introduction to electrodynamics 4th edition by David J Griffiths solutions manua

1. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad Instructor’s Solution Manual Introduction to Electrodynamics Fourth Edition David J. Gri?ths 2014 @solutionmanual1

2. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 2 Contents 1 Vector Analysis 4 2 Electrostatics 26 3 Potential 53 4 Electric Fields in Matter 92 5 Magnetostatics 110 6 Magnetic Fields in Matter 133 7 Electrodynamics 145 8 Conservation Laws 168 9 Electromagnetic Waves 185 10 Potentials and Fields 210 11 Radiation 231 12 Electrodynamics and Relativity 262 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. @solutionmanual1

3. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 3 Preface Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I would like to thank him particularly for all his help. If you find errors, please let me know (gri?th@reed.edu). David Gri?ths 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. @solutionmanual1

4. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ 3 CHAPTER 1. VECTOR ANALYSIS cl i ck h ere t o d ow nl oad 3 CHAPTER 1. VECTOR ANALYSIS 4 CHAPTER 1. VECTOR ANALYSIS 3 CHAPTER 1. VECTOR ANALYSIS Chapter 1 Chapter 1 Chapter 1 Vector Analysis Vector Analysis Chapter 1 Vector Analysis Vector Analysis Problem 1.1 ✒ ✣ } Problem 1.1 ✒ ✣ } (a) From the diagram, |B + C|cosθ3= |B|cosθ1+ |C|cosθ2. Multiply by |A|. |A||B + C|cosθ3= |A||B|cosθ1+ |A||C|cosθ2. So: A·(B + C) = A·B + A·C. (Dot product is distributive) Problem 1.1 (a) From the diagram, |B + C|cosθ3= |B|cosθ1+ |C|cosθ2. |A||B + C|cosθ3= |A||B|cosθ1+ |A||C|cosθ2. So: A·(B + C) = A·B + A·C. (Dot product is distributive) Problem 1.1 (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|ˆ n. |A||B + C|sin✓3ˆ n = |A||B|sin✓1ˆ n + |A||C|sin✓2ˆ 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 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). A⇥(B⇥C). Problem 1.3 Problem 1.3 A = +1ˆ x + 1ˆ y ? 1ˆ z; A =p3; B = 1ˆ x + 1ˆ y + 1ˆ z; B =p3. A·B = +1 + 1 ? 1 = 1 = AB cos✓ =p3p3cos✓ ) cos✓ =1 ✓ = cos?1?1 Problem 1.4 B + C ✒ ✣ B + C } |C|sinθ2 C |C|sinθ2 (a) From the diagram, |B + C|cosθ3= |B|cosθ1+ |C|cosθ2. |A||B + C|cosθ3= |A||B|cosθ1+ |A||C|cosθ2. So: A·(B + C) = A·B + A·C. (Dot product is distributive) Similarly: |B + C|sinθ3= |B|sinθ1+ |C|sinθ2. Mulitply by |A|ˆ n. |A||B + C|sinθ3ˆ n = |A||B|sinθ1ˆ n + |A||C|sinθ2ˆ 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) A×(B + C) = (A×B) + (A×C). (Cross product is distributive) C B + C ✯ |C|sinθ2 C θ2 θ3 θ2 }|B|sinθ1 ! Similarly: |B + C|sinθ3= |B|sinθ1+ |C|sinθ2. Mulitply by |A|ˆ n. |A||B + C|sinθ3ˆ n = |A||B|sinθ1ˆ n + |A||C|sinθ2ˆ n. If ˆ n is the unit vector pointing out of the page, it follows that ✯ B θ1 θ3 θ2 }|B|sinθ1 Similarly: |B + C|sinθ3= |B|sinθ1+ |C|sinθ2. Mulitply by |A|ˆ n. |A||B + C|sinθ3ˆ n = |A||B|sinθ1ˆ n + |A||C|sinθ2ˆ 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) Section 8 (cross product) ✲ A "# ✯ $ B θ1 θ3 }|B|sinθ1 ✲ A B θ1 ! "# ! "# $ ✲ A |C|cosθ2 |C|cosθ2 ! "# $ $ |B|cosθ1 ! "# $ ! "# $ |B|cosθ1 |B|cosθ1 |C|cosθ2 (b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and (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 Problem 1.2 Problem 1.2 C C 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). and has magnitude ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 6= 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, 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 ✲A = B ❂ B×C ✲A = B ❂ B×C ❂ ❄ A×(B×C) ❄ A×(B×C) ❄ A×(B×C) B×C z Problem 1.3 ✻ z Problem 1.3 ✻ z ✻ A = +1ˆ x + 1ˆ y − 1ˆ z; A =√3; B = 1ˆ x + 1ˆ y + 1ˆ z; A·B = +1 + 1 − 1 = 1 = AB cosθ =√3√3cosθ ⇒ cosθ. θ = cos−1%1 &≈ 70.5288◦ A = +1ˆ x + 1ˆ y − 1ˆ z; A =√3; B = 1ˆ x + 1ˆ y + 1ˆ z; B =√3. A·B = +1 + 1 − 1 = 1 = AB cosθ =√3√3cosθ ⇒ cosθ =1 θ = cos−1%1 A = +1ˆ x + 1ˆ y − 1ˆ z; A =√3; B = 1ˆ x + 1ˆ y + 1ˆ z; A·B = +1 + 1 − 1 = 1 = AB cosθ =√3√3cosθ ⇒ cosθ. θ = cos−1%1 ✣B ✣B ✣B θ ✲y ✲y ❲ A 3. 3. &≈ 70.5288◦ 3 θ θ ✲y ❲ A 3 ?⇡ 70.5288? &≈ 70.5288◦ ✰ x ❲ A x 3 3 ✰ ✰ Problem 1.4 x 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): 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): we might pick the base (A) and the left side (B): A = ?1ˆ x + 2ˆ y + 0ˆ z; B = ?1ˆ x + 0ˆ y + 3ˆ z. Problem 1.4 Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, 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 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. A = −1ˆ x + 2ˆ y + 0ˆ z; B = −1ˆ x + 0ˆ y + 3ˆ z. 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. 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. c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is c ⃝2005 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. @solutionmanual1

5. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 5 CHAPTER 1. VECTOR ANALYSIS ?????? ?????? ˆ x ˆ y ˆ z ?1 2 0 ?1 0 3 A⇥B = This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: |A⇥B| =p36 + 9 + 4 = 7. Problem 1.5 ?????? (I’ll just check the x-component; the others go the same way) = ˆ x(AyBxCy? AyByCx? AzBzCx+ AzBxCz) + ˆ y() +ˆ z(). B(A·C) ? C(A·B) = [Bx(AxCx+ AyCy+ AzCz) ? Cx(AxBx+ AyBy+ AzBz)]ˆ x + ()ˆ y + ()ˆ z = ˆ x(AyBxCy+ AzBxCz? AyByCx? AzBzCx) + ˆ y() +ˆ z(). They agree. Problem 1.6 = 6ˆ x + 3ˆ y + 2ˆ z. A⇥B ˆ n = |A⇥B|= 6 7ˆ x +3 7ˆ y +2 7ˆ z . ?????? ˆ x Ax ˆ y Ay ˆ z Az A⇥(B⇥C) = (ByCz? BzCy) (BzCx? BxCz) (BxCy? ByCx) = ˆ x[Ay(BxCy? ByCx) ? Az(BzCx? BxCz)] + ˆ y() +ˆ z() A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C)?C(A·B)+C(A·B)?A(C·B)+A(B·C)?B(C·A) = 0. So: A⇥(B⇥C) ? (A⇥B)⇥C = ?B⇥(C⇥A) = A(B·C) ? C(A·B). If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or one 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 = (4ˆ x + 6ˆ y + 8ˆ z) ? (2ˆ x + 8ˆ y + 7ˆ z) = 2ˆ x ? 2ˆ y + ˆ z r =p4 + 4 + 1 = 3 ˆ r =rr Problem 1.8 (a)¯Ay¯By+¯Az¯Bz= (cos?Ay+ sin?Az)(cos?By+ sin?Bz) + (?sin?Ay+ cos?Az)(?sin?By+ cos?Bz) = cos2?AyBy+ sin?cos?(AyBz+ AzBy) + sin2?AzBz+ sin2?AyBy? sin?cos?