy that flows through a particular surface per unit area per unit time. This law contains much of the essence of grad . Recall from Michaelmas Calculus that, for a given scalar function φ and a unit vector v, the directional derivative of φ in the direction v is ∇φ · v. In particular, φ increases fastest in the direction of ∇φ and decreases fastest in the direction −∇φ. So Fourier’s law, loosely put, states that at each point the heat energy moves towards the coolest nearby points. How about the physical motivation behind divergence? Example 30 Consider fluid particles in the xy-plane flowing according to the steady flow

Note how we have a source at (0, 0) from which the fluid emanates. However we have, at every (x, y) , that ∂x ∂y div u = + = 2. ∂x ∂y Suppose a particle starts off at (x0 , y0 ); then the particle’s position (x (t) , y (t)) at time t is determined by the differential equations

dx = x, dt 14

dy = y, dt

x (0) = x0 ,

y (0) = y0 DIV, GRAD AND CURL

and hence
(x (t) , y (t)) = x0 et , y0 et . If we consider the blob of fluid of area πε2 which at t = 0 occupies (x − 1)2 + y 2 < ε2 , then where will the particles which make up this blob have moved by time t? Initially the particles were at x0 = 1 + r cos θ, y = r sin θ where r < ε and they have moved to
(x (t) , y (t)) = et + ret cos θ, ret sin θ r < ε, which comprises a disc with centre (et , 0) and area A (t) = πε2 e2t . So throughout the motion we have had dA = (div u) A. dt

More generally, for a two dimensional flow, it is the case that   1 d area (Flow image of B (x, ε) at time t) , ∇·u (x) = lim ε→0 area (Flow image of B (x, ε) at time t) dt with the analogous result for 3D flows in terms of volume rather than area. In physical terms, then, the divergence of a vector field is a measure of the flow’s source density, the extent to which the vector field flow behaves like a source or a sink at a given point. If you choose to take the second year option in fluid dynamics you will meet: Example 31 Euler’s Equations for an ideal fluid (Second year course in Fluid Dynamics) Euler’s equations for an incompressible, inviscid fluid state that ∂u ∇p + (u · ∇) u = − + g, ∂t ρ

∇ · u = 0,

where u (x, t) is the flow velocity, p is the pressure, ρ is the density and g is acceleration due to gravity. It is the equation ∇·u=0 which encodes the incompressibility of the fluid.

PHYSICAL INTERPRETATION

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Roughly speaking the curl of a vector field relates to the rotation of the vector field, but very much in a local sense.

Example 32 (Rotation of a rigid body) Recall from the Michaelmas Dynamics course that if r = rer then ˙ θ. r˙ = re ˙ r + rθe In particular, if a rigid body is rotating with constant angular velocity ω about the z-axis then the distance of a point r from the z-axis will remain constant and θ˙ = ω so that r˙ = rωeθ = ωk∧ (rer ) = ωk ∧ r. We have already seen from Example 11 that curl r˙ = curl (ωk ∧ r) = 2ωk.

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DIV, GRAD AND CURL

Example 33 It is important to stress though that curl is very subtly a measure of local rotation. Consider the following two examples:

A simple calculation shows that curl u2 = −ck yet there appears to be no sense of global rotation, whilst in Exercise 5 on Sheet 1 you are asked to show that curl u1 = 0, yet there appears to be a clear sense of global rotation. Despite particles moving in circles in the flow u1 different circles are moving around the origin with different angular velocities and with a greater angular velocity towards the origin; in fact this differential is such that a circular paddle wheel inserted into the fluid would not spin as it moves around the origin. In contrast, with the shear flow u2 , whilst the particles are moving in lines, a small paddle wheel inserted into the flow would spin as the flow for greater y is moving faster. In general, the axis of this spinning is in the direction of curl . In the third year you may also meet div and curl as part of the Electromagnetism option. Example 34 Maxwell’s Equations in vacuo (Third year course in Electromagnetism) div B = 0;

div E = 0; 1 ∂E ∂B ; ∇∧B= 2 ; ∇∧E = − ∂t c ∂t where E is the electric field and B is the magnetic field. Recall that ∇ ∧ (∇ ∧ F) = ∇ (∇ · F) − ∇2 F so that   ∂ −1 ∂ 2 E ∂B , −∇ E = ∇ (∇ · E) − ∇ E =∇ ∧ (∇ ∧ E) = ∇ ∧ − = − (∇ ∧ B) = 2 ∂t ∂t c ∂t2   1 ∂E 1 ∂ −1 ∂ 2 B 2 2 −∇ B = ∇ (∇ · B) − ∇ B =∇ ∧ (∇ ∧ B) = ∇ ∧ 2 = 2 (∇ ∧ E) = 2 , c ∂t c ∂t c ∂t2 so that both E and B satisfy the wave equation 2

2

c2 ∇ 2 F =

PHYSICAL INTERPRETATION

∂ 2F . ∂t2 17

18

DIV, GRAD AND CURL

2. LINE INTEGRALS & CONSERVATIVE FIELDS 2.1 Curves Definition 35 By a curve we shall mean a piecewise smooth function γ : I → R3 defined on an interval I of R. Notice that order on I also gives the curve γ an orientation. We shall also use the term curve to describe the images of such maps γ. Given such an image then it will be the image of more than one such map γ and we will talk about parametrisations γ 1 and γ 2 of the curve. These parametrisations of the image come in two different possible orientations. Definition 36 We say a curve γ : [a, b] → R3 is simple if γ is 1-1, with the one possible exception that γ (a) = γ (b) may be true; this means that the curve does not cross itself except possibly by its endpoints meeting. Definition 37 We say a curve γ : [a, b] → R3 is closed if γ (a) = γ (b) . Example 38 The line through points p and q can be parameterised as γ (t) = p + t (q − p) . When 0 < t < 1 then γ (t) lies between p and q, for t > 1 beyond q and for t < 0 before p. Example 39 A curve of the form γ (t) = (a cos t, a sin t, ct)

(2.1)

is known as a circular helix. Example 40 Parameterise the parabola formed by intersecting the plane 3x + 4y + 5z = 1 with the cone x2 + y 2 = z 2 , z > 0. Solution.

A general point on the cone can be written as r (θ, z) = (z cos θ, z sin θ, z) where z > 0 and 0 6 θ < 2π. If this point also lies in the plane 3x + 4y + 5z = 1 then z (3 cos θ + 4 sin θ + 5) = 1 and so we see  γ (θ) =

sin θ 1 cos θ , , 3 cos θ + 4 sin θ + 5 3 cos θ + 4 sin θ + 5 3 cos θ + 4 sin θ + 5

LINE INTEGRALS & CONSERVATIVE FIELDS



19

lies on the parabola. What values should θ range through? We need 0 6= 3 cos θ + 4 sin θ + 5 = 5 cos (θ − α) + 5 where α = tan−1 43 . Hence   1 cos θ sin θ 1 γ (θ) = , , 5 cos (θ − α) + 1 cos (θ − α) + 1 cos (θ − α) + 1

α−π <θ <α+π

is a parameterisation.

Aside. Note in the above example, let 1 e1 = √ (−3, −4, 5), 50

1 e2 = (−4, 3, 0), 5

1 e3 = e1 ∧ e2 = √ (3, 4, 5) 50

be a set of basis vectors. With X = e1 · γ(θ),

Y = e2 · γ(θ))

on finds that Y 2 /X is constant for all θ, explicitly confirming we have a parabola in the plane spanned by e1 , e2 . Example 41 Show that the plane with equation Aa + Bb + Cc = D intersects the unit sphere x2 + y 2 + z 2 = 1 in a circle if and only if A2 + B 2 + C 2 > D2 . Parameterise the intersection of x + y + z = 1 with the unit sphere. Solution. The plane Aa + Bb + Cc = D has normal (A, B, C) and so the point closest to the origin is the point with position vector λ (A, B, C) which lies on the plane; by substituting this in we see λ = D/ (A2 + B 2 + C 2 ). This point is within unit distance of the origin if and only if √ |D| A2 + B 2 + C 2 D (A, B, C) = 1 > 2 A + B2 + C 2 A2 + B 2 + C 2 i.e. if and only if A2 + B 2 + C 2 > D2 . By this criterion the plane x + y + z = 1 intersects with the unit sphere. The centre of the circle which makes the intersection is at 12

1 (1, 1, 1) (1, 1, 1) = . 2 2 +1 +1 3

By Pythagoras’ Theorem the radius r of the circle satisfies r 2 (1, 1, 1) = 1 =⇒ r = 2 . r2 + 3 3 20

LINE INTEGRALS & CONSERVATIVE FIELDS

√ √ As e1 = (1, −1, 0) / 2 and e2 = (1, 1, −2) / 6 are two orthonormal vectors parallel to the plane then every point of the circle can be written in the form γ (t) for 0 6 t < 2π where r r 2 2 (1, 1, 1) + e1 cos t + e2 sin t γ (t) = 3 3  3  1 1 1 1 1 2 1 1 = + √ cos t + sin t, − √ cos t + sin t, − sin t . 3 3 3 3 3 3 3 3

2.2 Line Integrals Definition 42 Let C be a curve in R3 parameterised by γ : [a, b] → R3 , and let F be a vector field, whose domain includes C. Then we define the line integral of F along C as Z b Z F (γ (t)) · γ 0 (t) dt, F · dr = C

a

where · denotes the scalar product. Proposition 43 If oriented the same, the line integral of parametrisation.

R C

f · dr is independent of the choice

Proof. Suppose that γ 1 : [a1 , b1 ] → R3 and γ 2 : [a2 , b2 ] → R3 be two parametrisations of C with γ 1 (a1 ) = γ 2 (a2 ) and γ 1 (b1 ) = γ 2 (b2 ), so that γ 1 and γ 2 give C the same orientation. Then γ 2 = γ 1 ◦ ψ where ψ : [a2 , b2 ] → [a1 , b1 ] associates the γ 2 co-ordinates of points on C with their γ 1 co-ordinate. We now define I : [a1 , b1 ] → R and J : [a2 , b2 ] → R by Z t Z t 0 I (t) = F (γ 1 (s)) · γ 1 (s) ds, J (t) = F (γ 2 (s)) · γ 02 (s) ds. a1

a2

By the Fundamental Theorem of Calculus I 0 (t) = F (γ 1 (t)) · γ 01 (t) ,

J 0 (t) = F (γ 2 (t)) · γ 02 (t) .

Further, for ψ(t) a function of t we have, by the chain rule d I (ψ (t)) = ψ 0 (t) I 0 (ψ (t)) dt = ψ 0 (t) F (γ 1 (ψ (t))) · γ 01 (ψ (t)) . LINE INTEGRALS

21

Recall γ 2 (t) = γ 1 (ψ(t)). Thus γ 02 (t) = ψ 0 (t)γ 01 (ψ(t)) and hence d I (ψ (t)) = F (γ 2 (t)) · (γ 01 (ψ (t)) ψ 0 (t)) dt = F (γ 2 (t)) · γ 02 (t) = J 0 (t) . It follows that I (ψ (t)) and J (t) differ by a constant and, as they agree at t = a2 (when they are both zero), then I (ψ (t)) = J (t) and in particular when t = b2 we have Z

b1

F (γ 1 (s)) · a1

γ 01

Z

b2

(s) ds =

F (γ 2 (s)) · γ 02 (s) ds.

a2

Remark 44 If we parameterised C in the reverse orientation then the integral give negative what had been previously calculated.

R C

F · dr would

Example 45 Let F = c ∧ r where c is a constant vector and let C be the circular helix parameterised by r (t) = (cos t, sin t, t) , 0 6 t 6 2π. Then i j k c1 · (− sin t, cos t, 1) dt c c F · dr = 2 3 0 C cos t sin t t Z 2π − sin t cos t 1 c1 c2 c3 dt = 0 cos t sin t t Z 2π
= −c2 t sin t + c3 cos2 t + c1 sin t − c2 cos t + c3 sin2 t − c1 t cos t dt Z0 2π = (−c2 t sin t + c3 − c1 t cos t) dt

Z

Z

2π

0 2π = 2πc3 − c1 [t sin t + cos t]2π 0 − c2 [−t cos t + sin t]0 = 2π (c2 + c3 ) .

22

LINE INTEGRALS & CONSERVATIVE FIELDS

Definition 46 If φ is a scalar field defined on a curve C with parametrisation γ : [a, b] → R3 then we also define the line integral

Z

Z

b

φ ds = C

φ (γ (t)) |γ 0 (t)| dt.

a

If F = (f1 , f2 , f3 ) is a vector field defined on the curve C then we define

Z

Z F ds =

Z

Z

f1 ds,

C

f2 ds,

C

C

 f3 ds .

C

Remark 47 Note that if t is the unit tangent vector field along C, in the same direction as the parametrisation, then Z

Z (φt) · dr

φ ds = C

C

and so, by inheritance, we see that these line integrals also do not depend on the choice of parametrisation. Note further that the two types of integral defined above are also independent of the choice of orientation of C, as they are independent of the sign of γ 0 (t).

Definition 48 If C is a curve with parametrisation γ : [a, b] → R3 then the arc length of the curve is Z

Z ds =

C

LINE INTEGRALS

b

|γ 0 (t)| dt.

a

23

Example 49 Find the arc length of the circular helix r (t) = (cos t, sin t, t) where 0 6 t 6 2π. Solution. We have Z Z 2π |(− sin t, cos t, 1)| dt = s= 0

2π

Z p 2 2 1 + sin t + cos t dt =

0

2π

√ √ 2 dt = 2 2 π.

0

We could also parameterise the helix ”in reverse” by setting s (t) = r (2π − t) = (cos t, − sin t, 2π − t) where 0 6 t 6 2π. We would still find Z

2π

Z

2π

|(− sin t, − cos t, −1)| dt =

s= 0

Z p 2 2 1 + sin t + cos t dt =

0

2π

√

√ 2 dt = 2 2 π.

0

Notation 50 We will standardly use the notation dr = (dx, dy, dz)

and

ds = |dr| ,

even though this may seem a little non-rigorous and any analysis course would insist on such differentials only appearing as part of a limit or within an integral. Rest assured that these differentials can be rigorously defined, though doing so is not a primary concern of this course. Definition 51 With notation as in the Definition 42, if the vector field F represents a force on a particle then Z F · dr (2.2) C

is the work done by the force in moving the particle along C. This is a generalization of the formula Work = Force × Distance which applies to constant forces acting parallel to the direction of the motion. More generally, we have Work = Component of force in direction of travel × Distance if the force and movement are not parallel. The work integral (2.2) is just an integral of such infinitesimal contributions of work. Example 52 Show that the work done by gravity, F = −mgk, in moving a particle along a straight line from (x1 , y1 , z1 ) to (x2 , y2 , z2 ) equals mg (z1 − z2 ). 24

LINE INTEGRALS & CONSERVATIVE FIELDS

Solution. We can parameterise the line segment as r (t) = (x1 , y1 , z1 ) + t (x2 − x1 , y2 − y1 , z2 − z1 ) ,

0 6 t 6 1,

so that dr = (x2 − x1 , y2 − y1 , z2 − z1 ) dt. Then Z

Z

1

F · dr =

−mgk · (x2 − x1 , y2 − y1 , z2 − z1 ) dt

C

0

Z

1

−mg (z2 − z1 ) dt

= 0

= mg (z1 − z2 ) .

