4 CHAPTER 1. VECTOR ANALYSIS Chapter 1 Vector Analysis Problem 1.1 ✲ ✒ ✯ ✣ A B C B + C |B| cos θ1 |C| cos θ2 } |B| sin θ1 } |C| sin θ2 θ1 θ2 θ3 (a) From the diagram, |B + C| cos θ 3 = |B| cos θ 1 + |C| cos θ 2 . Multiply by |A|. |A||B + C| cos θ 3 = |A||B| cos θ 1 + |A||C| cos θ 2 . So: A·(B + C)= A·B + A·C. (Dot product is distributive) Similarly: |B + C| sin θ 3 = |B| sin θ 1 + |C| sin θ 2 . Mulitply by |A| ˆ n. |A||B + C| sin θ 3 ˆ n = |A||B| sin θ 1 ˆ n + |A||C| sin θ 2 ˆ n. If ˆ n is the unit vector pointing out of the page, it follows that A×(B + C)=(A×B)+(A×C). (Cross product is distributive) (b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product) Problem 1.2 ✲ A = B ✻ C ❂ B×C ❄ A×(B×C) The triple cross-product is not in general associative. For example, suppose A = B and C is perpendicular to A, as in the diagram. Then (B×C) points out-of-the-page, and A×(B×C) points down, and has magnitude ABC. But (A×B)= 0, so (A×B)×C = 0 = A×(B×C). Problem 1.3 ✲ y ✻ z ✰ x ✣ B ❲ A θ A = +1 ˆ x +1 ˆ y - 1 ˆ z; A = √ 3; B =1 ˆ x +1 ˆ y +1 ˆ z; B = √ 3. A·B = +1 + 1 - 1=1= AB cos θ = √ 3 √ 3 cos θ ⇒ cos θ = 1 3 . θ = cos -1 ( 1 3 ) ≈ 70.5288 ◦ Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, we might pick the base (A) and the left side (B): A = -1 ˆ x +2 ˆ y +0 ˆ z; B = -1 ˆ x +0 ˆ y +3 ˆ z. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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4 CHAPTER 1. VECTOR ANALYSIS
Chapter 1
Vector Analysis
Problem 1.1
CHAPTER 1. VECTOR ANALYSIS 3
Chapter 1
Vector Analysis
Problem 1.1
!
"
#
$
A
B
C
B+C
! "# $
|B| cos !1
! "# $
|C| cos !2
}|B| sin !1
}|C| sin !2
!1
!2
!3
(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)
Similarly: |B + C| sin !3 = |B| sin !1 + |C| sin !2. Mulitply by |A| n.|A||B + C| sin !3 n = |A||B| sin !1 n + |A||C| sin !2 n.If n is the unit vector pointing out of the page, it follows thatA!(B + C) = (A!B) + (A!C). (Cross product is distributive)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)
Problem 1.2
! A = B
%C
&B!C 'A!(B!C)
The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B!C) points out-of-the-page, and A!(B!C) points down,and has magnitude ABC. But (A!B) = 0, so (A!B)!C = 0 !=A!(B!C).
Problem 1.3
! y
%z
(x
!B
"A
!
A = +1 x + 1 y " 1 z; A =#
3; B = 1 x + 1 y + 1 z;
A·B = +1 + 1 " 1 = 1 = AB cos ! =#
3#
3 cos ! $ cos !.
! = cos!1%
13
&
% 70.5288"
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):
c#2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.
(a) From the diagram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multiply by |A|.|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)
Similarly: |B + C| sin θ3 = |B| sin θ1 + |C| sin θ2. Mulitply by |A| n.|A||B + C| sin θ3 n = |A||B| sin θ1 n + |A||C| sin θ2 n.If n is the unit vector pointing out of the page, it follows thatA×(B + C) = (A×B) + (A×C). (Cross product is distributive)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)
Problem 1.2
CHAPTER 1. VECTOR ANALYSIS 3
Chapter 1
Vector Analysis
Problem 1.1
!