(AyBz+ AzBy) + cos2?AzBz = (cos2? + sin2?)AyBy+ (sin2? + cos2?)AzBz= AyBy+ AzBz. X (b) (Ax)2+ (Ay)2+ (Az)2= ⌃3 i=1 ⇢1 if j = k Moreover, if R is to preserve lengths for all vectors A, then this condition is not only su?cient but also necessary. For suppose A = (1,0,0). Then ⌃j,k(⌃iRijRik)AjAk= ⌃iRi1Ri1, and this must equal 1 (since we want A Then we want 2 = ⌃j,k(⌃iRijRik)AjAk= ⌃iRi1Ri1+ ⌃iRi2Ri2+ ⌃iRi1Ri2+ ⌃iRi2Ri1. But we already know that the first two sums are both 1; the third and fourth are equal, so ⌃iRi1Ri2= ⌃iRi2Ri1= 0, and so on for other unequal combinations of j, k. X In matrix notation:˜RR = 1, where˜R is the transpose of R. = 3ˆ y +1 2 3ˆ x ?2 3ˆ z ?⌃3 ??⌃3 ?= ⌃j,k(⌃iRijRik)AjAk. i=1AiAi= ⌃3 j=1RijAj k=1RikAk ? This equals A2 x+ A2 y+ A2 zprovided ⌃3 i=1RijRik= 0 if j 6= k 2 x+A 2 y+A 2 z= 1). Likewise, ⌃3 i=1Ri2Ri2= ⌃3 i=1Ri3Ri3= 1. To check the case j 6= k, choose A = (1,1,0). 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. @solutionmanual1

6. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 6 CHAPTER 1. VECTOR ANALYSIS 5 CHAPTER 1. VECTOR ANALYSIS CHAPTER 1. VECTOR ANALYSIS 5 Problem 1.9 y ✻y ✻ z′ ✻ ❃ y ✻y ✻ z′ ✻ ✿ ❃ ✿ Looking down the axis: Looking down the axis: ❄ ✲x ✒ ❄ ✲x ✒ Looking down the axis: ■ sy′ ■ ✰ ✰ x′ sx ✠ z sy′ ✰ ✰ x′ sx ✠ z z z 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. 0 0 1 0 1 0 0 1 1 0 0 @ A R = 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 to behavior 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 vector and 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 pseudoscalar changes sign under inversion of coordinates. Problem 1.11 (a)rf = 2xˆ x + 3y2ˆ y + 4z3ˆ z (b)rf = 2xy3z4ˆ x + 3x2y2z4ˆ y + 4x2y3z3ˆ z (c)rf = exsiny lnz ˆ x + excosy lnz ˆ y + exsiny(1/z)ˆ z Problem 1.12 (a) rh = 10[(2y ? 6x ? 18)ˆ x + (2x ? 8y + 28)ˆ y]. rh = 0 at summit, so 2y ? 6x ? 18 = 0 2x ? 8y + 28 = 0 =) 6x ? 24y + 84 = 0 22y = 66 =) y = 3 =) 2x ? 24 + 28 = 0 =) x = ?2. Top is 3 miles north, 2 miles west, of South Hadley. ? 2y ? 18 ? 24y + 84 = 0. (b) Putting in x = ?2, y = 3: h = 10(?12 ? 12 ? 36 + 36 + 84 + 12) = 720 ft. (c) Putting in x = 1, y = 1: rh = 10[(2 ? 6 ? 18)ˆ x + (2 ? 8 + 28)ˆ y] = 10(?22ˆ x + 22ˆ y) = 220(?ˆ x + ˆ y). |rh| = 220p2 ⇡ 311 ft/mile; direction: northwest. 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. c ⃝2005 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. c ⃝2005 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. @solutionmanual1

7. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 7 CHAPTER 1. VECTOR ANALYSIS Problem 1.13 r = (x ? x0)ˆ x + (y ? y0)ˆ y + (z ? z0)ˆ z; r =p(x ? x0)2+ (y ? y0)2+ (z ? z0)2. (a) r(r2) = (b) r( = ?1 = ?()?3 (c) @z()ˆ z = 2(x?x0)ˆ x+2(y?y0)ˆ y+2(z?z0)ˆ z = 2 r . @y()?1 22(z ? z0)ˆ z @x[(x?x0)2+(y?y0)2+(z?z0)2]ˆ x+@ @ @x[(x ? x0)2+ (y ? y0)2+ (z ? z0)2]?1 22(x ? x0)ˆ x ?1 @y()ˆ y+@ @ @ @z()?1 1r) = 2()?3 2ˆ x + 2ˆ y + 2ˆ z @ 2()?3 2()?3 22(y ? y0)ˆ y ?1 2[(x ? x0)ˆ x + (y ? y0)ˆ y + (z ? z0)ˆ z] = ?(1/r3)r = ?