In fact it’s the case that the work done by gravity – which is positive if z1 > z2 , i.e. the particle has dropped – is the loss in gravitational potential energy which, commonly, will have been converted into kinetic energy. The work done is dependent only on the endpoints (x1 , y1 , z1 ) and (x2 , y2 , z2 ) and is independent of the path taken between these endpoints. This is common to certain types of field which are known as conservative.

2.3 Conservative Fields Proposition 53 If F = ∇φ and γ : [a, b] → S is any curve such that γ (a) = p, γ (b) = q then Z F (r) · dr = φ (q) − φ (p) . γ

In particular, the integral

R γ

F (r) · dr depends only on the endpoints of the curve γ.

Proof. If γ (t) = (x (t) , y (t) , z (t)) then Z

Z

b

F (γ (t)) · γ 0 (t) dt

F (r) · dr = γ

a

Z = = = = =

b

∇φ (γ (t)) · γ 0 (t) dt a  Z b ∂φ dx ∂φ dy ∂φ dz + + dt ∂x dt ∂y dt ∂z dt a  Z b d (φ (γ (t))) dt [by the chain rule] dt a φ (γ (b)) − φ (γ (a)) φ (q) − φ (p) .

CONSERVATIVE FIELDS

25

Definition 54 Let S be an open subset of R3 . A vector field F : S → R3 is said to be conservative if there exists a scalar field φ : S → R such that F = ∇φ. Definition 55 If F = ∇φ then φ is said to be a potential (or scalar potential) for F. Example 56 The gravitational force F = −mgk from Example 52 is conservative with potential φ = −mgz. Friction is an example of a force which is not conservative – if we push a coin across a table and back then work would be done against friction increases as the length of the overall path increases. Definition 57 A subset S of R3 is said to be path-connected if for any points p, q ∈ S there exists a curve γ : [a, b] → S such that γ (a) = p and γ (b) = q. Corollary 58 If ∇φ = 0 on a path-connected set S then φ is constant. In particular, if φ and ψ are potentials of the conservative field F, defined on S, then φ and ψ differ by a constant. Proof. If ∇φ = 0 then for any curve, with endpoints p, q, we have Z φ (q) − φ (p) = ∇φ · dr = 0. γ

Given a fixed point p in S, any q in S is connected to p by a curve, and so φ is constant. If F = ∇φ = ∇ψ then ∇ (φ − ψ) = 0 and the result follows.

Theorem 59 Let S be an open subset of R3 and let F : S → R3 be a vector field. Then the following three statements are equivalent. (i) F is conservative – i.e. F = ∇φ for some scalar field φ : S → R. (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral Z F (r) · dr γ

is independent of the choice of curve γ. (iii) For any simple closed curve γ then Z F (r) · dr = 0. γ

Further (iv) is a necessary consequence of each of (i), (ii), (iii): (iv) curl F = 0. But this is not generally equivalent – for example,   −x y , (x, y) 6= (0, 0) F= x2 + y 2 x2 + y 2 satisfies curl F = 0 but there is no φ on R2 − {(0, 0)} such that F = ∇φ. See Exercise Sheet 1, Question 5. 26

LINE INTEGRALS & CONSERVATIVE FIELDS

Proof. (i) =⇒ (ii) : This was just proved in Proposition 53 as φ (q) − φ (p) is dependent only on the endpoints p and q and not on the path taken. (ii) =⇒ (i) : Let p ∈ S be a fixed point and define for any q ∈ S Z q F (r) · dr φ (q) = p

which, by assumption, is independent of the curve taken and is a function only of q. As S is open there is an r > 0 small enough that q + ti is in S for 0 6 t 6 r. Then Z q+ti φ (q + ti) − φ (q) = F (r) · dr q

where the curve from q to q + ti can be taken to be a straight line. So Z s=t Z s=t F1 (q + si) ds F (q + si) · (i ds) = φ (q + ti) − φ (q) = s=0

s=0

where F = (F1 , F2 , F3 ). Hence ∂φ φ (q + ti) − φ (q) 1 (q) = lim = lim t→0 t→0 t ∂x t

Z

s=t

F1 (q + si) ds = F1 (q) s=0

by the continuity of F1 at q. Similarly ∂φ (q) = F2 (q) , ∂y

∂φ (q) = F3 (q) ∂z

and so ∇φ = F as required.

(ii) =⇒ (iii) : Suppose now that for any two points p and q the line integral independent of the curve taken.

Rq p

F (r) · dr is

Let γ be a simple closed curve in S and let p and q be two distinct points on γ. Then p and q split γ into two curves γ 1 and γ 2 , with γ 1 running from p to q and γ 2 running from q to p. So, by (ii) and Remark 44, Z Z Z Z q Z q F (r) · dr = F (r) · dr + F (r) · dr = F (r) · dr − F (r) · dr = 0 γ

γ1

γ2

p

p

(iii) =⇒ (ii) : Let p, q ∈ S and let γ 1 , γ 2 be two curves in S, starting at p and ending at q. If γ is the curve which follows γ 1 and comes back around γ 2 , so that we have returned to p then, by assumption and Remark 44, Z Z Z Z Z 0 = F (r) · dr = F (r) · dr − F (r) · dr =⇒ F (r) · dr = F (r) · dr γ

CONSERVATIVE FIELDS

γ1

γ2

γ1

γ2

27

and the line integral of F from p to q is independent of the choice of path taken. (i) =⇒ (iv) : If F = ∇φ then curl F = ∇ ∧ ∇φ = 0 by Proposition 24 as required. However curl F = 0 is not in general a sufficient condition for vector fields to be conservative. It is, however, sufficient for vector fields on R3 . Theorems 60 is a consequence of a result known as Poincar´e’s Lemma, that holds generally about n-dimensional Euclidean space. Theorem 60 Let F : R3 → R3 be such that curl F = 0. Then there exists φ : R3 → R such that F = ∇φ. (NB an easier and more general proof of this follows from Stokes’ Theorem.) Proof. For now we set Z

1

F (tr) · r dt.

φ (r) = 0

Note that this integral is simply the line integral of F along a line segment connecting the origin to r. Then Z 1 ∂φ ∂ (r) = F (tr) · r dt ∂x ∂x 0 Z 1 ∂ = (F1 (tx, ty, tz) , F2 (tx, ty, tz) , F3 (tx, ty, tz)) · (x, y, z) dt ∂x 0 Z 1 ∂ = {xF1 (tx, ty, tz) + yF2 (tx, ty, tz) + zF3 (tx, ty, tz)} dt ∂x 0  Z 1 ∂F1 ∂F2 ∂F3 = F1 (tx, ty, tz) + tx (tx, ty, tz) + ty (tx, ty, tz) + tz (tx, ty, tz) dt ∂x ∂x ∂x 0  Z 1 ∂F1 ∂F1 ∂F1 = F1 (tx, ty, tz) + tx (tx, ty, tz) + ty (tx, ty, tz) + tz (tx, ty, tz) dt ∂x ∂y ∂z 0 [as curl F = 0] Z 1 d = [tF1 (tx, ty, tz)] dt [by the chain rule] 0 dt = [tF1 (tx, ty, tz)]t=1 t=0 = F1 (x, y, z) .

By similar calculations it also follows that ∂φ (r) = F2 (x, y, z) , ∂y

∂φ (r) = F3 (x, y, z) . ∂z

Example 61 If we set F (x, y, z) = 2xy + z cos (zx) , x2 + ey−z , x cos (zx) − ey−z 28




LINE INTEGRALS & CONSERVATIVE FIELDS

it is not hard to check curl F = 0 or to see by inspection that F = ∇ψ where ψ (x, y, z) = x2 y + sin (zx) + ey−z , or we could have determined a potential φ by calculating Z 1 φ (p) = F (tp) · p dt. 0 Z 1   
 2
  2t xy + tz cos t2 zx x + t2 x2 + et(y−z) y + tx cos t2 zx − et(y−z) z dt = 0


t3
zet(y−z) 2t3 2 1 yet(y−z) 1 = x y + sin t2 zx + x2 y + + sin t2 zx − 3 2 3 y−z 2 y−z 2 y−z = x y + sin (zx) + e . 

1 0

Remark 62 Looking at the proof of Theorem 60 it might not be clear how the proof would breakdown if we were considering a vector field on a proper subset of R3 . Implicit in the proof is the fact that we can draw a straight line from some fixed point (the origin in this case) to a general point; we then generated a potential by considering the work done by the field along this line. Obviously if the domain of the field isn’t convex then we have to adapt the proof. If the domain is still path-connected then we might consider the work done along some curve from a fixed point to a general point. The problem that arises, for some domains, is that different work will be done along different curves – even though the endpoints are the same – and so our candidate function for a potential will not solely be a function of the general point. We shall return to this problem when we meet Stokes’ Theorem and we will see that the work done to a point will be independent of the path taken when the domain is simply connected.

CONSERVATIVE FIELDS

29

30

LINE INTEGRALS & CONSERVATIVE FIELDS

3. SURFACE AND VOLUME INTEGRALS 3.1 Volume Integrals Let φ be a scalar field on a subset S of R3 . We wish to define the volume integral ZZZ φ dV. S

For example, if φ is the density of a material inside S then the integral represents the total mass. Aside. For the regions S and functions φ we shall meet there will never be any issue as to whether the integral might exist or not and we will usually determine such integrals by calculating three definite integrals separately over co-ordinates x, y, z (and sometimes using other more appropriate co-ordinates). However there are sophisticated theories about what functions φ and what subsets S are measurable in three-dimensions. Doubters of this need only look towards the Banach-Tarski Paradox1 for a sense of the subtleties involved. Example 63 A cone of height h occupies the region x2 + y 2 6 z 2 ,

0 6 z 6 h.

and has density ρ (x, y, z) = (x2 + y 2 ) z at each point. Find the mass of the cone using Cartesian co-ordinates. Solution. Our first problem is in dividing up the cone properly and finding the correct limits. There are many different ways we might begin to take cross-sections of the cones, breaking the cone into two- and then one-dimensional sections, but poorly chosen co-ordinates, whilst they could be used in principle to determine the mass, are likely to lead to a very complicated set of integrals to calculate along the way. If we use z as our first variable to divide up the cone, then we see the cross-section of the cone with a plane z = z0 gives a disc x2 + y 2 6 (z0 )2 , z = z0 (at least if 0 6 z0 6 h). We can then, say, take cross-sections of such discs with the line y = y0 to produce horizontal slithers x2 6 (z0 )2 − (y0 )2 , y = y0 , z = z0 (at least if |y0 | 6 z0 ) 1 Banach and Tarski, two Polish mathematicians, proved in 1924, that it is possible to partition a solid sphere into as few as five pieces, move these pieces around using rigid motions (so no stretching going on) and then reassemble the pieces to form two solid spheres of the same size as the original. One of the reasons that this is possible is because the pieces involved are so pathological that is impossible to assign them any sensible volume – so volume need not be an invariant of the process. Perhaps surprisingly no such paradoxical decompositions of the disc are possible in two dimensions.

SURFACE AND VOLUME INTEGRALS

31

and calculating their contribution to the mass is a simple one-dimensional integral. So the mass M is given by the triple integral Z Z Z √ h

z 2 −y 2

z

M =

√ z=0 h

Z

y=−z

Z

z

x=−

Z √z2 −y2

=

√ z=0

y=−z

x=−

ρ (x, y, z) dx dy dz

z 2 −y 2


x2 + y 2 z dx dy dz.

z 2 −y 2

We need to calculate the internal integrals first. We have √z2 −y2   3 Z √z2 −y2
x + y2x x2 + y 2 z dx = z √ √ 3 z 2 −y 2

x=−

−

z 2 −y 2

3/2 p (z 2 − y 2 ) = 2z + y2 z2 − y2 3
p 2z 2 z2 − y2. = z + 2y 2 3

The next internal integral is then Z z y=−z

!


p 2z 2 z + 2y 2 z 2 − y 2 dy. 3

It very much seems like a substitution y = z sin t, where −π/2 6 t 6 π/2, will help here. In which case the integral becomes Z

π/2

t=−π/2 π/2

Z =

t=−π/2

2z 5 = 3

Z


p 2z 2 z + 2z 2 sin2 t z 2 − z 2 sin2 t (z cos t dt) 3
2z 3 1 + 2 sin2 t |z cos t| (z cos t dt) 3

π/2


1 + 2 sin2 t cos2 t dt.

t=−π/2

Now Z

π/2

Z

π/2

1 1 (1 + cos 2t) + sin2 2t dt 2 t=−π/2 2  π/2 1 1 3t 1 1 3π (1 + cos 2t) + (1 − cos 4t) dt = + sin 2t − sin 4t = . 2 4 4 4 8 4 −π/2 2

2

2

cos t + 2 sin t cos t dt = t=−π/2

Z

π/2

= t=−π/2

Finally then we have the external integral Z

h

z=0

32

 h 2z 5 3π π z6 πh6 × dz = = . 3 4 2 6 0 12 SURFACE AND VOLUME INTEGRALS

Solution. (Alternative method with planar polars) The mass M we just calculated could have been rewritten as Z h ZZ Z h ZZ
M= ρ (x, y, z) dx dy dz = x2 + y 2 z dx dy dz. z=0

z=0

x2 +y 2 6z 2

x2 +y 2 6z 2

We have already met in the MT calculus course how to change from Cartesian co-ordinates x, y to planar polar co-ordinates r, θ. The change of variable rule is ZZ ZZ ZZ φ dA = φ dx dy = φ (r dr dθ) where the r appears as it is the Jacobian ∂ (x, y) /∂ (r, θ) . Thus

ZZ

2

x +y

2




x2 +y 2 6z 2

Z

z

2π

z

r4 r dr = 2πz r z (r dθ dr) = 2πz 4 r=0 θ=0

Z

z dx dy = r=0

Z

2

3



z = 0

πz 5 . 2

Hence, finally Z

h

M= z=0

 6 h πz 5 πz πh6 dz = = . 2 12 0 12

This change of variable makes the calculation substantially simpler!