"
#
$
A
B
C
B+C
! "# $
|B| cos !1
! "# $
|C| cos !2
}|B| sin !1
}|C| sin !2
!1
!2
!3
(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2. Multiply by |A|.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)
Similarly: |B + C| sin !3 = |B| sin !1 + |C| sin !2. Mulitply by |A| n.|A||B + C| sin !3 n = |A||B| sin !1 n + |A||C| sin !2 n.If n is the unit vector pointing out of the page, it follows thatA!(B + C) = (A!B) + (A!C). (Cross product is distributive)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)
Problem 1.2
! A = B
%C
&B!C 'A!(B!C)
The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B!C) points out-of-the-page, and A!(B!C) points down,and has magnitude ABC. But (A!B) = 0, so (A!B)!C = 0 !=A!(B!C).
Problem 1.3
! y
%z
(x
!B
"A
!
A = +1 x + 1 y " 1 z; A =#
3; B = 1 x + 1 y + 1 z; B =#
3.
A·B = +1 + 1 " 1 = 1 = AB cos ! =#
3#
3 cos ! $ cos ! = 13.
! = cos!1%
13
&
% 70.5288"
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):
A = "1 x + 2 y + 0 z; B = "1 x + 0 y + 3 z.
c#2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material isprotected under all copyright laws as they currently exist. No portion of this material may bereproduced, in any form or by any means, without permission in writing from the publisher.
The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B×C) points out-of-the-page, and A×(B×C) points down,and has magnitude ABC. But (A×B) = 0, so (A×B)×C = 0 6=A×(B×C).
Problem 1.3
CHAPTER 1. VECTOR ANALYSIS 3
Chapter 1
Vector Analysis
Problem 1.1
!
"
#
$
A
B
C
B+C
! "# $
|B| cos !1
! "# $
|C| cos !2
}|B| sin !1
}|C| sin !2
!1
!2
!3
(a) From the diagram, |B + C| cos !3 = |B| cos !1 + |C| cos !2.|A||B + C| cos !3 = |A||B| cos !1 + |A||C| cos !2.So: A·(B + C) = A·B + A·C. (Dot product is distributive)
Similarly: |B + C| sin !3 = |B| sin !1 + |C| sin !2. Mulitply by |A| n.|A||B + C| sin !3 n = |A||B| sin !1 n + |A||C| sin !2 n.If n is the unit vector pointing out of the page, it follows thatA!(B + C) = (A!B) + (A!C). (Cross product is distributive)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) andSection 8 (cross product)
Problem 1.2
! A = B
%C
&B!C 'A!(B!C)
The triple cross-product is not in general associative. For example,suppose A = B and C is perpendicular to A, as in the diagram.Then (B!C) points out-of-the-page, and A!(B!C) points down,and has magnitude ABC. But (A!B) = 0, so (A!B)!C = 0 !=A!(B!C).
Problem 1.3
! y
%z
(x
!B
"A
!
A = +1 x + 1 y " 1 z; A =#
3; B = 1 x + 1 y + 1 z;
A·B = +1 + 1 " 1 = 1 = AB cos ! =#
3#
3 cos ! $ cos !.
! = cos!1%
13
&
% 70.5288"
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):
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A = +1 x + 1 y − 1 z; A =√
3; B = 1 x + 1 y + 1 z; B =√
3.
A·B = +1 + 1− 1 = 1 = AB cos θ =√
3√
3 cos θ ⇒ cos θ = 13 .
θ = cos−1(
13
)≈ 70.5288◦
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,we might pick the base (A) and the left side (B):
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by itslength:|A×B| =
√36 + 9 + 4 = 7. n = A×B
|A×B| = 67 x + 3
7 y + 27 z .
Problem 1.5
A×(B×C) =
∣∣∣∣∣∣x y zAx Ay Az
(ByCz −BzCy) (BzCx −BxCz) (BxCy −ByCx)
∣∣∣∣∣∣= x[Ay(BxCy −ByCx)−Az(BzCx −BxCz)] + y() + z()(I’ll just check the x-component; the others go the same way)= x(AyBxCy −AyByCx −AzBzCx +AzBxCz) + y() + z().