(1/r2)ˆ r . @x(rn) = n rn?1 @r 2 Problem 1.14 @x= n rn?1(1 1r2 rx) = n rn?1ˆ rx, so r(rn) = n rn?1ˆ r @ y = +y cos? + z sin?; multiply by sin?: y sin? = +y sin?cos? + z sin2?. z = ?y sin? + z cos?; multiply by cos?: z cos? = ?y sin?cos? + z cos2?. Add: y sin? + z cos? = z(sin2? + cos2?) = z. Likewise, y cos? ? z sin? = y. So @y @z= ?sin?;@z @y=@f @y (rf)z=@f Problem 1.15 @y @y= cos?; @y= sin?;@z @z= cos?. Therefore ) (rf)y=@f @y @y+@f @y @z+@f @z @y= +cos?(rf)y+ sin?(rf)z @z @z= ?sin?(rf)y+ cos?(rf)z So rf transforms as a vector. qed @z @z=@f @y @z (a)r·va= @x(x2) + @y(3xz2) + @ @z(?2xz) = 2x + 0 ? 2x = 0. @ @ (b)r·vb= @x(xy) + @y(2yz) + @ @z(3xz) = y + 2z + 3x. @ @ (c)r·vc= Problem 1.16 @x(y2) + @y(2xy + z2) + @ @z(2yz) = 0 + (2x) + (2y) = 2(x + y) @ @ h i x(x2+ y2+ z2)?3 z(x2+ y2+ z2)?3 @y(y r·v = + @y = ()?3 + z(?3/2)()?5 @x(x y(x2+ y2+ z2)?3 r3) + r3) + @ @z(z + @z r3) = h @ @ @ 2 @x h i i @ @ 2 2 2 + x(?3/2)()?5 22z = 3r?3?3r?5(x2+y2+z2) = 3r?3?3r?3= 0. 22x + ()?3 2 + y(?3/2)()?5 22y + ()?3 2 This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, r·v is infinite at that one point, and zero elsewhere, as we shall see in Sect. 1.5. Problem 1.17 vy= cos?vy+ sin?vz; vz= ?sin?vy+ cos?vz. @vy @y= = ⇣@vy cos? + ⌘ ⇣ ⌘ @vy @ycos? +@vz ⇣@vy @z= ?@vy = ? @y @y+@vy ⇣ @y sin? + @y @y+@vz ⌘ @vz @y @ysin? = cos? + sin?. Use result in Prob. 1.14: @vz @y @z @y @z @y @y @z @z ⌘ @ycos? +@vy @zsin? @vz @ycos? +@vz @z+@vy @z ⇣ @zsin? sin? + sin?. ⇣@vy ⌘ ⇣ ⌘ @y @y @z+@vz ⌘ @zsin? +@vz ⇣ @zcos? = ? @zcos? cos? @vz @z @z @z @z @z ⌘ ?@vy @ysin? +@vy @ysin? +@vz @zcos? cos?. So ?@vz 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. @solutionmanual1

8. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 8 CHAPTER 1. VECTOR ANALYSIS @vy @y+@vz @vy @ycos2? +@vy @zsin2? +@vy @ysin2? ?@vy @zsin?cos? +@vz ?@vz ?cos2? + sin2??+@vz ˆ z @ @z x23xz2?2xz ˆ x ˆ y ˆ z @ @x @y @z xy 2yz 3xz ˆ x ˆ y ˆ z @ @x @y @z y2(2xy + z2) 2yz Problem 1.19 @z= @ysin?cos? +@vz @zcos2? ?sin2? + cos2??= @zsin?cos? @ysin?cos? +@vz @vy @y @vy @y+@vz @z. X = @z Problem 1.18 ?????? ?????? ˆ x @ @x ˆ y @ @y (a) r⇥va= = ˆ x(0 ? 6xz) + ˆ y(0 + 2z) +ˆ z(3z2? 0) = ?6xz ˆ x + 2z ˆ y + 3z2ˆ z. ?????? ?????? (b) r⇥vb= @ @ = ˆ x(0 ? 2y) + ˆ y(0 ? 3z) +ˆ z(0 ? x) = ?2y ˆ x ? 3z ˆ y ? xˆ z. ?????? As we go from point A to point B (9 o’clock to 10 o’clock), x increases, y increases, vxincreases, and vydecreases, so @vx/@y > 0, while @vy/@y < 0. On the circle, vz = 0, and there is no dependence on z, so Eq. 1.41 says ?????? (c) r⇥vc= @ @ = ˆ x(2z ? 2z) + ˆ y(0 ? 0) +ˆ z(2y ? 2y) = 0. y v v B ✓@vy ◆ v @x?@vx r ⇥ v = ˆ z @y A x points in the negative z direction (into the page), as the right hand rule would suggest. (Pick any other nearby points on the circle and you will come to the same conclusion.) [I’m sorry, but I cannot remember who suggested this cute illustration.] z v Problem 1.20 v = y ˆ x + xˆ y; or v = yz ˆ x + xz ˆ y + xyˆ z; or v = (3x2z ? z3)ˆ x + 3ˆ y + (x3? 3xz2)ˆ z; or v = (sinx)(coshy)ˆ x ? (cosx)(sinhy)ˆ y; etc. Problem 1.21 ⇣ = f + g ⌘ ⇣ ⌘ ⇣ ⌘ (i) r(fg) =@(fg) @xˆ x +@(fg) ⇣ @ @x(AyBz? AzBy) + = Ay@Bz +Ax = Bx ˆ y +@(fg) f@g ⌘ @ @z(AxBy? AyBx) @x+ Az@Bx @z? Ay@Bx + By @x ?? Az ˆ x + @z @x f@g @x+ g@f @y+ g@f = f(rg) + g(rf). f@g @z+ g@f qed ˆ z = ˆ x + ˆ y + ˆ z @y @z ⌘ @x @y @z ⇣ @ @y(AzBx? AxBz) + @By @x? By@Az @z? Bx ?