Remark 64 In the above we essentially used cylindrical polar co-ordinates appreciating that the volume element dV is given by dV = dx dy dz = r dr dθ dz. The equivalent expression for spherical polar co-ordinates is dV = r2 sin θ dr dθ dφ. We postpone a proof and discussion of this till later – see Example 86 – but make use of this in the following example. Definition 65 The centre of mass of a body occupying a region R and with density ρ (r) at the point with position vector r is ZZZ 1 r ρ (r) dV ¯ r = (¯ x, y¯, z¯) = M R

where M is the total mass of the body. VOLUME INTEGRALS

33

Example 66 Find the centre of mass of a uniform hemisphere  (x, y, z) : x2 + y 2 + z 2 < a2 , z > 0 . Solution. By symmetry, the centre of mass lies at a point (0, 0, z¯) on the z-axis. Say ρ is the density of the hemisphere, so that its mass is 32 πa3 ρ. Then Z a Z 2π Z θ=π/2
3 2 z¯ = (r cos θ) ρr sin θ dθ dφ dr 2πa3 ρ r=0 φ=0 θ=0 Z θ=π/2 Z a 3 3 cos θ sin θ dθ = r dr × 2πρ 2πa3 ρ θ=0 r=0  4 a  π/2 3 r − cos 2θ = 3× × a 4 0 4 0 4 a 1 3 × = 3× a 4 2 3a . = 8

Example 67 Find the volume of the cone and centre of mass of a uniform cone of height h and with base radius a. Solution. Cylindrical polar co-ordinates seem to be the most natural co-ordinate system for this problem. If we consider the cone to be pointing down the z-axis with its apex at the origin then we see that the cross-section of the cone with the plane z = z0 (where 0 6 z0 6 h) is the disc r 6 z0 (a/h) . So Z z=h Z r=az/h Z 2π Z z=h Z r=az/h Z z=h  2 az/h r dz V = r dθ dr dz = 2π r dr dz = 2π 2 0 z=0 r=0 θ=0 z=0 r=0 z=0  2 3 z=h Z z=h 2 2 az az πa2 h = π dz = π = . h2 3h2 z=0 3 z=0 By symmetry we have x¯ = y¯ = 0. So we only need to calculate Z z=h Z r=az/h Z 2π Z z=h Z r=az/h Z z=h  2 az/h 6 1 2πρ zr z¯ = z ρ r dθ dr dz = 1 2 z r dr dz = 2 dz M z=0 r=0 a h z=0 2 0 πa hρ z=0 r=0 θ=0 3  h Z z=h 2 3 az 3 a2 z 4 3 a2 h4 3h 3 dz = × = × × = . = 2 2 2 2 2 2 a h z=0 h ah h 4 0 ah h 4 4 Hence ¯ r=

34

3h k. 4

SURFACE AND VOLUME INTEGRALS

Example 68 Find ZZZ


x2n + y 2n + z 2n dV.

x2 +y 2 +z 2 61

Solution. By symmetry ZZZ

ZZZ

2n

x dV = x2 +y 2 +z 2 61

and so

y

2n

ZZZ

x2 +y 2 +z 2 61

ZZZ x

2n

2n

+y

+z

2n

z 2n dV

dV = x2 +y 2 +z 2 61

ZZZ




dV = 3

z 2n dV.

x2 +y 2 +z 2 61

x2 +y 2 +z 2 61

The choice of co-ordinate z here is because the formula for z in standard spherical polar coordinates is much easier than the formulae for x and y. So, making this change, we see ZZZ z x2 +y 2 +z 2 61

2n

Z dV

1

Z

π

Z

2π

= r=0

Z

θ=0 1

Z

(r cos θ)2n r2 sin θ dφ dθ dr

φ=0 π

r2n+2 cos2n θ sin θ dθ dr r=0 θ=0  2n+3 1  π r − cos2n+1 θ = 2π 2n + 3 0 2n + 1 0 2n+1
1 − (−1) 1 = 2π × × 2n + 3 2n + 1 4π . = (2n + 3) (2n + 1) = 2π

Example 69 Find the volume of the region R that lies above the paraboloid z = x2 + y 2 and beneath the plane x + y + z = 1. Solution.

VOLUME INTEGRALS

35

The x, y co-ordinates of a point (x, y, z) on the top planar surface of R satisfy 1 − x − y 6 x2 + y 2 and so the top planar surface projects vertically down to the set  W = (x, y) ∈ R2 : x + y + x2 + y 2 6 1 ( )  2  2 1 1 3 = (x, y) ∈ R2 : x + + y+ 6 2 2 2 p which is a disc, centre (−1/2, −1/2) and radius 3/2. Hence we can determine the volume of the region R as  Z Z Z 1−x−y V = dz dA z=x2 +y 2 (x,y)∈W

ZZ =


1 − x − y − x2 − y 2 dA

(x,y)∈W

 2  2 ! 1 1 3 − x+ − y+ dA 2 2 2

ZZ = (x,y)∈W

Natural co-ordinates for parametrising W are r, θ where 1 x = − + r cos θ, 2

1 y = − + r sin θ, 2

0 6 θ < 2π, 0 6 r 6

p 3/2.

Hence we have Z √3/2 Z

V

= = = = =

36

2π



3 − r2 cos2 θ − r2 sin2 θ 2 r=0 θ=0  Z √3/2  3r 3 2π − r dr 2 r=0  2 √3/2 4 3r r 2π − 4 4 0   9 9 2π − 8 16 9π . 8

 (r dθ dr)

SURFACE AND VOLUME INTEGRALS

3.2 Parameterised Surfaces We likely all feel we know a smooth surface in R3 when we see one, and this instinct for what a surface is will largely be satisfactory for the purposes of this course. For example: Example 70 (Examples of Quadrics) The standard equations of quadrics are • Sphere: x2 + y 2 + z 2 = a2 ; • Ellipsoid: x2 /a2 + y 2 /b2 + z 2 /c2 = 1; • Hyperboloid of One Sheet: x2 /a2 + y 2 /b2 − z 2 /c2 = 1; • Hyperboloid of Two Sheets: x2 /a2 − y 2 /b2 − z 2 /c2 = 1; • Paraboloid: z = x2 + y 2 ; • Hyperbolic Paraboloid: z = x2 − y 2 ; • Cone: x2 + y 2 = z 2 .

We recognise all the examples above as smooth surfaces, with the exception of the cone at (0, 0, 0) and should be comfortable working out their tangent planes and normals – they are all level surfaces and the level surface f (x, y, z) = c at (p, q, r) has normal ∇f (p, q, r). In the main we will be happy to work with surfaces without a formal definition, but for those seeking a more rigorous treatment of the topic, the following might provide a suitable working definition. Definition 71 A smooth parameterised surface is a map r, known as the parametrisation r : U → R3 : (u, v) 7→ (x (u, v) , y (u, v) , z (u, v)) from a subset U ⊆ R2 to R3 such that PARAMETERISED SURFACES

37

• x, y, z have continuous partial derivatives with respect to u and v of all orders • r is a bijection with both r and r−1 being continuous • at each point the vectors ∂r ∂r and ∂u ∂v are linearly independent (i.e. are not scalar multiples of one another). Equivalently ∂r ∂r ∧ 6= 0. ∂u ∂v We will not be looking to treat this definition with any generality. Rather we shall just look to parameterise some of the ”standard” surfaces previously described. We define: Definition 72 Let r : U → R3 be a smooth parameterised surface and let p be a point on the surface. The plane containing p and which is parallel to the vectors ∂r (p) ∂u

and

∂r (p) ∂v

is called the tangent plane to r (U ) at p. Because these vectors are independent the tangent plane is well-defined. Definition 73 Any vector in the direction ∂r ∂r (p) ∧ (p) ∂u ∂v is said to be normal to the surface at p. There are two unit normals of length one which we denote as n and −n.

3.3 Surface Integrals Let r : U → R3 be a smooth parameterised surface with r (u, v) = (x (u, v) , y (u, v) , z (u, v)) and consider the small element of the plane that is bounded by the co-ordinate lines u = u0 and u = u0 + δu and v = v0 and v = v0 + δv. Then r maps this to a small region of the surface r (U ) and we are interested in calculating the surface area of this small region. Note

38

r (u + δu, v) − r (u, v) ≈

∂r (u, v) δu, ∂u

r (u, v + δv) − r (u, v) ≈

∂r (u, v) δv. ∂v SURFACE AND VOLUME INTEGRALS

Recall that the area of a parallelogram with sides a and b is |a ∧ b| . So the element of surface area we are considering is approximately ∂r ∂r ∂r ∂r δu ∧ δu δv. δv = ∧ ∂u ∂v ∂u ∂v This motivates the following definitions: Definition 74 Let r : U → R3 be a smooth parameterised surface. Then the surface area (or simply area) of r (U ) is defined to be ZZ ∂r ∂r ∂u ∧ ∂v du dv. U

Definition 75 We will often write ∂r ∂r du dv ∧ dS = ∂u ∂v to denote an infinitesimal part of surface area. We will also write dS =

∂r ∂r ∧ du dv. ∂u ∂v

This is also commonly written as n dS. Proposition 76 The surface area of r (U ) is independent of the choice of parametrisation. Proof. Let Σ = r (U ) = s (W ) be two different parametrisations of a surface X; take u, v as the co-ordinates on U and p, q as the co-ordinates on W . Let f = (f1 , f2 ) : U → W be the co-ordinate change map – i.e. for any (u, v) ∈ U we have r (u, v) = s (f (u, v)) = s (f1 (u, v) , f2 (u, v)) = s(p, q). Then

∂s ∂f1 ∂s ∂f2 ∂r = + , ∂u ∂p ∂u ∂q ∂u

∂r ∂s ∂f1 ∂s ∂f2 = + . ∂v ∂p ∂v ∂q ∂v

Hence

∂r ∂r ∂s ∂f1 ∂s ∂f2 ∂s ∂f2 ∂s ∂f1 ∧ = ∧ + ∧ ∂u ∂v ∂p ∂u ∂q ∂v ∂q ∂u ∂p ∂v   ∂f1 ∂f2 ∂f1 ∂f2 ∂s ∂s = − ∧ ∂u ∂v ∂v ∂u ∂p ∂q ∂ (p, q) ∂s ∂s = ∧ . ∂ (u, v) ∂p ∂q SURFACE INTEGRALS

39

Finally ZZ ZZ ∂r ∂r ∂u ∧ ∂v du dv = U U ZZ =

∂ (p, q) ∂s ∂s ∂ (u, v) ∂p ∧ ∂q du dv ∂s ∂s ∂ (p, q) ∧ ∂p ∂q ∂ (u, v) du dv U ZZ ∂s ∂s ∧ dp dq = ∂p ∂q W

by the two-dimensional rule for the change of variables in integrals (as lectured in MT).)

Example 77 Let 0 < a < b. Find the area of the torus obtained by revolving the circle (x − b)2 + z 2 = a2 in the xz-plane about the z-axis. Solution. We can parameterise the torus as r (θ, φ) = ((b + a sin θ) cos φ, (b + a sin θ) sin φ, a cos θ)

0 6 θ, φ 6 2π.

We have rθ = (a cos θ cos φ, a cos θ sin φ, −a sin θ) ,

rφ = (− (b + a sin θ) sin φ, (b + a sin θ) cos φ, 0)

and i j k a cos θ cos φ a cos θ sin φ −a sin θ = − (b + a sin θ) sin φ (b + a sin θ) cos φ 0 i j k = a (b + a sin θ) cos θ cos φ cos θ sin φ − sin θ − sin φ cos φ 0 = a (b + a sin θ) (sin θ cos φ, sin θ sin φ, cos θ) .

rθ ∧ rφ



The surface area of the torus is Z 2π Z 2π q a (b + a sin θ) sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ dφ dθ θ=0

Z

φ=0 2π

= 2πa

(b + a sin θ) dθ θ=0

= 4π 2 ab.

40

SURFACE AND VOLUME INTEGRALS

Example 78 Let z = f (x, y) denote the graph of a function f defined on a subset S of the xy-plane. Show that the graph has surface area ZZ q 1 + (fx )2 + (fy )2 dx dy. S

Deduce that a sphere of radius a has surface area 4πa2 . Solution. We can parameterise the surface as r (x, y) = (x, y, f (x, y)) Then

i j k rx ∧ ry = 1 0 fx 0 1 fy

(x, y) ∈ S.

= (−fx , −fy , 1) .

Hence the graph has surface area ZZ ZZ q 1 + (fx )2 + (fy )2 dx dy. |rx ∧ ry | dx dy = S

S

We can calculate the area of a hemisphere of radius a by setting p f (x, y) = a2 − x2 − y 2 x2 + y 2 < a2 . We then have fx = p

−x a2 − x2 − y 2

−y fy = p a2 − x 2 − y 2

,

and so the hemisphere’s area is ZZ

s 1+

x2 + y 2 dx dy a2 − x 2 − y 2

x2 +y 2
ZZ =

a p dx dy 2 a − x2 − y 2

x2 +y 2
a r dθ dr − r2 r=0 θ=0 h √ ia = 2π −a a2 − r2 =

√

a2

0

= 2πa2 . Hence the area of the whole sphere is 4πa2 . Example 79 Calculate the area of a sphere of radius a using spherical polar co-ordinates. SURFACE INTEGRALS

41

Solution. We can parameterise the sphere by r (θ, φ) = (a sin θ cos φ, a sin θ sin φ, a cos θ)

0 6 θ 6 π, 0 6 φ 6 2π.

Then i j k ∂x ∂y ∂z ∂r ∂r ∧ = ∂θ ∂θ ∂θ ∂θ ∂φ ∂x ∂y ∂z ∂φ ∂φ ∂φ i j k = a cos θ cos φ a cos θ sin φ −a sin θ −a sin θ sin φ a sin θ cos φ 0
2 2 2 = a sin θ cos φ, sin θ sin φ, sin θ cos θ = a2 sin θ (sin θ cos φ, sin θ sin φ, cos θ) .



Hence dS = a2 sin θ (− sin θ cos φ, sin θ sin φ, cos θ) dθ dφ = a2 |sin θ| dθ dφ. Finally Z

π

Z

2π

a2 |sin θ| dφ dθ

A = θ=0

= 2πa2

Zφ=0 π |sin θ| dθ θ=0

= 4πa2 .

Definition 80 If F and φ are a vector field and scalar field defined on a surface Σ, parameterised as r (U ), then we may define the following surface integrals:   ZZ ZZ ∂r ∂r F · dS = F (r (u, v)) · ∧ du dv; ∂u ∂v Σ U ZZ ZZ ∂r ∂r du dv; F dS = F (r (u, v)) ∧ ∂u ∂v Σ U   ZZ ZZ ∂r ∂r φ dS = φ (r (u, v)) ∧ du dv; ∂u ∂v Σ U ZZ ZZ ∂r ∂r du dv. φ dS = φ (r (u, v)) ∧ ∂u ∂v Σ

U

We will most commonly meet integrals of the first type ZZ F · dS Σ

which are known as flux integrals. 42

SURFACE AND VOLUME INTEGRALS

Definition 81 The solid angle is the angle an object subtends at a point in three-dimensional space. More precisely, half-lines from a fixed point (or observer) will either intersect with the object in question or not; those lines of sight that are blocked by the object represent a subset of the unit sphere centred on the observer. The solid angle is area of this subset (strictly it is the area of this subset divided by the unit of length squared to ensure the solid angle is dimensionless). The unit of solid angle is the steradian. Given that the surface area of a sphere is 4π (radius)2 then a whole solid angle is 4π steradian.

If Σ is a surface then the solid angle Ω subtended at O by Σ equals, by definition, ZZ ZZ er · dS r · dS = , Ω= 2 r r3 Σ

Σ

where r = |r|. Example 82 Find the solid angle at the apex of a right pyramid with square base of side 2d and height h. Solution.