B(A·C)−C(A·B) = [Bx(AxCx +AyCy +AzCz)− Cx(AxBx +AyBy +AzBz)] x + () y + () z= x(AyBxCy +AzBxCz −AyByCx −AzBzCx) + y() + z(). They agree.
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, orone is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)
Conclusion: A×(B×C) = (A×B)×C⇐⇒ either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
r = (4 x + 6 y + 8 z)− (2 x + 8 y + 7 z) = 2 x− 2 y + z
}Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but alsonecessary. For suppose A = (1, 0, 0). Then Σj,k (ΣiRijRik)AjAk = ΣiRi1Ri1, and this must equal 1 (since wewant A
2
x+A2
y+A2
z = 1). Likewise, Σ3i=1Ri2Ri2 = Σ3
i=1Ri3Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).Then we want 2 = Σj,k (ΣiRijRik)AjAk = ΣiRi1Ri1 + ΣiRi2Ri2 + ΣiRi1Ri2 + ΣiRi2Ri1. But we alreadyknow that the first two sums are both 1; the third and fourth are equal, so ΣiRi1Ri2 = ΣiRi2Ri1 = 0, and soon for other unequal combinations of j, k. X In matrix notation: RR = 1, where R is the transpose of R.
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Looking down the axis:
CHAPTER 1. VECTOR ANALYSIS 5
! x
"y
#z
$%
Looking down the axis:
"y
&x
'z
"z!
&y!
'x!
(
)
*
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A 120◦ rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az,Ay = Ax, Az = Ay.
R =
0 0 11 0 00 1 0
Problem 1.10
(a) No change. (Ax = Ax, Ay = Ay, Az = Az)
(b) A −→ −A, in the sense (Ax = −Ax, Ay = −Ay, Az = −Az)
(c) (A×B) −→ (−A)×(−B) = (A×B). That is, if C = A×B, C −→ C . No minus sign, in contrast tobehavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A×B) −→ (A)×(B) =(A×B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vectorand a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.Angular momentum (L = r×p) and torque (N = r×F) are pseudovectors.
(d) A·(B×C) −→ (−A)·((−B)×(−C)) = −A·(B×C). So, if a = A·(B×C), then a −→ −a; a pseudoscalarchanges sign under inversion of coordinates.Problem 1.11
(a)∇f = 2x x + 3y2 y + 4z3 z
(b)∇f = 2xy3z4 x + 3x2y2z4 y + 4x2y3z3 z
(c)∇f = ex sin y ln z x + ex cos y ln z y + ex sin y(1/z) z
= −()−32 [(x− x′) x + (y − y′) y + (z − z′) z] = −(1/r 3)r = −(1/r 2) r .
(c) ∂∂x (r n) = n r n−1 ∂ r
∂x = n r n−1( 12
1r 2 r x) = n r n−1 r x, so ∇(r n) = n r n−1 rProblem 1.14
y = +y cosφ+ z sinφ; multiply by sinφ: y sinφ = +y sinφ cosφ+ z sin2 φ.z = −y sinφ+ z cosφ; multiply by cosφ: z cosφ = −y sinφ cosφ+ z cos2 φ.Add: y sinφ+ z cosφ = z(sin2 φ+ cos2 φ) = z. Likewise, y cosφ− z sinφ = y.So ∂y
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from theorigin. How, then, can ∇·v = 0? The answer is that ∇·v = 0 everywhere except at the origin, but at theorigin our calculation is no good, since r = 0, and the expression for v blows up. In fact, ∇·v is infinite atthat one point, and zero elsewhere, as we shall see in Sect. 1.5.Problem 1.17
∣∣∣∣∣∣ = x(0− 6xz) + y(0 + 2z) + z(3z2 − 0) = −6xz x + 2z y + 3z2 z.
(b) ∇×vb =
∣∣∣∣∣∣x y z∂∂x
∂∂y
∂∂z
xy 2yz 3xz
∣∣∣∣∣∣ = x(0− 2y) + y(0− 3z) + z(0− x) = −2y x− 3z y − x z.