@Ax ⇣ 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. @g @xˆ x +@g @yˆ y +@g @f @xˆ x +@f @yˆ y +@f @zˆ z @zˆ z (iv) r·(A⇥B) = @Ay @x? Az @By @z+ By@Ax @y?@Ay ?Ay @x+ Bz @y+ Bx@Az @Ay @z ?+ Bz @y = B·(r⇥A) ? A·(r⇥B). ⌘ @y? Ax@Bz ⌘ @y? Bz@Ax ⇣ @y ⇣ ⌘ ⇣@Ay ⌘ @Bz @y?@By @Az @z?@Az ⇣@By @(fAx) @x?@Ax ? Ax @z @y @z ⌘ ?@Bx @z qed @z?@Bz ⌘ @x?@Bx @x ⇣ ⇣@(fAy) ⌘ ?@(fAy) @(fAz) @y ?@(fAz) ?@(fAx) (v) r⇥(fA) = ˆ y + ˆ z @x @y @solutionmanual1

9. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 9 CHAPTER 1. VECTOR ANALYSIS ⇣ ⌘ ⇣ ⌘ @y? f@Ay f@Ay ⌘ Ay@f @y+ Az@f + h⇣ ? = f (r⇥A) ? A⇥(rf). ⇣ + Ax@Bz @z? Ay@f @x? f@Ax ˆ x +?@Ax @y @z+ Ax@f ˆ z ⇣@Ay @z @x? Az@f = ˆ x + ˆ y f@Az f@Ax ⌘ @x?@Ax @x? Ax@f @z? f@Az @z @x ⇣ h⇣ @x+ Ay@f @y? Ax@f ?ˆ y + ˆ x + Az@f qed @y ⌘ i Ax@f @Az @y?@Ay = f ˆ z ⇣ @z?@Az ⌘ @z @x @y ˆ y + ⇣ ⌘ ⌘ i @z? Az@f @y? Ay@f ˆ z @x Problem 1.22 ⌘ ⇣ ⌘ @By @x+ Ay ⌘ @By @y+ Az @By @z (a) (A·r)B = @x+ Ay@Bx ⇣ x2+y2+z2. Let’s just do the x component. [(ˆ r·r)ˆ r]x= =1 r x + x(?1 =1 r r3 Same goes for the other components. Hence: (ˆ r·r)ˆ r = 0 . (c) (va·r)vb= = x2(y ˆ x + 0ˆ y + 3zˆ z) + 3xz2(xˆ x + 2z ˆ y + 0ˆ z) ? 2xz (0ˆ x + 2y ˆ y + 3xˆ z) =?x2y + 3x2z2?ˆ x +?6xz3? 4xyz?ˆ y +?3x2z ? 6x2z?ˆ z @y+ Az@Bx ˆ x + ˆ y Ax@Bx Ax @z @x+ Ay@Bz @y+ Az@Bz ˆ z. @z x ˆ x+y ˆ y+z ˆ z p 1 p n (b) ˆ r =r r= ⇣ ⌘ @x+ y@ @y+ z@ x@ h r? x p @z x2+y2+z2 i h i h io 2) )32x + yx )32y ?x2+ y2+ z2? =1 + zx )32z ?x 1 1 ?1 r? 1 ?1 1 p (p (p (p 2 2 ?x ?x3+ xy2+ xz2? =1 ?x ?= 0. 1 x r3 r?x r r r ⇣ ⌘ @x+ 3xz2 @ @y? 2xz@ (xy ˆ x + 2yz ˆ y + 3xzˆ z) x2 @ @z = x2?y + 3z2?ˆ x + 2xz?3z2? 2y?ˆ y ? 3x2zˆ z Problem 1.23 @x+@Ay ?? Az @y+ Az@Bx @By @x+@Az ? (ii) [r(A·B)]x= [A⇥(r⇥B)]x= Ay(r⇥B)z? Az(r⇥B)y= Ay [B⇥(r⇥A)]x= By [(A·r)B]x=?Ax So [A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A]x = Ay +Ax@Bx = Bx@Ax +Bz = [r(A·B)]x(same for y and z) (vi) [r⇥(A⇥B)]x= = @x(AxBx+ AyBy+ AzBz) =@Ax @xBx+ Ax@Bx ?@By @x @x+ Ay@Bx @xBy+ Ay ?@Bx @xBz+ Az@Bz @ @x @x?@Bx ? @z?@Bz @y @x ?@Ay ?? Bz @z ?@Ax @x?@Ax @x+ Ay @x+ By@Ax @z?@Az ?Bx= Ax@Bx @z @y @y+ Az @y+ Bz@Ax @ @ @ @z [(B·r)A]x= Bx@Ax @By @x? Ay@Bx @x+ Ay@Bx @x+ Ax@Bx ??@Ax @Ay @x? By@Ax @y+ Bz@Ax ?+ Ay @x+ By @x+ By@Ax @y / +@Ax ?+ Az @y(A⇥B)z? @Ax @yBy+ Ax @z+ Az@Bz @z+ Bx@Ax ?@Ay @z+ Bz@Az @y? Az@Bx @y+ Az@Bx @x+ By @z / +@Az @y? Bz@Ax @z ?@By @x ? @y / @y / +@Bx @y / @x?@Ax @x?@Bx @z / ??@Bx @z(A⇥B)y= @By @y?@Ay ? @z / @z / +@Bz @x+@Ax @x+@Bx @y(AxBy? AyBx) ? @yBx? Ay@Bx @ @z(AzBx? AxBz) @z+@Ax @ @ @ @zBz+ Ax@Bz @y?@Az @zBx? Az@Bx @z [(B·r)A ? (A·r)B + A(r·B) ? B(r·A)]x = Bx@Ax ?@Bx ??Bx ?@Ax ? @x+@By @x+@Ay @x+By@Ax @y+Bz@Ax @z+Ax @y+@Bz @y+@Az @z?Ax@Bx @x?Ay@Bx @y?Az@Bx @z @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. @solutionmanual1

10. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 10 CHAPTER 1. VECTOR ANALYSIS ??@Bx ?+ Bx ?@Ax ? @x / +@Bx ?+ Az @x / +@By ??@Bx @x / ?@Ax @x / ?@Ay = By@Ax + Ay = [r⇥(A⇥B)]x(same for y and z) Problem 1.24 r(f/g) = = g2 = g2 g @y+ Ax ??@Bx @y+@Bz ?+ Bz @y?@Az @z @z ?@Ax ? @y @z @z @x(f/g)ˆ x + g@f h @ @x(Ax/g) + = = g2 g @y(f/g)ˆ y + g@f g2 @xˆ x +@f @ @z(f/g)ˆ z ˆ y + ⌘ @ @z(Az/g) @y?Ay g2 @x+@Ay @ @ @y?f@g @x?f@g @z?f@g g2 ? f g@f ˆ x + ˆ z @y @x @z ⇣ ⇣ ⌘i =grf?frg @f @yˆ y +@f @g @xˆ x +@g @yˆ y +@g qed @zˆ z @zˆ z 1 . g2 r·(A/g) = @y(Ay/g) + + @ g@Ay @g @y g@Ax @x?Ax h @y(Az/g) ? = = g2 g = g@Az @z?Az ⇣ @g @x @g @x + g2 ⇣ @ g2 @x+ Ay@g ⌘ ⌘i =gr·A?A·rg Ax@g @y+ Az@g @y+@Az qed @Ax 1 ? . g2 @z @z [r⇥(A/g)]x= @ @z(Ay/g) g@Ay @z?Ay ⌘ g@Az @y?Az h @g @y @g @z ? g2 ⇣ g2 ? (same for y and z). ⇣ ⌘i @Az @y?@Ay @g @y? Ay@g 1 Az @z @z g(r⇥A)x+(A⇥rg)x g2 qed Problem 1.25 ?????? ?????? ˆ x x 3y ?2x 0 @ @x(6xz) + ⇣ @ @y(0) ? ˆ y 2y 3z ˆ z (a) A⇥B = = ˆ x(6xz) + ˆ y(9zy) +ˆ z(?2x2? 6y2) r·(A⇥B) = r⇥A = ˆ x r⇥B = ˆ x @y(9zy) + @z(2y) @ @z(?2x2? 6y2) = 6z + 9z + 0 = 15z + ˆ y?@ @ ⌘ ⇣ ⌘ @x(3z)?+ˆ z @y(3z) ? @z(x) ? @z(3y) ? @x(2y) ? ⇣ @y(x) = 0; B·(r⇥A) = 0 @y(3y) = ?5ˆ z; A·(r⇥B) = ?15z @ @ @ @ @ ⇣ ?= B·(r⇥A) ? A·(r⇥B) = 0 ? (?15z) = 15z. X (b) A·B = 3xy ? 4xy = ?xy ; r(A·B) = r(?xy) = ˆ x@ ?????? (A·r)B = (B·r)A = A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A = ?10y ˆ x + 5xˆ y + 6y ˆ x ? 2xˆ y + 3y ˆ x ? 4xˆ y = ?y ˆ x ? xˆ y = r·(A·B). X (c) r⇥(A⇥B) = ˆ x = ˆ x(?12y ? 9y) + ˆ y(6x + 4x) +ˆ z(0) = ?21y ˆ x + 10xˆ y r·A = ⌘ ⌘ + ˆ y?@ @x(0)?+ˆ z @ @z(?2x) @x(?2x) ? @ @ @ r·(A⇥B) @x(?xy) + ˆ y@ @y(?xy) = ?y ˆ x ? xˆ y ?????? ˆ x ˆ y x 2y 3z 0 0 ?5 x@ ˆ z A⇥(r⇥B) = = ˆ x(?10y) + ˆ y(5x); B⇥(r⇥A) = 0 ⇣ ⌘ @x+ 2y@ 3y@ @y+ 3z@ ⌘ (3y ˆ x ? 2xˆ y) = ˆ x(6y) + ˆ y(?2x) (xˆ x + 2y ˆ y + 3zˆ z) = ˆ x(3y) + ˆ y(?4x) @z ⇣ @x? 2x@ @y ⇣ ⌘ ⇣ ⌘ + ˆ y?@ @x(?2x2? 6y2)?+ˆ z @x(3y) + @y(?2x2? 6y2) ? @ @z(9zy) @z(6xz) ? @x(9zy) ? @y(6xz) @ @ @ @ @x(x) + @y(2y) + @ @z(3z) = 1 + 2 + 3 = 6; r·B = @y(?2x) = 0 @ @ @ @ 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. @solutionmanual1

11. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 11 CHAPTER 1. VECTOR ANALYSIS (B·r)A ? (A·r)B + A(r·B) ? B(r·A) = 3y ˆ x ? 4xˆ y ? 6y ˆ x + 2xˆ y ? 18y ˆ x + 12xˆ y = ?21y ˆ x + 10xˆ y = r⇥(A⇥B). X Problem 1.26 (a)@2Ta @2Ta @y2 =@2Ta @x2 = 2; @z2 = 0 ) r2Ta= 2. @y2 =@2Tb (b)@2Tb @x2 =@2Tb @z2 = ?Tb ) r2Tb= ?3Tb= ?3sinxsiny sinz. @x2 = 25Tc;@2Tc (c)@2Tc @y2 = ?16Tc;@2Tc @y2 =@2vx @2vy @y2 = 0 ;@2vy @x2 =@2vz Problem 1.27 @z2 = ?9Tc ) r2Tc= 0. @z2 = 0 ) r2vx= 2 @z2 = 6x ) r2vy= 6x @y2 =@2vz 9 > ?+ @z @x? @2vx @x2 = 2 ;@2vx @2vy @x2 = @2vz (d) > ; = r2v = 2ˆ x + 6xˆ y. @z2 = 0 ) r2vz= 0 ⇣ @y @x ⌘ ⇣@vy @x@z ⌘ ?@vx @z @y @vz @y?@vy @2vz r·(r⇥v) = = + @z?@vz ⌘ @x?@vx @2vy @ @ @ @z @x @z ⇣ @y @x ⇣ @y ⇣ ⌘ ⌘ @2vy @2vz @x@y? @2vx @y @z?@2vx + + = 0, by equality of cross-derivatives. From Prob. 1.18: r⇥va= ?6xz ˆ x+2z ˆ y+3z2ˆ z ) r·(r⇥va) = Problem 1.28 ?????? In Prob. 1.11(b), rf = 2xy3z4ˆ x + 3x2y2z4ˆ y + 4x2y3z3ˆ z, so ?????? Problem 1.29 (a) (0,0,0) ?! (1,0,0). x : 0 ! 1,y = z = 0;dl = dxˆ x;v · dl = x2dx;Rv · dl =R1 (1,1,0) ?! (1,1,1). x = y = 1,z : 0 ! 1;dl = dzˆ z;v · dl = y2dz = dz;Rv · dl =R1 (b) (0,0,0) ?! (0,0,1). x = y = 0,z : 0 ! 