Consider placing the apex of the pyramid at the origin and orientating the axis of the pyramid along the positive z-axis so that the square base of the pyramid has vertices (±d, ±d, h) . By symmetry the solid angle at the apex is 4 times the solid angle subtended by the smaller square with vertices (0, 0, h) , (0, d, h) , (d, d, h) (0, d, h) . Hence the solid angle is Z d Z Ω=4 y=0

SURFACE INTEGRALS

d

(x, y, h) · k dx dy

x=0 (x2

+

y2

+

h2 )3/2

Z

d

= 4h y=0

Z

d

x=0 (x2

dx dy + y 2 + h2 )3/2

. 43

If we set x =

p  p y 2 + h2 tan t and tan τ = d/ y 2 + h2 then we have Z

d

Ω = 4h y=0

Z

d

t=0

y=0 d

t=0 τ

Z

= 4h y=0 d

Z = 4h

y=0 d

Z = 4h

y=0

3/2

((y 2 + h2 ) tan2 t + y 2 + h2 ) p y 2 + h2 sec2 t dt dy

τ

Z

= 4h Z

p y 2 + h2 sec2 t dt dy

τ

Z

t=0

(y 2 + h2 )3/2 (tan2 t + 1) cos t dt dy (y 2 + h2 )

3/2

sin τ dy (y 2 + h2 ) dy p (y 2 + h2 ) y 2 + h2 + d2 −1/2

as sin τ = d (y 2 + h2 + d2 )

.

√ √ If we make a similar substitution again, namely y = h2 + d2 tan φ and tan α = d/ h2 + d2 then we see that √ Z α h2 + d2 sec2 φ dφ √ Ω = 4h 2 2 2 2 2 2 φ=0 ((h + d ) tan φ + h ) h + d sec φ Z

α

= 4h φ=0 α

Z = 4h

φ=0 α

Z = 4h

φ=0

cos φ dφ
(h2 + d2 ) sin2 φ + h2 cos2 φ cos φ dφ
2 2 (h + d ) sin2 φ + h2 cos2 φ cos φ dφ . 2 d sin2 φ + h2

√ Our final substitution is u = sin φ so that sin−1 α = d/ h2 + 2d2 and Z

sin−1 α

Ω = 4h u=0

  sin−1 α   du 4h d du d2 −1 −1 √ = tan = 4 tan . d2 u2 + h2 d h h h h2 + 2d2 0

Example 83 Evaluate the integral ZZ F ∧ dS Σ

44

SURFACE AND VOLUME INTEGRALS

where in each case Σ is the closed hemispherical surface made up of points (x, y, z) such that either x2 + y 2 + z 2 = 1 and z > 0, or x2 + y 2 6 1 and z = 0;. orient Σ so that dS is in the direction of the outward-pointing normal and F = (zx, zy, z 2 ). Solution. We could proceed to parameterise the two parts of Σ, the upper part of the hemispherical surface Σ1 and the planar disc Σ2 . These can be respectively parameterised as r1 (θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ) 0 6 φ 6 2π, 0 6 θ 6 π/2; r2 (r, θ) = (r cos θ, r sin θ, 0) 0 6 θ 6 2π, 0 6 r 6 1. With some further calculation we would find ∂r1 ∂r1 ∧ = sin θ (sin θ cos φ, sin θ sin φ, cos θ) , ∂θ ∂φ

∂r1 ∂r1 ∧ = (0, 0, r) . ∂r ∂θ

In order to get the correct outward-pointing direction we need to set dS = sin θ (sin θ cos φ, sin θ sin φ, cos θ) dθ dφ,

dS = −rk,

respectively. However if we stop to think a little we can straight away see that ...

On Σ1 we have F = zr and n = r so that F ∧ dS = zr ∧ n dS = zr ∧ r dS = 0 on all of Σ1 , and further as z = 0 on all of Σ2 then it’s also true that F = 0 on all of Σ2 . Moral of the story: take some brief time to consider the nature of your function and the region, what integrals actually need calculating and what co-ordinates are best.

Remark 84 Two points of notation in relation to sheet 2. In question 2 ∂Σ refers to the boundary of Σ, which is formed from three curves which can be parameterised C1 = {(z, 0, z) : 0 6 z 6 2} , C2 = {(2 cos t, 2 sin t, 2) : 0 6 t 6 π} , C3 = {(z, 0, −z) : 0 6 z 6 2} . R Be sure to orient the curves consistently in a loop to find ∂Σ f dr (which is defined only up to sign). In question 5 ∂R refers to the boundary of R, which is formed from three surfaces namely  Σ1 = (x, y, z) : x2 + y 2 6 4, z = 0 ,  Σ2 = (x, y, z) : x2 + y 2 = 4, 0 6 z 6 4 ,  Σ3 = (x, y, z) : x2 + y 2 6 4, z = x2 + y 2 . SURFACE INTEGRALS

45

ZZ To ensure equality, when calculating

F·dS then dS should be in the direction of the outward ∂R

pointing normal. This is in fact a standard convention; for a surface enclosing a 3d shape, it is standard to use the outward pointing normal.

3.4 Changing Co-ordinates in Volume Integrals Recall the one and two-dimensional substitution rules you have already met, namely: • if f : [c, d] → [a, b] is an increasing differentiable bijection and g is a continuous function on [a, b] then Z b Z d g (x) dx = g (f (t)) f 0 (t) dt; a

c

notice that the one-dimensional Jacobian associated with the change of variable x = f (t) is dx ∂ (x) = = f 0 (t) . ∂ (t) dt • if f : W → U is a bijection between subsets of R2 which is differentiable and has differentiable inverse then the substitution formula associated with the change of variable (u1 , u2 ) = f (w1 , w2 ) is given by ZZ ZZ ∂ (u1 , u2 ) dw1 dw2 , g (u1 , u2 ) du1 du2 = g (f (w1 , w2 )) ∂ (w1 , w2 ) U

W

where g is a continuous function on U . Does the pattern continue for changes of variable in a volume integral? Let f : W → U be a bijection between subsets of R3 which is differentiable and has differentiable inverse. Further take co-ordinates ui on U and wi on W related by the formula (u1 , u2 , u3 ) = f (w1 , w2 , w3 ). Consider a small cuboid [w1 , w1 + δw1 ] × [w2 , w2 + δw2 ] × [w3 , w3 + δw3 ] ⊆ W which has volume δw1 δw2 δw3 . Under the map f this small volume maps to a deformed parallelepiped with sides

∂f δw1 ; ∂w1 ∂f δw2 ; f (w1 , w2 + δw2 , w3 ) − f (w1 , w2 , w3 ) ≈ ∂w2 ∂f f (w1 , w2 , w3 + δw3 ) − f (w1 , w2 , w3 ) ≈ δw3 . ∂w3 f (w1 + δw1 , w2 , w3 ) − f (w1 , w2 , w3 ) ≈

46

SURFACE AND VOLUME INTEGRALS

The volume of a parallelepiped with sides a, b, c is [a, b, c] = a · (b ∧ c) and so, as a first approximation, 

δu1 δu2 δu3

 ∂f ∂f ∂f = , , δw1 δw2 δw3 ∂w1 ∂w2 ∂w3 ∂ (u1 , u2 , u3 ) δw1 δw2 δw3 . = ∂ (w1 , w2 , w3 )

Hence we also have for a change in three variables ZZ

ZZ g (u1 , u2 , u3 ) du1 du2 du3 =

U

∂ (u1 , u2 , u3 ) dw1 dw2 dw3 . g (f (w1 , w2 , w3 )) ∂ (w1 , w2 , w3 )

W

If the previous motivation for this formula seems somewhat non-rigorous do note that the chain rule for Jacobians does ensure that it is impossible that we might ever determine two different answers for a volume integral by using different sets of variables.

Example 85 (Cylindrical Polar Co-ordinates) x = r cos θ,

y = r sin θ,

z = z.

Then

∂ (x, y, z) = ∂ (r, θ, z)

∂x ∂r ∂y ∂r ∂z ∂r

∂x ∂θ ∂y ∂θ ∂z ∂θ

∂x ∂z ∂y ∂z ∂z ∂r

cos θ −r sin θ 0 = sin θ r cos θ 0 = cos θ −r sin θ sin θ r cos θ 0 0 1

= r.

Example 86 (Spherical Polar Co-ordinates) x = r sin θ cos φ,

y = r sin θ sin φ,

CHANGING CO-ORDINATES IN VOLUME INTEGRALS

z = r cos θ. 47

Then ∂x ∂x ∂x ∂r ∂θ ∂φ ∂ (x, y, z) ∂y ∂y = ∂y ∂r ∂θ ∂φ ∂z ∂ (r, φ, θ) ∂z ∂r ∂z ∂θ ∂φ sin θ cos φ r cos θ cos φ −r sin θ sin φ = sin θ sin φ r cos θ sin φ r sin θ cos φ cos θ −r sin θ 0 r cos θ cos φ −r sin θ sin φ sin θ cos φ −r sin θ sin φ = cos θ + r sin θ sin θ sin φ r sin θ cos φ r cos θ sin φ r sin θ cos φ

= cos θ r2 sin θ cos θ cos2 φ + sin2 φ + r2 sin3 θ cos2 φ + sin2 φ
= r2 sin θ cos2 θ + sin2 θ = r2 sin θ.



Example 87 Notice that if co-ordinates (X, Y, Z) are related to (x, y, z) by the matrix transformation (X, Y, Z)T = A (x, y, z)T then ∂ (X, Y, Z) = det A. ∂ (x, y, z) In particular, when A is an orthogonal matrix, then ∂ (X, Y, Z) ∂ (x, y, z) = 1. Example 88 Evaluate the integral ZZZ x dV R

√ over the intersection of the unit ball x2 + y 2 + z 2 6 1 with the half-space x + y > 3 2/5. Solution. If we take as an orthonormal basis 1 e1 = √ (1, −1, 0) , e2 = (0, 0, 1) , 2 with corresponding co-ordinates X, Y, Z so that

1 e3 = √ (1, 1, 0) , 2

Xe1 + Y e2 + Ze3 = xi + yj + zk, √ √ then we have x = (X + Z) / 2 and x + y = Z 2.

So we are left considering the integral ZZZ



X +Z √ 2

 dV.

X 2 +Y 2 +Z 2 61 Z > 3/5

48

SURFACE AND VOLUME INTEGRALS

By symmetry ZZZ X dV = 0, X 2 +Y 2 +Z 2 61 Z > 3/5

so we are left to calculate ZZZ 1 √ Z dV 2

1 = √ 2

Z

2π

φ=0

Z

cos−1 (3/5)

θ=0

Z

1

(r cos θ) r2 sin θ dr dθ dφ

r=(3/5) sec θ

X 2 +Y 2 +Z 2 61 Z > 3/5

= = = =

= = =

Z −1 Z 2π cos (3/5) 1 √ r3 cos θ sin θ dr dθ 2 θ=0 r=(3/5) sec θ   √ Z cos−1 (3/5) r4 1 π 2 cos θ sin θ dθ 4 (3/5) sec θ θ=0 √ Z −1   π 2 cos (3/5) 81 4 sec θ cos θ sin θ dθ 1− 4 625 θ=0 √ Z −1   π 2 cos (3/5) 81 sin θ cos θ sin θ − dθ 4 625 cos3 θ θ=0 √  cos−1 (3/5) 81 1 π 2 1 2 − cos θ − 4 2 1250 cos2 θ 0 √   9 81 25 1 81 π 2 1 − × + + − × 4 2 25 1250 9 2 1250 √  √  π 2 128 32π 2 . = 4 625 625

Example 89 By a suitable rotation of co-ordinates, or otherwise, evaluate ZZZ (ax + by + cz)4 dV taken over the region x2 + y 2 + z 2 6 1, where a, b, c are positive constants. Solution. Consider instead this integral as ZZZ

((a, b, c) · r)4 dV.

x2 +y 2 +z 2 61

We can rotate the axes to create new co-ordinates X, Y, Z in such a way that the Z axis points in the direction of vector (a, b, c) and under this change of co-ordinates the integral becomes ZZZ ZZZ √ 4
2 2 2 2 a + b + c e3 · r dV = a2 + b2 + c2 Z 4 dV X 2 +Y 2 +Z 2 61

CHANGING CO-ORDINATES IN VOLUME INTEGRALS

X 2 +Y 2 +Z 2 61

49

where e3 is the unit vector pointing down the Z-axis; notice that we are still integrating over the unit sphere as we have simply rotated the axes about the origin. Now, by a change to spherical polar co-ordinates, ZZZ
2 a2 + b2 + c2 Z 4 dV X 2 +Y 2 +Z 2 61

=

2

2

a +b +c


2 2

Z

1

Z

r=0

2π

φ=0

Z

π


(r cos θ)4 r2 sin θ dθ dφ dr

θ=0

 7 1  π
cos5 θ 2 2 2 2 r = 2π a + b + c − 7 0 5 0
1 2 2 2 2 2 = 2π a + b + c × × 7 5
4π 2 2 2 2 = a +b +c . 35

Example 90 Let a, b, c ∈ R. Determine ZZZ cos (ax + by + cz) dx dy dz. I= x2 +y 2 +z 2 61

Show that your answer is consistent with the fact the volume of the unit sphere is 4π/3. Solution. We can rewrite the integral as ZZZ I=

cos (a · r) dx dy dz

x2 +y 2 +z 2 61

where a = (a, b, c) .

The vector

a (a, b, c) =√ 2 |a| a + b2 + c2 is of unit length and so we can extend it to an orthonormal basis e1 , e2 , e3 with associated co-ordinates X, Y, Z. In terms of these co-ordinates e3 =

a · r = (0, 0, |a|) · (X, Y, Z) = |a| Z. The unit sphere x2 +y 2 +z 2 6 1 is still given as X 2 +Y 2 +Z 2 6 1 as e1 , e2 , e3 are an orthonormal basis and dV = dX dY dZ. So ZZZ I= cos (|a| Z) dX dY dZ. X 2 +Y 2 +Z 2 61

50

SURFACE AND VOLUME INTEGRALS

If we now change to spherical polar co-ordinates r, θ, φ associated with X, Y, Z we have Z 1 Z π Z 2π cos (|a| r cos θ) r2 sin θ dφ dθ dr I = r=0

Z

θ=0 φ=0 1 Z π

cos (|a| r cos θ) r2 sin θ dθ dr θ=0 π Zr=0 1  − sin (|a| r cos θ) 2 2π r dr |a| r r=0 0 Z 1 r 2π (2 sin (|a| r)) dr r=0 |a| ( ) 1 Z 1 −r cos (|a| r) cos (|a| r) 4π dr + |a| |a| |a| r=0 r=0 1  4π −r cos (|a| r) sin (|a| r) + |a| |a| |a|2 r=0   4π − cos |a| sin (|a|) + |a| |a| |a|2 4π (sin |a| − |a| cos |a|) . |a|3

= 2π = = = = = =

If we wish to determine the volume of the unit sphere we need to set a = b = c = 0. So letting |a| → 0 we see ! !! 4π 4π |a|3 |a|2 3
4
(sin |a| − |a| cos |a|) = |a| − + O |a| − |a| 1 − + O |a| 6 2 |a|3 |a|3 ! 4π |a|3 4
= + O |a| 3 |a|3 =

4π 4π + O (|a|) → as |a| → 0. 3 3

CHANGING CO-ORDINATES IN VOLUME INTEGRALS

51

52

SURFACE AND VOLUME INTEGRALS

4. STOKES’ AND DIVERGENCE THEOREMS 4.1 Stokes’ Theorem Recall Green’s Theorem from the Hilary Term course which states that:

Theorem 91 (Green’s Theorem) Let C be a simple closed positive-oriented curve in R2 which bounds a region R. Further let L and M be differentiable scalar fields defined in and on C. Then  Z ZZ  ∂L ∂M − L dx + M dy = dA. ∂x ∂y C R

If we wish to rewrite this in the language ofR line and surface integrals then we might set F = (L, M, 0) in which case the LHS becomes C F · dr and we further see  i j k  ∂M ∂L ∂ ∂ ∂ − k, curl F = ∂x ∂y ∂z = ∂x ∂y L (x, y) M (x, y) 0 and so the RHS of Green’s Theorem can be rephrased as  ZZ  ZZ ZZ ∂M ∂L − dA = curl F · (k dA) = curl F · dS. ∂x ∂y R

R

R

Aside. Note that we need to take the normal in the direction of the positive z-axis if equality is to hold. If we’d taken the reverse orientation on R then the two sides would be negative one another. The identity Z

ZZ F · dr =

C

STOKES’ AND DIVERGENCE THEOREMS

curl F · dS, R

53

for regions of the plane, is purely a rewriting of Green’s Theorem, but it is part of a larger identity known as Stokes’ Theorem which applies more generally to surfaces in R3 with boundary. Before we can properly understand the identity in its fullest generality, we need to appreciate how to consistently orient a curve C which bounds a surface Σ. Definition 92 At each point of a smooth surface there are two unit normal vectors ±n. Given a smooth surface Σ, then by an orientation of Σ we shall mean a continuous choice of unit normal n for the whole surface. For the bounding curve, which we shall denote as ∂Σ, there are two possible orientations. Given a certain orientation of ∂Σ and an oriented tangent vector t along ∂Σ then the vector t ∧ n lies in the tangent plane of Σ at the boundary point and is normal to the bounding curve ∂Σ. So t ∧ n points either towards the surface Σ or away from Σ. We shall always choose to orient a surface Σ and its boundary ∂Σ in such a way that t ∧ n points away from Σ. Note that this is consistent in the plane with taking positively oriented curves and normal k.