(c) ∇×vc =
∣∣∣∣∣∣x y z∂∂x
∂∂y
∂∂z
y2 (2xy + z2) 2yz
∣∣∣∣∣∣ = x(2z − 2z) + y(0− 0) + z(2y − 2y) = 0.
Problem 1.19
A x
y
z
v
v
v
vB
As we go from point A to point B (9 o’clock to 10 o’clock), xincreases, y increases, vx increases, and vy decreases, so ∂vx/∂y >0, while ∂vy/∂y < 0. On the circle, vz = 0, and there is nodependence on z, so Eq. 1.41 says
∇× v = z(∂vy
∂x− ∂vx
∂y
)points in the negative z direction (into the page), as the righthand rule would suggest. (Pick any other nearby points on thecircle and you will come to the same conclusion.) [I’m sorry, but Icannot remember who suggested this cute illustration.]
Problem 1.20
v = y x + x y; or v = yz x + xz y + xy z; or v = (3x2z − z3) x + 3 y + (x3 − 3xz2) z;or v = (sinx)(cosh y) x− (cosx)(sinh y) y; etc.
(iii) da = dx dz y, y = 2. v·da = 4z dx dz.∫v·da =
∫∫4z dx dz = 16.
(iv) da = −dx dz y, y = 0. v·da = 0.∫v·da = 0.
(v) da = dx dy z, z = 2. v·da = 6x dx dy.∫v·da = 24.
(vi) da = −dx dy z, z = 0. v·da = 0.∫v·da = 0.
⇒∫v·da = 8 + 16 + 24 = 48 X
Problem 1.34
∇×v = x(0− 2y) + y(0− 3z) + z(0− x) = −2y x− 3z y − x z.da = dy dz x, if we agree that the path integral shall run counterclockwise. So(∇×v)·da = −2y dy dz.∫
(∇×v)·da =∫ {∫ 2−z
0(−2y)dy
}dz
↪→ y2∣∣2−z
0= −(2− z)2
= −∫ 2
0(4− 4z + z2)dz = −
(4z − 2z2 + z3
3
)∣∣∣20
= −(8− 8 + 8
3
)= − 8
3 -
6z
y
@@
@@
@@
y =2−z
Meanwhile, v·dl = (xy)dx+ (2yz)dy + (3zx)dz. There are three segments.
-
6z
y
@@
@@
@@
-(1)
@@I (2)
?
(3)
(1) x = z = 0; dx = dz = 0. y : 0→ 2.∫v·dl = 0.
(2) x = 0; z = 2− y; dx = 0, dz = −dy, y : 2→ 0. v·dl = 2yz dy.∫v·dl =
∫ 0
22y(2− y)dy = −
∫ 2
0(4y − 2y2)dy = −
(2y2 − 2
3y3)∣∣2
0= −
(8− 2
3 · 8)
= − 83 .
(3) x = y = 0; dx = dy = 0; z : 2→ 0. v·dl = 0.∫v·dl = 0. So
∮v·dl = − 8
3 . X
Problem 1.35
By Corollary 1,∫(∇×v)·da should equal 4
3 . ∇×v = (4z2 − 2x)x + 2z z.
(i) da = dy dz x, x = 1; y, z : 0→ 1. (∇×v)·da = (4z2 − 2)dy dz;∫(∇×v)·da =
∫ 1
0(4z2 − 2)dz
= (43z
3 − 2z)∣∣10
= 43 − 2 = − 2
3 .
(ii) da = −dx dy z, z = 0; x, y : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.
(iii) da = dx dz y, y = 1; x, z : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.
(iv) da = −dx dz y, y = 0; x, z : 0→ 1. (∇×v)·da = 0;∫(∇×v)·da = 0.
(v) da = dx dy z, z = 1; x, y : 0→ 1. (∇×v)·da = 2 dx dy;∫(∇×v)·da = 2.
(a) Use the product rule ∇×(fA) = f(∇×A)−A× (∇f) :∫Sf(∇×A) · da =
∫S
∇×(fA) · da +∫S[A× (∇f)] · da =
∮PfA · dl +
∫S[A× (∇f)] · da. qed
(I used Stokes’ theorem in the last step.)
(b) Use the product rule ∇·(A×B) = B · (∇×A)−A · (∇×B) :∫V
B · (∇×A)dτ =∫V
∇·(A×B) dτ +∫V
A · (∇×B) dτ =∮S(A×B) · da +
∫V
A · (∇×B) dτ. qed
(I used the divergence theorem in the last step.)
Problem 1.37 r =√x2 + y2 + z2; θ = cos−1
(z√
x2+y2+z2
); φ = tan−1
(yx
).
Problem 1.38
There are many ways to do this one—probably the most illuminating way is to work it out by trigonometryfrom Fig. 1.36. The most systematic approach is to study the expression:
r = x x + y y + z z = r sin θ cosφ x + r sin θ sinφ y + r cos θ z.
If I only vary r slightly, then dr = ∂∂r (r)dr is a short vector pointing in the direction of increase in r. To make
it a unit vector, I must divide by its length. Thus:
r =∂r∂r∣∣ ∂r∂r
∣∣ ; θ =∂r∂θ∣∣ ∂r∂θ
∣∣ ; φ =∂r∂φ∣∣ ∂r∂φ
∣∣ .∂r∂r = sin θ cosφ x + sin θ sinφ y + cos θ z;
∣∣ ∂r∂r
∣∣2 = sin2 θ cos2 φ+ sin2 θ sin2 φ+ cos2 θ = 1.∂r∂θ = r cos θ cosφ x + r cos θ sinφ y − r sin θ z;
∣∣ ∂r∂θ
∣∣2 = r2 cos2 θ cos2 φ+ r2 cos2 θ sin2 φ+ r2 sin2 θ = r2.∂r∂φ = −r sin θ sinφ x + r sin θ cosφ y;
So F3 can be written as the gradient of a scalar (F3 = ∇U3) and as the curl of a vector (F3 = ∇×A3). Infact, U3 = xyz does the job. For the vector potential, we have
∂Az
∂y −∂Ay
∂z = yz, which suggests Az = 14y
2z + f(x, z); Ay = − 14yz
2 + g(x, y)∂Ax
∂z −∂Az
∂x = xz, suggesting Ax = 14z
2x+ h(x, y); Az = − 14zx
2 + j(y, z)∂Ay
∂x −∂Ax
∂y = xy, so Ay = 14x
2y + k(y, z); Ax = − 14xy
2 + l(x, z)
Putting this all together: A3 = 1
4
{x(z2 − y2
)x + y
(x2 − z2
)y + z
(y2 − x2
)z}
(again, not unique).
Problem 1.51(d) ⇒ (a): ∇×F = ∇×(−∇U) = 0 (Eq. 1.44 – curl of gradient is always zero).(a) ⇒ (c):
∮F · dl =
∫(∇×F) · da = 0 (Eq. 1.57–Stokes’ theorem).
(c) ⇒ (b):∫ b
a IF · dl−
∫ b
a IIF · dl =
∫ b
a IF · dl +
∫ a
b IIF · dl =
∮F · dl = 0, so∫ b
a I
F · dl =∫ b
a II
F · dl.
(b) ⇒ (c): same as (c) ⇒ (b), only in reverse; (c) ⇒ (a): same as (a)⇒ (c).