In practice it is fairly clear how to consistently orient a surface and its boundary. If one imagines an observer moving along the boundary ∂Σ (in the direction of the orientation) and upright (in the sense of the surface normal n) then the surface Σ will be to his/her left. Example 93 Given the hemisphere Σ = {(x, y, z) : x2 + y 2 + z 2 = 1, z > 0} with boundary ∂Σ = {(x, y, 0) : x2 + y 2 = 1} then we can either orient Σ as n (x, y, z) = (x, y, z)

or

− n (x, y, z) = (−x, −y, −z) .

We can parameterise ∂Σ as (cos θ, sin θ, 0). So one choice of unit tangent is t = (− sin θ, cos θ, 0) . At (cos θ, sin θ, 0) we have t ∧ n = (− sin θ, cos θ, 0) ∧ (cos θ, sin θ, 0) = (0, 0, −1) which points down, away from the hemisphere. The two consistent ways to orient the surface and boundary are (n on Σ and t on ∂Σ) or (−n on Σ and − t on ∂Σ) . 54

STOKES’ AND DIVERGENCE THEOREMS

Example 94 Not all surfaces have an orientation, such surfaces being called non-orientable. Examples of non-orientable surfaces are the M¨ obius strip and the Klein bottle. We shall only be interested in orientable surfaces though. Theorem 95 (Stokes’ Theorem c. 1850. The first known appearance of this result though is in fact in a letter from Kelvin to Stokes.) Let Σ be a smooth oriented surface in R3 whose boundary is the curve ∂Σ. Let F be a smooth vector field defined on Σ ∪ ∂Σ. Then Z ZZ F · dr = curl F · dS. (4.1) ∂Σ

Σ

Remark 96 The following proof applies only to patches of surface – we shall see in later Examples (102 and 104) that Stokes’ Theorem applies generally to more complicated surfaces and their boundaries. In any event the proof of Stokes’ Theorem is not examinable. Proof. Suppose that Σ is parameterised as r (u, v) where (u, v) ranges over some S ⊆ R2 , with ∂Σ being the image of ∂S under r. We shall first prove Stokes Theorem for the vector field F = (F, 0, 0), noting that similar proofs apply for fields (0, F, 0) and (0, 0, F ) and then the more general (4.1) follows by linearity. So, if F = (F, 0, 0), then (4.1) reads as Z ZZ F dx = (0, Fz , −Fy ) · dS. (4.2) ∂Σ

Σ

Now using the parametrisation r (u, v) we have ZZ ZZ (0, Fz , −Fy ) · dS = (0, Fz , −Fy ) · (ru ∧ rv ) du dv Σ

S

ZZ = Z SZ

i j k (0, Fz , −Fy ) · xu yu zu du dv xv yv zv [Fz (zu xv − xu zv ) − Fy (xu yv − yu xv )] du dv.

= S

STOKES’ THEOREM

55

On the other hand if we apply Green’s Theorem to the LHS of (4.2) we have Z

Z F dx = ∂Σ

F (xu du + xv dv) Z∂S Z [(F xv )u − (F xu )v ] du dv

= Z SZ

[Fu xv + F xvu − Fv xu − F xuv ] du dv

= Z SZ

[Fu xv − Fv xu ] du dv

= Z SZ

[(Fx xu + Fy yu + Fz zu ) xv − (Fx xv + Fy yv + Fz zv ) xu ] du dv

= Z SZ

[Fz (zu xv − xu zv ) − Fy (xu yv − yu xv )] du dv.

= S

The result follows.

Remark 97 For an intuitive appreciation of why Stokes’ Theorem is true, recall that curl F, when physically interpreted, is a measure of the local spin of the field F. So one might envisage summing up all the curl over the surface Σ with most contributions to this total spin cancelling out, in a manner akin to intermeshing cogs. The only contributions that wouldn’t be cancelled out are those at the boundary, these remaining contributions making up the integral around the boundary. Note that Stokes’ Theorem is a scalar identity, that is both sides of the identity are scalar quantities. There is also a vector identity related to Stokes’ Theorem. But firstly we note: Lemma 98 If c · v = c · w for all c ∈ R3 then v = w. 56

STOKES’ AND DIVERGENCE THEOREMS

Proof. We have c · (v − w) = 0 for all c and so in particular this is true when c = v − w, so that

=⇒ =⇒ =⇒ =⇒

(v − w) · (v − w) = 0, |v − w|2 = 0, |v − w| = 0, v − w = 0, v = w.

Corollary 99 Let Σ be a smooth oriented surface in R3 whose boundary is the curve ∂Σ. Let ψ be a smooth scalar field defined on Σ ∪ ∂Σ. Then ZZ Z ∇ψ ∧ dS = − ψ dr. ∂Σ

Σ

Proof.

Let c be an arbitrary constant vector and set F = ψc. In this case, and using the curl product rule, Stokes Theorem reads ZZ ZZ Z ψc · dr. curl (ψc) · dS = (ψ curl c + ∇ψ ∧ c) · dS = Σ

∂Σ

Σ

As c is constant then curl c = 0 and we have   Z ZZ ZZ   c· − ∇ψ ∧ dS = ∇ψ ∧ c · dS = c · Σ

∂Σ

Σ

As c is arbitrary then by Lemma 98 we have ZZ Z − ∇ψ ∧ dS = Σ

 ψ dr .

ψ dr

∂Σ

and the result follows.

Example 100 Verify Stokes’ Theorem when F = (y, z, x) on the hemisphere  Σ = (x, y, z) : x2 + y 2 + z 2 = a2 , z > 0 . STOKES’ THEOREM

57

Solution. We can parameterise the sphere as r (θ, φ) = (a sin θ cos φ, a sin θ sin φ, a cos θ)

0 6 θ 6 π, 0 6 φ 6 2π.

Then i j k ∂r ∂r ∧ = a cos θ cos φ a cos θ sin φ −a sin θ ∂θ ∂φ −a sin θ sin φ a sin θ cos φ 0

= a2 sin θ (sin θ cos φ, sin θ sin φ, cos θ) .

Now curl F = (−1, −1, −1) and hence Z π/2 Z 2π ZZ (−1, −1, −1) · a2 sin θ (sin θ cos φ, sin θ sin φ, cos θ) dφ dθ curl F · dS = Σ

θ=0

= −a

2

φ=0

Z

π/2

Z

2π

sin θ (sin θ cos φ + sin θ sin φ + cos θ) dφ dθ θ=0 2

Z

φ=0 π/2

= −2πa

sin θ cos θ dθ θ=0

2

= −2πa



sin2 θ 2

π/2 0

2

= −πa . On the other hand, once we parameterise and orient ∂Σ as r (θ) = (a cos θ, a sin θ, 0), we have Z 2π Z (a sin θ, 0, a cos θ) · (−a sin θ, a cos θ, 0) dθ F · dr = 0 ∂Σ Z 2π 2 sin2 θ dθ = −a 0 2

= −πa .

Example 101 Verify Corollary 99 when ψ = x and Σ is the upper hemisphere of the unit sphere. Solution. The bounding curve ∂Σ is the circle x2 + y 2 = a2 in the z = 0 plane. So Z Z 2π − ψ dr = − a cos θ (−a sin θ, a cos θ, 0) dθ ∂Σ θ=0 Z 2π
2 = a sin θ cos θ, − cos2 θ, 0 dθ θ=0 2π  1 2 −θ sin 2θ 2 = a sin θ, − ,0 2 2 4 0 = −πa2 j. 58

STOKES’ AND DIVERGENCE THEOREMS

On the other hand dS = a2 sin θ (sin θ cos φ, sin θ sin φ, cos θ) dθ dφ,

∇ψ = ∇x = i,

and Z

ZZ

π/2

2π

Z

i ∧ a2 sin θ (sin θ cos φ, sin θ sin φ, cos θ) dθ dφ

∇ψ ∧ dS = φ=0

θ=0

Σ

= a

2

Z

π/2

θ=0

= 2πa2

Z

Z

2π


0, − sin θ cos θ, sin2 θ sin φ dθ dφ

φ=0 π/2

(0, − sin θ cos θ, 0) dθ θ=0

 π/2 − sin2 θ = 2πa 0, ,0 2 0 = −πa2 j. 2

Example 102 Let 0 < a < b. Verify Stokes’ Theorem when F = (y, z, x) and Σ is the top half of the torus generated by rotating the circle (x − b)2 + z 2 = a2 about the z-axis. Solution. The boundary ∂Σ consists of two circles in the z = 0 plane with equations x2 + y 2 = (b − a)2 ,

x2 + y 2 = (b + a)2 .

If we take the upward pointing normal to Σ then we need to orient the outer circle anti-clockwise and the inner circle clockwise. So Z Z 2π F · dr = ((b + a) sin t, 0, (b + a) cos t) · (− (b + a) sin t, (b + a) cos t, 0) dt ∂Σ 0 Z 2π − ((b − a) sin t, 0, (b − a) cos t) · (− (b − a) sin t, (b − a) cos t, 0) dt 0 Z 2π Z 2π 2 2 2 sin2 t dt = − (b + a) sin t dt + (b − a) 0 0 2 2
= π (b − a) − (b + a) = −4πab. On the other hand curl F = (−1, −1, −1) , and we can parameterise the torus as r (θ, φ) = ((b + a sin θ) cos φ, (b + a sin θ) sin φ, a cos θ)

0 6 φ 6 2π, −π/2 6 θ 6 π/2.

Then rθ ∧ rφ

STOKES’ THEOREM

i j k a cos θ cos φ a cos θ sin φ −a sin θ = − (b + a sin θ) sin φ (b + a sin θ) cos φ 0 = a (b + a sin θ) (sin θ cos φ, sin θ sin φ, cos θ) .



59

Hence ZZ curl F · dS Z

Σ π/2

Z

2π

(−1, −1, −1) · a (b + a sin θ) (sin θ cos φ, sin θ sin φ, cos θ) dφ dθ

= θ=−π/2 Z π/2

φ=0 Z 2π

(b + a sin θ) (− sin θ cos φ − sin θ sin φ − cos θ) dφ dθ

= a θ=−π/2

Z

π/2

φ=0

Z

2π

b (− cos θ) + a (− sin θ cos θ) dφ dθ

= a θ=−π/2 Z π/2

φ=0

b (− cos θ) + a (− sin θ cos θ) dθ

= 2πa θ=−π/2

h iπ/2 a = 2πa −b sin θ + cos 2θ 4 −π/2 = −4πab.

Example 103 Verify Stokes’ Theorem when F = (2y, 3x, x2 + y 2 + z 2 ) and Σ is the lower half of the ellipsoid x2 /4 + y 2 /9 + z 2 /27 = 1. Solution. Here ∂Σ is the ellipse x2 /4+y 2 /9 = 1 in the xy-plane. If ∂Σ is oriented anti-clockwise then we need to take the upward pointing normal on Σ. We may parameterise ∂Σ by r (t) = (2 cos t, 3 sin t, 0) and so Z 2π Z
F · dr = 6 sin t, 6 cos t, 4 cos2 t + 9 sin2 t · (−2 sin t, 3 cos t, 0) dt ∂Σ Z0 2π
= −12 sin2 t + 18 cos2 t dt Z0 2π = ((−6 − 6 cos 2t) + (9 + 9 cos 2t)) dt 0

= 6π. On the other hand we can parameterise the lower half of the ellipsoid as   √ r (θ, φ) = 2 sin θ cos φ, 3 sin θ sin φ, 3 3 cos θ . So rθ ∧ rφ

60

i j √k = 2 cos θ cos φ 3 cos θ sin φ −3 3 sin θ −2 sin θ sin φ 3 sin θ cos φ 0  √  √ = 9 3 sin2 θ cos φ, 6 3 sin2 θ sin φ, 6 sin θ cos θ , STOKES’ AND DIVERGENCE THEOREMS

but this points downwards, so we will take its negative. Also i j k ∂ ∂ ∂ = (2y, −2x, 1) . curl F = ∂x ∂y ∂z 2y 3x x2 + y 2 + z 2 Finally, then ZZ curl F · dS Σ π

Z

θ=π/2 π

Z

Z

2π

= Z

φ=0 2π

= θ=π/2

Z

 √ √ 2 2 (6 sin θ sin φ, −4 sin θ cos φ, 1) · −9 3 sin θ cos φ, −6 3 sin θ sin φ, −6 sin θ cos θ dφ dθ 



 √ √ −54 3 sin2 θ sin φ cos φ + 24 3 sin3 θ cos φ sin φ − 6 sin θ cos θ dφ dθ

φ=0 π

= −2π 6 sin θ cos θ dθ θ=π/2  π 3 cos 2θ = 2π 2 π/2 = 3π (1 + 1) = 6π. Example 104 Verify Stokes’ Theorem for F = (x2 + 3yz, −2xz, 7y) on Σ, where Σ is that part of the cylindrical surface x2 + y 2 = a2 between the planes x + y + z = 0 and x + y + z = h, where h > 0. Solution. We may parameterise the cylinder as r (θ, z) = (a cos θ, a sin θ, z) and orient it with outward pointing normal dS = (cos θ, sin θ, 0) a dθ dz. So i j k ∂ ∂ ∂ curl F = ∂x ∂y ∂z = (7 + 2x, 3y, −5z) . x2 + 3yz −2xz 7y

ZZ

Z

2π

Z

h−a cos θ−a sin θ

curl F · dS =

(7 + 2a cos θ, 3a sin θ, −5z) · (cos θ, sin θ, 0) a dz dθ θ=0

Σ

Z

2π

z=−a cos θ−a sin θ

Z

h−a cos θ−a sin θ

= θ=0 2π

Z =


7 cos θ + 2a cos2 θ + 3a sin2 θ a dz dθ

z=−a cos θ−a sin θ


7 cos θ + 2a cos2 θ + 3a sin2 θ ah dθ

θ=0

= (2aπ + 3aπ) ah = 5πa2 h. STOKES’ THEOREM

61

On the other side, we have a disconnected boundary ∂Σ made up of C1 = {(a cos θ, a sin θ, −a cos θ − a sin θ) : 0 6 θ 6 2π} ; C2 = {(a cos θ, a sin θ, h − a cos θ − a sin θ) : 0 6 θ 6 2π} .

As we used the outward pointing normal to orient Σ then we need to orient C1 in the direction of increasing θ but orient C2 in the direction of decreasing θ. If we write x (θ) = a cos θ, y (θ) = a sin θ, z (θ) = −a cos θ − a sin θ then we have Z Z 2π
F · dr = x (θ)2 + 3y (θ) z (θ) , −2x (θ) z (θ) , 7y (θ) · dr (θ) C 0 Z 1 Z 2π
F · dr = − x (θ)2 + 3y (θ) [z (θ) + h] , −2x (θ) [z (θ) + h] , 7y (θ) · dr (θ) . C2

0

Hence Z

Z

2π

F · dr = ∂Σ

(−3hy (θ) , 2hx (θ) , 0) · dr (θ) 0 2π

Z

(−3a sin θ, 2a cos θ, 0) · (− sin θ, cos θ, sin θ − cos θ) dθ

= ah 0 2

2π

Z


3 sin2 θ + 2 cos2 θ dθ

= ah 0

= 5πa2 h.

Example 105 Let C be a closed curve bounding a smooth surface Σ. Show that Z ZZ r ∧ dr = 2 dS. C

Σ

ZZ Hence, or otherwise, evaluate

dS where Σ is the surface Σ

  x2 y 2 2 2 Σ = (x, y, z) : z = 2 + 2 , x + y < 1 . a b [You may assume that ∇ ∧ (u ∧ v) = (v · ∇) u − (∇ · u) v + (∇ · v) u − (u · ∇) v.] 62

STOKES’ AND DIVERGENCE THEOREMS

Solution.

If we set F = c ∧ r where c is an arbitrary constant vector, then we have ∇ ∧ (c ∧ r) = = = =

(r · ∇) c − (∇ · c) r + (∇ · r) c − (c · ∇) r (∇ · r) c − (c · ∇) r [as c is constant] 3c − c 2c

as 

∂ ∂ ∂ (c · ∇) r = c1 + c2 + c3 ∂x ∂y ∂z

 (x, y, z) = (c1 , c2 , c3 ) = c.

So, by Stokes’ Theorem, ZZ

Z (c ∧ r) · dr =

2c · dS

C

 Σ   ZZ r ∧ dr = c·  2dS .

Z =⇒ c· C

Σ

As c is arbitrary then by Lemma 98 Z

ZZ r ∧ dr = 2

C

dS.

(4.3)

Σ

The bounding curve of the given surface Σ is

z=

x2 y 2 + 2, a2 b

x2 + y 2 = 1

which can naturally be parameterised as  r (θ) = STOKES’ THEOREM

cos2 θ sin2 θ cos θ, sin θ, + 2 a2 b

 ,

0 6 θ 6 2π. 63

By (4.3) we have (writing c = cos θ, s = sin θ)      ZZ Z  1 2π c2 s 2 −1 1 dS = c, s, 2 + 2 ∧ −s, c, + 2 2cs dθ 2 0 a b a2 b Σ  Z  1 2π cs2 (−2cs2 − c3 ) c2 s (−s3 − 2c2 s) = + , 2 + , 1 dθ 2 0 b2 a2 a b2  Z  1 2π cs2 (−cs2 − c) c2 s (−s − c2 s) = + , 2 + , 1 dθ 2 0 b2 a2 a b2  3  2π s 1 (−s3 /3 − s) −c3 (c + c3 /3) = + , , 2 + ,θ 2 3b2 a2 3a b2 0 1 [(0, 0, 2π)] = 2 = πk.

64

STOKES’ AND DIVERGENCE THEOREMS

4.2 The Divergence Theorem Theorem 106 (Divergence Theorem – Lagrange 1762, Gauss 1813, Green 1825) Let R be a region of R3 with a piecewise smooth boundary ∂R, and let F be a differentiable vector field on R. Then ZZZ ZZ F · dS,

div F dV = R

∂R

where dS is oriented in the direction of the outward pointing normal from R. Definition 107 We say that a region R is convex if for any p, q ∈ R the line segment connecting p and q is contained in R. We will only prove the Divergence Theorem for convex regions, though the proof extends with very little extra work to any region that can be decomposed into convex regions. Proof. (Not examinable.) Let W be a subset of the xy-plane and α, β continuous functions on W .

Let  R = (x, y, z) ∈ R3 : (x, y) ∈ W, α (x, y) 6 z 6 β (x, y) and F =F k is a differentiable vector field on R. (That is R is z-convex in the sense that the intersection of R with any line parallel with the z-axis is an interval.) The Divergence Theorem in this case reads as ZZ ZZZ ∂F dV = F k · dS. ∂z R

∂R

Note that ∂R is made up of three surfaces:  Σ+ = (x, y, z) ∈ R3 : (x, y) ∈ W, z = β (x, y) ,  Σ = (x, y, z) ∈ R3 : (x, y) ∈ ∂W, α (x, y) 6 z 6 β (x, y) ,  Σ− = (x, y, z) ∈ R3 : (x, y) ∈ W, z = α (x, y) . THE DIVERGENCE THEOREM

65

Then ZZZ

∂F dV = ∂z

ZZ Z

α(x,y)

W

R

β(x,y)

∂F dz dA = ∂z

ZZ (F (x, y, β (x, y)) − F (x, y, α (x, y))) dA. W

Regarding the boundary integral:

on Σ+ :

i j k dS = 1 0 β x 0 1 βy


dx dy = −β x , −β y , 1 dx dy;

on Σ− :

i j k dS = − 1 0 αx 0 1 αy

dx dy = (αx , αy , −1) dx dy;

k · dS = 0.

on Σ :

Hence we have ZZ

ZZ F k · dS =

+

F (x, y, β (x, y)) dx dy; W

ZΣZ

ZZ F k · dS = −

−

F (x, y, α (x, y)) dx dy; W

ZΣZ

F k · dS = 0; Σ

and we have shown

ZZZ

∂F dV = ∂z

R

ZZ F k · dS ∂R

for the given region R. However, any convex body has the property of being in the form of R when viewed from each of the x, y, z directions. Hence we have shown ZZZ ZZ div F dV = F · dS R

∂R

for any convex body. Remark 108 Stokes’ Theorem and the Divergence Theorem are in fact statements of a more general theorem, also known as Stokes Theorem, which applies in all dimensions. It more generally reads: Z Z dω = ω M

66

∂M

STOKES’ AND DIVERGENCE THEOREMS

where M is a (compact) oriented n-dimensional manifold (i.e. the n-dimensional equivalent of a surface) with boundary ∂M and ω is a differential form of degree n − 1. A proper appreciation of this result is approximately at fourth year undergraduate level. Note, though, that when n = 1 Stokes’ Theorem reads Z b f 0 (x) dx = f (b) − f (a) , a

when M = [a, b] and ∂M = {a, b} oriented to sum positively the contribution at b and negatively at a. Corollary 109 Let φ be a smooth scalar field defined on a region R ⊆ R3 with a piecewise smooth boundary. Then ZZZ ZZ ∇φ dV = φ dS. R

∂R

Proof.

Let c be a constant vector and F = cφ. Then div (cφ) = c · ∇φ + φ div c = c · ∇φ as c is constant. By the Divergence Theorem  ZZZ

ZZ c · ∇φ dV =

R





ZZZ cφ · dS, =⇒ c · 

∇φ dV  = c ·  R

∂R

 ZZ φdS . ∂R

As c is an arbitrary vector then ZZZ

ZZ ∇φ dV =

R

φ dS ∂R

by Lemma 98.

Example 110 Verify the Divergence Theorem when R is the finite region bounded by the paraboloid z = x2 + y 2 and the plane x + y + z = 1, and F = (x, y, z) . THE DIVERGENCE THEOREM

67

Solution.

We already met this region in Example 69. We showed that the top surface of R projects vertically down to the set  W = (x, y) ∈ R2 : (x + 1/2)2 + (y + 1/2)2 6 3/2 p which is a disc, centre (−1/2, −1/2) and radius 3/2. Further we calculated the volume of the region to be 9π/8. As div F = 3 then ZZZ 27π . div F dV = 3 Vol (R) = 8 R

The boundary ∂R is made up of two surfaces  Σ1 = (x, y, z) : z > x2 + y 2 , x + y + z = 1 = {(x, y, z) : (x, y) ∈ W, x + y + z = 1} ;   Σ2 = (x, y, z) : z = x2 + y 2 , x + y + z 6 1 = (x, y, z) : (x, y) ∈ W, z = x2 + y 2 .

We can parameterise Σ1 and Σ2 as r1 (u, v) = (u, v, 1 − u − v) ; Note


r2 (u, v) = u, v, u2 + v 2 .

i j k (r1 )u ∧ (r1 )v = 1 0 −1 0 1 −1

= (1, 1, 1)

which is outward-pointing from R, and i j k (r2 )u ∧ (r2 )v = 1 0 2u 0 1 2v

= (−2u, −2v, 1)

which is inward-pointing and so we will take its negative. So ZZ ZZ ZZ p 2 3/2 = 3π/2. F · dS = (u, v, 1 − u − v) · (1, 1, 1) du dv = du dv = π Σ1

68

W

W

STOKES’ AND DIVERGENCE THEOREMS

For the second surface ZZ ZZ
u, v, u2 + v 2 · (2u, 2v, −1) dA F · dS = W

Σ2

ZZ


u2 + v 2 dA

= W √

Z

3/2

Z

2π

= r=0 √ 3/2

Z

√

Z = 2π = 2π



r2 − r cos θ − r sin θ + 1/2 r dθ dr

θ=0 3/2 n

r=0



(−1/2 + r cos θ)2 + (−1/2 + r sin θ)2 r dθ dr

θ=0 2π

Z

= r=0



ro dθ dr r + 2 √

r4 r2 + 4 4

3

3/2

0

= 2π (9/16 + 3/8) =

15π . 8

Hence, in total, we have ZZ 3π 15π 12π 15π 27π F · dS = + = + = . 2 8 8 8 8 ∂R

Example 111 The twice continuously differentiable function f (x, y, z) is homogeneous of degree n, so that f (tx, ty, tz) = tn f (x, y, z) for all t ∈ R. (4.4) Prove Euler’s Theorem, which states that r · ∇f = nf. Let B be the unit ball x2 + y 2 + z 2 < 1 and let ∂B be its boundary. Show that ZZZ ZZZ 1 ∇2 f dV, f dV = n (n + 3) B

and hence show that

B

ZZZ

(x − y + z)4 dV =

36π . 35

B

THE DIVERGENCE THEOREM

69

Solution. If we differentiate the identity (4.4) with respect to t then we find x

∂f ∂f ∂f (tx, ty, tz) + y (tx, ty, tz) + z (tx, ty, tz) = ntn−1 f (x, y, z) . ∂x ∂y ∂z

If we set t = 1 then we arrive at Euler’s Theorem r · ∇f = nf. By the Divergence Theorem, with B as the unit ball, and ∂B its boundary, then ZZZ ZZZ ZZ ZZ ZZ 2 ∇ f dV = ∇ · ∇f dV = ∇f · n dS = ∇f · r dS = n f dS B

B

∂B

∂B

∂B

as n = r on ∂B. On the other hand ∇ · (f r) = (∇f ) · r + f (∇ · r) = nf + 3f = (n + 3) f. and so, by the Divergence Theorem again, ZZZ ZZZ ZZ ZZ (n + 3) f dV = ∇ · (f r) dV = f r · n dS = f dS B

B

as ∂B is the unit sphere. Hence ZZZ

1 f dV = n (n + 3)

∂B

ZZZ

∂B

∇2 f dV,

B

B

The function (x − y + z)4 is homogeneous of degree 4 in x, y, z and so that ZZZ (x − y + z)4 dV B

1 = 4×7 1 = 4×7

ZZZ Z ZBZ

∇2 (x − y + z)4 dV 36 (x − y + z)2 dV

[which is again homogeneous]

B

36 1 × = 4×7 2×5 =

36 1 × 4×7 2×5

ZZZ

∇2 (x − y + z)2 dV

Z ZBZ 6 dV B

36 6 4π × × 4×7 2×5 3 36π = . 35 =

70

STOKES’ AND DIVERGENCE THEOREMS

Example 112 The vector field F (R) is defined by Z F (R) = |r − R|2 dr C

where r lies on the simple closed curve C. Show that there are constant vectors A and B such that F (R) = R ∧ A + B. Deduce that ZZ ∇ ∧ F = −4 dS S

where S is any smooth surface spanning C. Solution. We have from Corollary 99 ZZ

Z ∇φ ∧ dS = −

S

So

Z

ZZ

2

|r − R| dr = −

F (R) =

φ dr. C

C


∇ |r − R|2 ∧ dS,

S

where ∇ refers to the variable r and we think of R as a constant vector.