Problem 1.52(d) ⇒ (a): ∇·F = ∇·(∇×W) = 0 (Eq 1.46—divergence of curl is always zero).(a) ⇒ (c):
is any constant vector—in particular, it could be be x, or y, or z—so each component of the integral on leftequals corresponding component on the right, and hence∫
∇T dτ =∫T da. qed
(b) Let v → (v × c) in divergence theorem. Then∫
∇·(v × c)dτ =∫
(v × c) · da. Product rule #6 ⇒∇·(v × c) = c · (∇×v) − v · (∇×c) = c · (∇×v). (Note: ∇×c = 0, since c is constant.) Meanwhile vectorindentity (1) says da · (v × c) = c · (da× v) = −c · (v × da). Thus
∫c · (∇×v) dτ = −
∫c · (v × da). Take c
outside, and again let c be x, y, z then:∫(∇×v) dτ = −
(d) Rewrite (c) with T ↔ U :∫ (U∇2T + (∇T ) · (∇U)
)dτ =
∫(U∇T ) ·da. Subtract this from (c), noting
that the (∇U) · (∇T ) terms cancel:∫ (T∇2U − U∇2T
)dτ =
∫(T∇U − U∇T ) · da. qed
(e) Stokes’ theorem:∫
(∇×v) · da =∮
v · dl. Let v = cT . By Product Rule #(7): ∇×(cT ) = T (∇×c)−c× (∇T ) = −c× (∇T ) (since c is constant). Therefore, −
∫(c× (∇T )) · da =
∮Tc · dl. Use vector indentity
#1 to rewrite the first term (c× (∇T )) · da = c · (∇T × da). So −∫
c · (∇T × da) =∮
c ·T dl. Pull c outside,and let c→ x, y, and z to prove: ∫
∇T × da = −∮T dl. qed
Problem 1.62(a) da = R2 sin θ dθ dφ r. Let the surface be the northern hemisphere. The x and y components clearly integrateto zero, and the z component of r is cos θ, so
a =∫R2 sin θ cos θ dθ dφ z = 2πR2 z
∫ π/2
0
sin θ cos θ dθ = 2πR2 zsin2 θ
2
∣∣∣π/2
0= πR2 z.
(b) Let T = 1 in Prob. 1.61(a). Then ∇T = 0, so∮da = 0. qed
(c) This follows from (b). For suppose a1 6= a2; then if you put them together to make a closed surface,∮da = a1 − a2 6= 0.
(d) For one such triangle, da = 12 (r × dl) (since r × dl is the area of the parallelogram, and the direction is
perpendicular to the surface), so for the entire conical surface, a = 12
∮r× dl.
(e) Let T = c · r, and use product rule #4: ∇T = ∇(c · r) = c × (∇×r) + (c ·∇)r. But ∇×r = 0, and(c ·∇)r = (cx ∂
∂x + cy∂∂y + cz
∂∂z )(x x + y y + z z) = cx x + cy y + cz z = c. So Prob. 1.61(e) says∮
Evidently there is no delta function at the origin.
∇× (rn r) =1r2
∂
∂r
(r2rn
)=
1r2
∂
∂r
(rn+2
)=
1r2
(n+ 2)rn+1 = (n+ 2)rn−1
(except for n = −2, for which we already know (Eq. 1.99) that the divergence is 4πδ3(r)).
(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero.To be certain there is no lurking delta function here, we integrate over a sphere of radius R, usingProb. 1.61(b): If ∇×(rn r) = 0, then
∫(∇×v) dτ = 0 ?= −
∮v × da. But v = rn r and da =
R2 sin θ dθ dφ r are both in the r directions, so v × da = 0. X
Problem 1.64(a) Since the argument is not a function of angle, Eq. 1.73 says
D = − 14π
1r2
d
dr
[r2(−1
2
)2r
(r2 + ε2)3/2
]=
14πr2
d
dr
[r3
(r2 + ε2)3/2
]=
14πr2
[3r2
(r2 + ε2)3/2− 3
2r3 2r
(r2 + ε2)5/3
]=
14πr2
3r2
(r2 + ε2)5/2
(r2 + ε2 − r2
)=
3ε2
4π(r2 + ε2)5/2. X
(b) Setting r → 0:
D(0, ε) =3ε2
4πε5=
34πε3
,
which goes to infinity as ε→ 0. X(c) From (a) it is clear that D(r, 0) = 0 for r 6= 0. X(d) ∫
D(r, ε) 4πr2 dr = 3ε2∫ ∞
0
r2
(r2 + ε2)5/2dr = 3ε2
(1
3ε2
)= 1. X
(I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for δ3(r).