Now
∇ |r − R|2 = ∇ ((r − R) · (r − R)) = ∇ (r · r) − 2∇ (r · R) but
∇ (r · r) = ∇ x2 + y 2 + z 2 = (2x, 2y, 2z) = 2r, ∇ (r · R) = ∇ (R1 x + R2 y + R3 z) = (R1 , R2 , R3 ) = R, so that ZZ F (R) = −

2


∇ |r − R|

ZZ ∧ dS = −

S

(2r − 2R) ∧ dS = R ∧ A + B, S

where

ZZ A=2

ZZ dS,

S

B = −2

r ∧ dS. S

Now, with R as the variable of F, we have ∇ ∧ F = ∇ ∧ (R ∧ A + B) = ∇ ∧ (R ∧ A) . Recall that ∇ ∧ (R ∧ A) = R (∇ · A) − A (∇ · R) + (A · ∇) R − (R · ∇) A = −A (∇ · R) + (A · ∇) R THE DIVERGENCE THEOREM

71

as A is constant. Further ∇ · R = 3 and (A · ∇) R = A, so that ZZ dS. ∇ ∧ F = −3A + A = −2A = −4 S

4.3 Applications Definition 113 A region R ⊆ R3 is said to be simply connected if every simple closed curve C can be continuously deformed to a single point. Specifically, if r : [0, 1] → R is a simple closed curve beginning and ending in p then there exists a map H : [0, 1] × [0, 1] → R such that H (t, 1) = r (t) and H (t, 0) = p. Equivalently any continuous map from the unit circle S 1 to R can be extended to a map from the unit disc D to R. Example 114 R3 is simply connected, as is any plane or line. R2 −{0} is not simply connected, nor is the cylinder x2 + y 2 = a2 , nor a torus. R3 − {0} is simply connected though. Corollary 115 (Existence of a Potential) Let R be a simply connected region and let F be a smooth vector field on R for which curl F = 0. Then there exists a potential φ on R such that F = ∇φ. Remark 116 Note that this corollary extends Theorem 60 to a much more general collection of domains. Proof. Let p be a fixed point in R and define for any q ∈ R, Z φ (q) = F · dr C1

where C1 is a curve in S from p to q. If C2 is a second such curve then there is a simple closed curve C formed by C1 followed by C2 in reverse orientation. As R is simply connected then C is the boundary of a surface Σ. By Stokes’ Theorem then Z ZZ F · dr = curl F · dS = 0. C

This means that

Σ

Z F · dr =

φ (q) = C1

72

Z F · dr C2

STOKES’ AND DIVERGENCE THEOREMS

and so φ is a well-defined function on R, dependent only on the variable q and not on the choice of curve from p to q. Further, arguing as in Theorem 59, we can show that ∇φ = F. The Divergence Theorem also has important implications for certain boundary-value problems. Corollary 117 (Uniqueness of the Dirichlet Problem) Let R ⊆ R3 be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Suppose that φ1 and φ2 are such that ∇2 φ1 = 0 = ∇2 φ2 in R; φ1 = f = φ2 on ∂R. Then φ1 = φ2 in R. Proof. Let ψ = φ1 − φ2 , so that ∇2 ψ = 0 in R,

ψ = 0 on ∂R.

If we consider the function ψ 2 then, by the Divergence Theorem and as ∇2 = ∇ · ∇,

ZZ ∇ ψ

2




∇ ψ

2




· dS =


∇2 ψ 2 dV.

R

∂R

Looking at the LHS, ZZ

ZZZ

ZZ · dS =

∂R

2ψ∇ψ · dS = 0

as ψ = 0 on ∂R.

∂R

On the RHS we have
∇2 ψ 2 = 2ψ∇2 ψ + 2∇ψ · ∇ψ = 2 |∇ψ|2 Hence

ZZZ

as ∇2 ψ = 0 in R.

2 |∇ψ|2 dV = 0 =⇒ ∇ψ = 0 in R.

R

As ∇ψ = 0 in R and R is path-connected then ψ is constant throughout R. But ψ = 0 on ∂R and so by continuity ψ = 0 throughout R.

APPLICATIONS

73

Corollary 118 (Uniqueness, up to a constant, of the Neumann Problem) Let R ⊆ R3 be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Suppose that φ1 and φ2 are such that ∇2 φ1 = 0 = ∇2 φ2 in R; ∂φ1 ∂φ2 = f= on ∂R. ∂n ∂n Then φ1 − φ2 is constant in R. Proof. Let ψ = φ1 − φ2 , so that ∇2 ψ = 0 in R,

∂ψ = ∇ψ · n = 0 on ∂R. ∂n

If we consider the function ψ 2 , then by the Divergence Theorem and as ∇2 = ∇ · ∇, ZZ ZZZ

2 ∇ ψ · dS = ∇2 ψ 2 dV. R

∂R

Looking at the LHS, ZZ ZZ
2 ∇ ψ · dS = 2ψ∇ψ · n dS = 0 ∂R

as ∇ψ · n = 0 on ∂R.

∂R

On the RHS we have
∇2 ψ 2 = 2ψ∇2 ψ + 2∇ψ · ∇ψ = 2 |∇ψ|2 Hence

ZZZ

as ∇2 ψ = 0 in R.

2 |∇ψ|2 dV = 0 =⇒ ∇ψ = 0 in R.

R

As ∇ψ = 0 in R then ψ is constant throughout R. Corollary 119 (Robin (or Mixed) Boundary Problem) Let R ⊆ R3 be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Further let β be a real constant. Suppose that φ1 and φ2 are such that ∇2 φ1 = 0 = ∇2 φ2 in R; ∂φ ∂φ βφ1 + 1 = f = βφ2 + 2 on ∂R. ∂n ∂n (i) If β > 0 then φ1 = φ2 in R. (ii) If β = 0 then φ1 − φ2 is constant in R. (iii) If β < 0 then the solution need not be unique, even up to a constant. 74

STOKES’ AND DIVERGENCE THEOREMS

Proof. Let ψ = φ1 − φ2 , so that ∇2 ψ = 0 in R,

βψ +

∂ψ = βψ + ∇ψ · n = 0 on ∂R. ∂n

If we consider the function ψ 2 then, by the Divergence Theorem and as ∇2 = ∇ · ∇, ZZ ZZZ

2 ∇ ψ · dS = ∇2 ψ 2 dV. R

∂R

Looking at the LHS and noting βψ + ∇ψ · n = 0 on ∂R we have ZZ ZZ ZZ ZZ
2 2ψ (∇ψ · n) dS = 2ψ (−βψ) dS = −2β ψ 2 dS. ∇ ψ · dS = ∂R

∂R

∂R

∂R

On the RHS we have as before
∇2 ψ 2 = 2ψ∇2 ψ + 2∇ψ · ∇ψ = 2 |∇ψ|2 Hence

ZZ −2β

2

ZZZ

ψ dS = ∂R

as ∇2 ψ = 0 in R.

2 |∇ψ|2 dV.

R

If β > 0 then the LHS is non-positive and the RHS is non-negative – consequently both are zero and ∇ψ = 0 in R and ψ = 0 on ∂R. Hence ψ = 0 throughout R. If β = 0 then the Robin Boundary Problem is just the Neumann Problem, with which we have already dealt.

The following example shows that solutions need not be unique if β < 0. Example 120 Let R be the interior of the cylinder x2 + y 2 = 1. Find two solutions to the Robin Boundary Problem ∇2 ψ = 0 in R,

ψ=

∂ψ on ∂R. ∂n

Solution. Laplace’s equation in cylindrical polars reads   1 ∂ ∂ψ 1 ∂ 2ψ ∂ 2ψ 2 ∇ψ= r + 2 2 + 2 = 0. r ∂r ∂r r ∂θ ∂z We know for ψ = rn sin nθ, where n is a non-negative integer, that ∇2 ψ = 0 and on the boundary r = 1 we have ∂ψ ∂ψ = = nrn−1 sin nθ = n sin nθ; ∂n ∂r APPLICATIONS

ψ = sin nθ. 75

Hence when n = 1, we see r sin θ satisfies the given boundary problem. Other solutions are 0, r cos θ and r sin (θ − α) for arbitrary α so we see this solution is far from unique. Finally, as a last application, we use the Divergence Theorem to rederive the heat equation. We need to following lemma first. Lemma 121 If φ : R3 → R is a continuous scalar field such that ZZZ φ dV = 0 R

for all bounded subsets R, then φ ≡ 0. Proof.

Suppose, for a contradiction that φ (p) 6= 0 for some point p. Without any loss of generality let’s suppose φ (p) > 0. Let ε = 21 φ (p) > 0. Then, by the continuity of φ at p, there exists δ > 0 such that |φ (x) − φ (p)| < ε when |x − p| < δ. In particular, φ (x) > 12 φ (p) on all of B (p, δ) = {x ∈ R3 : |x − p| < δ} and so ZZZ 1 φ dV > φ (p) × Vol (B (p, δ)) > 0 2 B(p,δ)

which is a contradiction.

Theorem 122 Let T (x, t) denote the temperature at position x and at time t in an isotropic medium R with thermal conductivity k, density ρ, specific heat c with heat flow determined by Fourier’s Law which states that q = −k∇T where q is the heat flux. Then T satisfies the heat equation ∂T k 2 = ∇ T. ∂t ρc Proof. Let S ⊆ R be an arbitrary subset of R. The total heat energy in S equals ZZZ ρcT dV. S

76

STOKES’ AND DIVERGENCE THEOREMS

The amount of heat flowing out of S is given by the flux integral ZZ q · dS. ∂S

So if no heat is generated within S then the only heat loss or gain is via the boundary ∂S. Hence we have d dt

ZZ

ZZZ ρcT dV

= −

q · dS

S

∂S

ZZ k∇T · dS

=

[by Fourier’s Law]

Z∂S ZZ ∇ · (k∇T ) dV

= Z ZSZ =

[by the Divergence Theorem]

k ∇2 T dV.

S

So ZZZ ∂T dV = k ∇2 T dV, ρc ∂t S S  ZZZ  ∂T 2 =⇒ ρc − k ∇ T dV = 0. ∂t ZZZ

S

As this is true for an arbitrary subset S of R then it follows, from Lemma 121, that ρc

APPLICATIONS

∂T − k ∇2 T = 0 ∂t

throughout R.

77

78

STOKES’ AND DIVERGENCE THEOREMS

5. GRAVITY Definition 123 (Newton’s Law of Gravity) The gravitational field associated with a mass M at the origin O is GM GM f = − 3 r = − 2 er . r r The gravitational force on a particle of mass m at position vector r is F = mf = −

GM m er . r2

Here G is the gravitational constant 6.67300 × 10−11 m3 kg−1 s−2 . Proposition 124 The gravitational field f is conservative with gravitational potential φ=

GM . r

Recall that potentials are determined by the field only up to a constant. This choice of potential is such that φ → 0 as r → ∞. Proof. For a spherically symmetric function f (r) then ∇f = f 0 (r) er and so we have   GM GM ∇φ = ∇ = − 2 er r r as required. So we have that curl f = 0, but it is also the case that      1 d 1 d −GM 2 2 dφ 2 div f = ∇ φ = 2 r = 2 r = 0, r dr dr r dr r2

r 6= 0,

and ∇2 φ = 0

for gravitational potential in vacuo.

This is physically intuitive as, aside from at the origin, there is no other mass contributing to the gravitational field. Proposition 125 The gravitational potential φ = GM/r is the amount of work gravity does per unit mass to bring an object from ∞ to r. Proof. The work done by gravity bring a unit mass from ∞ to r is Z Z r 1 r F · dr = f · dr = [φ (r) − φ (∞)] = φ (r) . m ∞ ∞

GRAVITY

79

Definition 126 The gravitational potential energy of a mass m at r equals −mφ (r). Remark 127 In many texts the gravitational potential Φ is instead defined to be the gravitational energy per unit mass; in such texts then the potential used is Φ = −φ and f = −∇Φ. Example 128 A particle of mass M is placed at the origin (and stays there). A second particle of mass m is free to move along the x-axis and is released from rest at distance a from O. Find the time that it takes to hit the origin. Solution. The force on the second mass is F = m¨ x = −GM m/x2 and so we have the differential equation d2 x −GM = 2 dt x2

If we multiply both sides by dx/dt then we have −GM dx d2 x dx = 2 dt dt x2 dt and integrate this with respect to t then we have  2 1 dx GM + constant. = 2 dt x Notice that this is essentially the energy equation  2 dx 1 GM m m + − = E. 2 dt x } | {z | {z } kinetic energy

potential energy

As the particle starts out at rest from x = a then we have  2 GM GM GM (a − x) 1 dx = − = . 2 dt x a ax So we have

r dx 2GM (a − x) =− , dt ax taking the negative square root as the particle moves in the direction of decreasing x. This gives r Z x=0 r Z t=T a x − dx = dt = T. 2GM x=a a−x t=0 So r Z a r a x T = dx 2GM x=0 a − x r Z a r a a sin2 θ (2a sin θ cos θ) dθ [setting x = a sin2 θ] = 2GM x=0 a cos2 θ r r Z π/2 a πa a 2 = 2a sin θ dθ = . 2GM 0 2 2GM 80

GRAVITY

If we were now faced with n particles of mass Mi at points with position vectors ri then the gravitational potential φ and field f would be given by n X GMi φ (p) = , |p − r | i i=1

f (p) = −

n X GMi (p − ri ) i=1

|p − ri |3

.

Suppose instead, in the continuous case, we have matter of density ρ (r) occupying a region R; then the potential φ and field f are given by ZZZ ZZZ Gρ (r) dV Gρ (r) (p − r) dV φ (p) = , f (p) = − . |p − r| |p − r|3 R

R

Example 129 A uniform wire of mass M occupies {(x, 0, 0) : −a 6 x 6 a}. Find the gravitational potential at (x, 0, 0) where a < x. Find the gravitational field at this point. Solution. The gravitational potential φ (x) is given by the integral Z a Z a Z a Gρ ds Gρ ds Gρ ds = = , φ (x) = −a |x − s| −a x − s −a |(x, 0, 0) − (s, 0, 0)| where M = 2ρa. Hence

φ (x) = [−Gρ log (x − s)]a−a = Gρ (log (x + a) − log (x − a))   GM x+a = log . 2a x−a By symmetry, the y and z component of the gravitational field are zero at (x, 0, 0). Then the x component of the field is given by ex · f = ex · ∇φ =

∂φ ∂x

evaluated at (x, 0, 0). Thus       GM 1 1 GM −2a −GM f= − ex = ex = ex . 2a x+a x−a 2a x 2 − a2 x 2 − a2 at (x, 0, 0).

GRAVITY

81

Example 130 A uniform disc of mass M and radius a occupies {(x, y, 0) : x2 + y 2 6 a2 } . Find the gravitational field at (0, 0, z). Solution. Let the density of the disc be σ per unit area so that M = πa2 σ. Then a

2π

Gσ ((0, 0, z) − (r cos θ, r sin θ, 0)) 3 r dθ dr r=0 θ=0 |(0, 0, z) − (r cos θ, r sin θ, 0)| Z a Z 2π ((r cos θ, r sin θ, −z)) Gσ r dθ dr (r2 + z 2 )3/2 r=0 θ=0 Z a r dr −2πGσzk 3/2 r=0 (r 2 + z 2 ) h
−1/2 ia −2πGσzk − r2 + z 2  0 1 1 −√ −2πGσzk . z a2 + z 2 Z

Z

f (z) = f (0, 0, z) = − = = = =

Example 131 A uniform ball mass M occupies {(x, y, z) : x2 + y 2 + z 2 6 a2 }. Find the gravitational potential at (0, 0, z) where a < z. What is the gravitational potential at the point (0, 0, z) where 0 < z < a? Solution. Let ρ denote the density of the ball so that M = 4πa3 ρ/3. Using spherical polar co-ordinates we see that the potential is given by a

Z

r=0 a

Z

Z

2π

Z

φ=0 2π

Z

π

φ (p) = Z

θ=0 π

= r=0

φ=0

= 2πGρ

= = = = 82

Gρ r2 sin θ q dθ dφ dr r2 sin2 θ + (z − r cos θ)2

π

r2 sin θ dθ dr r2 − 2zr cos θ + z 2 r=0 θ=0 √ 2 π Z a r − 2zr cos θ + z 2 2 2πGρ r dr zr r=0 θ=0 Z √ 2πGρ a √ 2 r r + 2zr + z 2 − r r2 − 2zr + z 2 dr z Zr=0 2πGρ a r (r + z) − r |r − z| dr z r=0 Z 2πGρ a r (r + z) − r (z − r) dr [as z > a > r] z r=0 Z 2πGρ a 4πGρa3 GM 2r2 dr = = . z 3z z r=0 Z

=

a

θ=0

Gρ r2 sin θ dθ dφ dr |(0, 0, z) − (r sin θ cos φ, r sin θ sin φ, r cos θ)|

Z

√

GRAVITY

If we are inside the sphere at (0, 0, z) then much of the above calculation applies, until we arrive at the line

φ (p) = = = =

Z 2πGρ a r (r + z) − r |r − z| dr z r=0 Z a  Z z Z a 2πGρ r (r + z) dr − r |r − z| dr − r |r − z| dr z r=0 r=0 r=z Z a  Z z Z a 2πGρ r (r + z) dr − r (z − r) dr − r (r − z) dr z r=0 r=0 r=z   GM 2πGρ  2 z2 2 3a − z = 3− 2 . 3 2a a

By symmetry, then, for a uniform sphere of radius a, we have    GM r2   3− 2 r6a 2a a φ (r) = where r = |r| .   GM r>a r In particular we have shown: Proposition 132 The gravitational potential (and similarly the gravitational field) outside a uniform sphere of mass M and radius a are equal to the gravitational potential (respectively field) of a point mass M placed at the centre of the sphere. In the previous example note that 4πGρa3 4πGρa2 = , 3a 3 4πGρa2 2πGρ  2 φ (a− ) = 3a − a2 = , 3 3 φ (a+ ) =

so we have continuity in the potential φ across the boundary r = a. Further  4πGρr ∂φ − 3 r6a f (r) = ∇φ = = 4πGρa3 ∂r r>a − 3r2 and so f (a+ ) = − GRAVITY

4πGρa , 3

f (a− ) = −

4πGρa3 4πGρa =− , 2 3a 3 83

and we likewise have continuity in the field f across the boundary r = a. Note also       1 d  4πGρr3 − 3 r6a 1 d −4πGρ r 6 a r2 dr 2 2 dφ   r = ∇ φ= 2 = 3 0 r > a.  12 d − 4πGρa r dr dr r>a r dr

3

This is our first view of Poisson’s Equation ∇2 φ = −4πGρ which applies in both regions r < a and r > a. What follows is a somewhat informal (and off-syllabus) motivation for the more general Poisson equation. Note that, for a ball B radius a centred at the origin, and with f (r) = r/r3 = er /r2 , ZZ ZZ ZZ 1 4πa2 er · (e dS) = = 4π. dS = f · dS = r a2 a2 a2 ∂B

∂B

∂B

A direct calculation shows that ∇·

r = 0 for r 6= 0, r3

but the Divergence Theorem (improperly applied to a discontinuous function) ”should” give ZZZ ZZ r r ∇ · 3 dV = · dS = 4π, r r3 B

∂B

and this would be true for any size sphere (or indeed region) containing the origin. So, for a general region R, we have that  ZZZ r 4π if 0 ∈ R, ∇ · 3 dV = 0 if 0 6∈ R. r R

The shorthand for this is to write r ∇ · 3 = 4πδ (r) r where δ is the Dirac Delta Function. This function has the important filtering property that ZZZ φ (r) δ (r − a) dV = φ (a) . R

(All the above can, in fact, be made rigorous as part of the theory of distributions or generalized functions which were developed by Schwartz in the 1940s.) 84

GRAVITY

Theorem 133 (Poisson’s Equation) Let φ be the gravitational potential associated with a non-uniform material of density ρ (r) occupying a region R. Then ∇2 φ = −4πGρ (r) . Proof. (Non-examinable for the general case.) The gravitational field associated with this matter is given by ZZZ Gρ (r) (p − r) dV. f (p) = ∇φ (p) = − |p − r|3 R

Then ∇2 φ (p) equals   ZZZ ZZZ (p − r) ∇ · f (p) = − Gρ (r) ∇ · dV = − Gρ (r) δ (r − p) dV = −4πGρ (p) . |p − r|3 R

R

Example 134 A spherical shell, centred at O and with internal and external radii a and b respectively, has matter distribution with density ρ given by  1/r 0 < a 6 r 6 b, ρ (r) = 0 otherwise. Find the gravitational potential at all points in space. [You may assume that the spherically symmetric form of Laplace’s equation is such that ∇2 φ (r) =
1 d r2 dφ .] r2 dr dr Solution. Method One – Poisson’s Equation. By Poisson’s equation gravitational potential φ satisfies     0 r < a, 1 d dφ −4πG/r a < r < b, ∇2 φ = 2 r2 =  r dr dr 0 b < r. Solving ∇2 φ = 0 in the region r < a we get

=⇒ =⇒

  1 d 2 dφ r = 0, r2 dr dr dφ dφ −A r2 = −A =⇒ = 2 , dr dr r A φ = + B. r

Similarly φ = E/r + B in the region r > b for constants E and F . GRAVITY

85

In the middle region a < r < b we have

=⇒ =⇒ =⇒

  −4πG 1 d 2 dφ r = 2 r dr dr r   d dφ dφ r2 = −4πGr =⇒ r2 = −2πGr2 − C dr dr dr dφ C = −2πG − 2 dr r C φ = −2πGr + + D. r

So, for constants A, B, C, D, E, F, we have  A +B r < a,  r −2πGr + Cr + D a < r < b, φ (r) =  E +F b < r. r We can use certain conditions now to determine these constants: (i) because of physical considerations φ is finite at r = 0, (ii) φ is both continuous and differentiable at both r = a and at r = b, (iii) φ is unique only up to addition by a constant – to specify φ uniquely we typically stipulate that φ (r) → 0 as r → ∞.

From condition (i) we have that A = 0 and from (iii) that F = 0. The two boundary conditions at r = a, namely that φ (a− ) = φ (a+ ) and φ0 (a− ) = φ0 (a+ ) , tell us that C + D, a C 0 = −2πG − 2 . a

B = −2πGa +

Similar conditions at r = b give us that C E +D = , b b C E −2πG − 2 = − 2 . b b

−2πGb +

We have four linear equations in the four unknowns B, C, D, E in terms of the radii a, b and the gravitational constant G. The second equation immediately gives C and thus the fourth equation gives E. Then the third equation gives D and the first finally gives B. Thus one can then deduce: 86

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  

4πG (b − a) r < a, 2πGa2 −2πGr − r + 4πGb a < r < b, φ (r) =  2πG(b2 −a2 )  b < r. r

Because of the symmetry of the above problem, we can take another approach using Gauss’ Flux Theorem. Theorem 135 (Gauss’ Flux Theorem) For a smooth and bounded region R, which contains matter of total mass M , then ZZ f · dS = −4πGM. ∂R

This result is in fact equivalent to Poisson’s equation. Proof. Poisson’s equation gives us that ∇2 φ = −4πGρ where ρ (r) is the density of the matter. Applying the Divergence Theorem we have ZZ

ZZZ f · dS =

ZZZ ∇ · f dV =

R

∂R

ZZZ

2

∇ φ dV = −4πG R

ρ dV = −4πGM. R

Conversely suppose that we know ZZ f · dS = −4πGM ∂R

for any bounded region R. Then ZZZ

2

ZZZ

∇ φ dV = −4πG R

and so

ZZZ

ρ dV R


∇2 φ + 4πGρ dV = 0 for any bounded region R.

R

Hence (at least if ∇2 φ and ρ are piecewise continuous) we have ∇2 φ + 4πGρ ≡ 0.

GRAVITY

87

Solution. (to Example 134) Method Two - Gauss’ Flux Theorem. Alternatively we may use Gauss’ Flux Theorem applied to concentric spheres centred on the shell’s centre. Now φ is only dependent on r and so is constant on the sphere r = R, which has surface area 4πR2 . So if we apply the flux theorem to the region r 6 R we have ZZ ZZ ZZ 0 φ0 (R) dS = 4πR2 φ0 (R) = −4πGM (R) φ (R) er · dS = ∇φ · dS = r=R

r=R

r=R

where M (R) is the total mass within the region r 6 R and as the . For a 6 R 6 b we have Z

R

Z

π

Z

2π

1 2 r sin θ dα dθ dr r=a θ=0 α=0 r Z R π = 2π × [− cos θ]0 × r dr r=a
= 2π R2 − a2 .

M (R) =

Hence φ0 (R) =

        

R < a,  02  a 2πG −1 a < R < b, R2 2πG (a2 − b2 ) b < R. R2

If we integrate we have

φ (R) =

        

A 

R < a,  a2 B − 2πG +R a < R < b, R 2πG (b2 − a2 ) +C b < R. R

As φ (∞) = 0 then C = 0. As φ (b− ) = φ (b+ ) then  2  a 2πG (b2 − a2 ) B − 2πG +b = =⇒ B = 4πGb. b b Finally as φ (a− ) = φ (a+ ) then  A = 4πGb − 2πG

a2 +a a

 = 4πG (b − a) ,

and we have the same expressions for φ as were achieved by the previous method.

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Example 136 (i) A thin disc of uniform mass per unit area σ has radius R. Find the gravitational potential at a height z directly above the centre of the disc. Hence, or otherwise, calculate the gravitational field at this point due to the disc. (ii) Deduce that the gravitational field at the apex of a cone of uniform density ρ, height h and semiangle α has magnitude 2πGρh (1 − cos α). (iii) Consider the Earth to be a sphere of uniform density ρ, mass M0 and radius R0 . A mountain on the surface of the Earth is modelled by a cone of height h, density 2ρ and semiangle α. Take h to be small compared to R0 and show that to first order in h the gravitational field at the top of the mountain has magnitude

GM0 2GM0 − F = R02 R03



 3 1 − (1 − cos α) h. 2

As h increases does F become stronger or weaker? [You may assume without proof that the gravitational field external to a uniform sphere is as if all the mass were concentrated at the centre of the sphere.]

Solution. √ (i) The distance from (0, 0, z) to an infinitesimal piece of mass σr dr dθ at (r, θ) in the disc is r2 + z 2 . Hence the gravitational potential at (0, 0, z) is

Z

R

Z

2π

G (σr dθ dr) √ r2 + z 2 r=0 θ=0 Z R r dr √ = 2πGσ r2 + z 2 r=0 ir=R h√ = 2πGσ r2 + z 2 r=0 √  = 2πGσ R2 + z 2 − z .

φ (0, 0, z) =

The gravitational field at this point is ∇φ, but as ∂φ/∂x = ∂φ/∂y = 0 due to the symmetry of the disc, then ∂φ f = ∇φ = k = 2πGσ ∂z GRAVITY



 z √ − 1 k. R2 + z 2 89

(ii)

Recall the field at point p is given by ZZZ Gρ (r) (p − r) dV f (p) = − . |p − r|3 R

By symmetry, the field at the apex of the cone with uniform density ρ, height h and semiangle α is in the k direction. The field can be found by integrating infinitesimal contribution from the horizontal cross-sections at distance z away (0 6 z 6 h) of radius z tan α. Hence the field is given by

ZZZ f (p) = −k

Gρ (r) k · (p − r) dV |p − r|3

R

ZZZ

z

= Gρk R

Z

h

(z 2 + r2 )3/2 

Z

h

r dθ dr dz = 2πGρk

Z

z tan α

z z=0

r=0

r (z 2 + r2 )3/2

! dr dz



z q − 1 dz 2 z=0 (z tan α) + z 2  Z h  1 √ = 2πGρk − 1 dz 2 tan α + 1 z=0   1 = 2πGρhk √ −1 sec2 α = 2πGρh (1 − cos α) (−k) . = 2πGρk

So the magnitude of the field is 2πGρh (1 − cos α) in a vertically downward direction. (iii) Noting the density of the mountain is 2ρ the strength of the gravitational field at its apex due to its own mass is given by 4πGρh (1 − cos α) 90

GRAVITY

. The field strength associated with the rest of the planet is GM0 , (R0 + h)2 as it can be considered as a single point mass at distance R0 + h away. So the combined field is

GM0 + 4πGρh (1 − cos α) (R0 + h)2  −2   GM0 h 3M0 4πρR03 = = M0 ] 1 + + 4πG h (1 − cos α) [as R02 R0 4πR03 3      2 ! 2h h GM0 3M0 = 1− +G h (1 − cos α) + O 2 3 R0 R0 R0 R0    2 ! GM0 2GM0 3 h − = 1 − (1 − cos α) h + O . 2 3 R0 R0 2 R0

f =

Note 2GM0 − R03 provided α < cos−1

1 3

  3 1 − (1 − cos α) < 0 2

≈ 70◦ and then the field is less at the mountian apex.

Shield volcanos have very large footprints and very shallow slopes even though they can reach mountainous heights (e.g. Hawaii’s Mauna Loa). For most mountain geometries the field at the summit would be less given the assumptions made.

Example 137 A solid hemisphere, radius a, has density λr where λ is a positive constant and r is the distance from the centre, O, of its plane face. Taking the z-axis along the line joining O to a point on the rim, find the gravitational potential at the point (0, 0, c), where c > a. Find also the potential if 0 < c < a. Find the gravitational field in the direction of the z-axis at the point (0, 0, 2a) . GRAVITY

91

Solution.

Assume, without loss of generality, the sphere to be in the half-space x > 0. The distance from (0, 0, c) to an infinitesimal piece of mass (λr) r2 sin θ dr dθ dφ at (r, θ, φ) in the hemisphere is q (r sin θ cos φ)2 + (r sin θ sin φ)2 + (c − r cos θ)2 p r2 sin2 θ + c2 − 2rc cos θ + r2 cos2 θ = √ = r2 − 2rc cos θ + c2 . Hence the gravitational potential at (0, 0, c) is Z

r=a

r=0

Z

θ=π

θ=0

φ=π/2

G (λr) r2 sin θ dφ dθ dr √ r2 − 2rc cos θ + c2 φ=−π/2 Z r=a Z θ=π r3 sin θ dθ dr √ = πGλ r2 − 2rc cos θ + c2 r=0 θ=0 θ=π Z r=a  √ 1 2 2 = πGλ r − 2rc cos θ + c r3 dr rc θ=0 Zr=0  √ πGλ r=a √ 2 = r + 2rc + c2 − r2 − 2rc + c2 r2 dr c Zr=0 πGλ r=a = (r + c − |r − c|) r2 dr. c r=0

Z

If c > a > r then |r − c| = c − r and so the potential at (0, 0, c) equals

Z πGλ r=a ((r + c) − (c − r)) r2 dr c Zr=0 πGλ r=a 3 = 2r dr c r=0 πGλa4 = . 2c

92

GRAVITY

If a > c > 0 then |r − c| = c − r for 0 6 r 6 c and |r − c| = r − c for c 6 r 6 a. So the potential at (0, 0, c) equals πGλ c

Z

r=a

r=0

(r + c − |r − c|) r2 dr Z r=c  Z r=a πGλ 2 2 (r + c − (c − r)) r dr + (r + c − (r − c)) r dr = c r=0 r=c  Z r=c Z r=a πGλ 2 3 = 2cr dr 2r dr + c r=c r=0  3  2a 1 3 = πGλ − c . 3 6

The z-component of the field equals ∇φ · k = ∂φ/∂c. For the range c > a this is given by ∂φ d = ∂c dc



πGλa4 2c

 =

−πGλa4 . 2c2

When c = 2a this equals −πGλa2 . 8

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93