1;dl = dzˆ z;v · dl = y2dz = 0;Rv · dl = 0. (0,1,1) ?! (1,1,1). x : 0 ! 1,y = z = 1;dl = dxˆ x;v · dl = x2dx;Rv · dl =R1 @x(?6xz)+@ @y(2z)+@ @z(3z2) = ?6z+6z = 0. @ ?????? ˆ x @ @x @t @x ˆ y @ @y @t @y ˆ z @ @z @t @z = ˆ x? ?+ ˆ y? ?+ˆ z? ? @2t @y @z? @2t @z @y @2t @z @x? @2t @x@z @2t @x@y? @2t @y @x r⇥(rt) = = 0, by equality of cross-derivatives. ?????? ˆ x @ @x ˆ y @ @y ˆ z @ @z r⇥(rf) = 2xy3z43x2y2z44x2y3z3 = ˆ x(3 · 4x2y2z3? 4 · 3x2y2z3) + ˆ y(4 · 2xy3z3? 2 · 4xy3z3) +ˆ z(2 · 3xy2z4? 3 · 2xy2z4) = 0. X 0x2dx = (x3/3)|1 0= 1/3. (1,0,0) ?! (1,1,0). x = 1,y : 0 ! 1,z = 0;dl = dy ˆ y;v · dl = 2yz dy = 0;Rv · dl = 0. Total: 0dz = z|1 0= 1. Rv · dl = (1/3) + 0 + 1 = 4/3. (0,0,1) ?! (0,1,1). x = 0,y : 0 ! 1,z = 1;dl = dy ˆ y;v·dl = 2yz dy = 2y dy;Rv·dl =R1 Total: 02y dy = y2|1 0= 1/3. 0= 1. 0x2dx = (x3/3)|1 Rv · dl = 0 + 1 + (1/3) = 4/3. (c) x = y = z : 0 ! 1;dx = dy = dz;v · dl = x2dx + 2yz dy + y2dz = x2dx + 2x2dx + x2dx = 4x2dx; Rv · dl =R1 (d)Hv · dl = (4/3) ? (4/3) = 0. 04x2dx = (4x3/3)|1 0= 4/3. 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. @solutionmanual1

12. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 12 CHAPTER 1. VECTOR ANALYSIS Problem 1.30 x,y : 0 ! 1,z = 0;da = dxdyˆ z;v · da = y(z2? 3)dxdy = ?3y dxdy;Rv · da = ?3R2 xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total flux for the cube is 20 + 12 = 32. Problem 1.31 RT d⌧ =Rz2dxdy dz. You can do the integrals in any order—here it is simplest to save z for last: z2 0dxR2 0y dy = 0)(y2 ?3(x|2 0) = ?3(2)(2) = ?12. In Ex. 1.7 we got 20, for the same boundary line (the square in the 2|2 Z ✓Z ◆ ? dx = 1?y?z. For a given z, y ranges from 0 to 0 = (1?z)2?[(1?z)2/2] = (1?z)2/2 = 0(z2 Z dx dy dz. The sloping surface is x+y+z = 1, so the x integral isR(1?y?z) 0 (1/2) ? z + (z2/2). Finally, the z integral isR1 Problem 1.32 0 1?z, so the y integral isR(1?z) 1 6?1 (1?y?z)dy = [(1?z)y?(y2/2)]|(1?z) 0z2(1 2)dz =R1 2? z +z2 2? z3+z4 2)dz = (z3 6?z4 4+z5 10)|1 0= 4+ 10= 1/60. 1 T(b) = 1 + 4 + 2 = 7; T(a) = 0. ) T(b) ? T(a) = 7. rT = (2x + 4y)ˆ x + (4x + 2z3)ˆ y + (6yz2)ˆ z; rT·dl = (2x + 4y)dx + (4x + 2z3)dy + (6yz2)dz (a) Segment 1: x : 0 ! 1, y = z = dy = dz = 0.RrT·dl =R1 Segment 3: z : 0 ! 1, x = y = 1, dx = dy = 0.RrT·dl =R1 Segment 2: y : 0 ! 1, x = 0, z = 1, dx = dz = 0.RrT·dl =R1 = (x2+ 4x)??1 rT·dl = (2x + 4x)dx + (4x + 2x6)dx + (6xx4)2xdx = (10x + 14x6)dx. Rb Problem 1.33 r·v = y + 2z + 3x R(r·v)d⌧ =R(y + 2z + 3x)dxdy dz =RRnR2 =RnR2 ,!⇥y2+ (4z + 6)y⇤2 =R2 Numbering the surfaces as in Fig. 1.29: 9 > > > 0(2x)dx = x2??1 0= 1. > ; > > = 9 > Rb Segment 2: y : 0 ! 1, x = 1, z = 0, dx = dz = 0.RrT·dl =R1 (b) Segment 1: z : 0 ! 1, x = y = dx = dy = 0.RrT·dl =R1 Segment 3: x : 0 ! 1, y = z = 1, dy = dz = 0.RrT·dl =R1 (c) x : 0 ! 1, y = x, z = x2, dy = dx, dz = 2xdx. arT·dl = 7. X 0(4)dy = 4y|1 0= 4. 0= 2. 0(6z2)dz = 2z3??1 0(2)dy = 2y|1 0(2x + 4)dx 0(0)dz = 0. > ; = 0= 2. Rb arT·dl = 7. X 0= 1 + 4 = 5. 0(10x + 14x6)dx = (5x2+ 2x7)??1 arT·dl =R1 0= 5 + 2 = 7. X o 0(y + 2z + 3x)dx ,!⇥(y + 2z)x +3 0= 4 + 2(4z + 6) = 8z + 16 dy dz 2x2⇤2 0= 2(y + 2z) + 6 o 0(2y + 4z + 6)dy dz 0(8z + 16)dz = (4z2+ 16z)??2 0= 16 + 32 = 48. 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. @solutionmanual1

13. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 13 CHAPTER 1. VECTOR ANALYSIS (i) da = dy dz ˆ x,x = 2. v·da = 2y dy dz.Rv·da =RR2y dy dz = 2y2??2 (iv) da = ?dxdz ˆ y,y = 0. v·da = 0.Rv·da = 0. (vi) da = ?dxdyˆ z,z = 0. v·da = 0.Rv·da = 0. Problem 1.34 0= 8. (ii) da = ?dy dz ˆ x,x = 0. v·da = 0.Rv·da = 0. (v) da = dxdyˆ z,z = 2. v·da = 6xdxdy.Rv·da = 24. )Rv·da = 8 + 16 + 24 = 48 X r⇥v = ˆ x(0 ? 2y) + ˆ y(0 ? 3z) +ˆ z(0 ? x) = ?2y ˆ x ? 3z ˆ y ? xˆ z. da = dy dz ˆ x, if we agree that the path integral shall run counterclockwise. So (r⇥v)·da = ?2y dy dz. R(r⇥v)·da =RnR2?z ,! y2??2?z = ??8 ? 8 +8 Meanwhile, v·dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments. (iii) da = dxdz ˆ y,y = 2. v·da = 4z dxdz.Rv·da =RR4z dxdz = 16. o 6 @ (?2y)dy dz = ?(2 ? z)2 z 0 y = 2 ? z @ ⌘??? ⇣ 0 = ?R2 2 4z ? 2z2+z3 @ 0(4 ? 4z + z2)dz = ? ?= ?8 3 @ 0 @@ - y 3 3 6 @ z I(2) @ @ (3) @ @ @ ? @@ - - y (1) Rv·dl = 0. (1) x = z = 0; dx = dz = 0. y : 0 ! 2. (2) x = 0; z = 2 ? y; dx = 0, dz = ?dy, y : 2 ! 0. v·dl = 2yz dy. Rv·dl =R0 Problem 1.35 By Corollary 1,R(r⇥v)·da should equal4 (i) da = dy dz ˆ x, x = 1; y,z : 0 ! 1. (r⇥v)·da = (4z2? 2)dy dz; = (4 3. (ii) da = ?dxdyˆ z, z = 0; x,y : 0 ! 1. (r⇥v)·da = 0; (iii) da = dxdz ˆ y, y = 1; x,z : 0 ! 1. (r⇥v)·da = 0; (iv) da = ?dxdz ˆ y, y = 0; x,z : 0 ! 1. (r⇥v)·da = 0; (v) da = dxdyˆ z, z = 1; x,y : 0 ! 1. (r⇥v)·da = 2dxdy; )R(r⇥v)·da = ?2 3y3???2 0(4y ? 2y2)dy = ??2y2?2 0= ??8 ?2 3· 8?= ?8 22y(2 ? y)dy = ?R2 3. Rv·dl = 0. SoHv·dl = ?8 3. r⇥v = (4z2? 2x)ˆ x + 2zˆ z. 3. X (3) x = y = 0; dx = dy = 0; z : 2 ! 0. v·dl = 0. R(r⇥v)·da =R1 0(4z2? 2)dz 3z3? 2z)??1 0=4 3? 2 = ?2 R(r⇥v)·da = 0. R(r⇥v)·da = 0. R(r⇥v)·da = 0. R(r⇥v)·da = 2. 3. X 3+ 2 =4 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. @solutionmanual1

14. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 14 CHAPTER 1. VECTOR ANALYSIS Problem 1.36 (a) Use the product rule r⇥(fA) = f(r⇥A) ? A ⇥ (rf) : Z (I used Stokes’ theorem in the last step.) Z Z I Z f(r⇥A) · da = r⇥(fA) · da + [A ⇥ (rf)] · da = fA · dl + [A ⇥ (rf)] · da. qed S S S P S (b) Use the product rule r·(A ⇥ B) = B · (r⇥A) ? A · (r⇥B) : Z (I used the divergence theorem in the last step.) Z Z I Z B · (r⇥A)d⌧ = r·(A ⇥ B)d⌧ + A · (r⇥B)d⌧ = (A ⇥ B) · da + A · (r⇥B)d⌧. qed V V V S V ✓ ◆ px2+ y2+ z2; ? = tan?1?y ?. Problem 1.37 r = ✓ = cos?1 ; z p x x2+y2+z2 Problem 1.38 There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry from Fig. 1.36. The most systematic approach is to study the expression: r = xˆ x + y ˆ y + zˆ z = rsin✓cos?ˆ x + rsin✓sin?ˆ y + rcos✓ˆ z. If I only vary r slightly, then dr = it a unit vector, I must divide by its length. Thus: @ @r(r)dr is a short vector pointing in the direction of increase in r. To make @r @? ??@r @r @r @r @✓ ??;ˆ✓ = @r ??@r ??;ˆ? = ˆ r = ??@r ??@r ??. @r ??@r ??2= r2sin2✓sin2? + r2sin2✓cos2? = r2sin2✓. @✓ @? ??2= sin2✓cos2? + sin2✓sin2? + cos2✓ = 1. @r @r= sin✓cos?ˆ x + sin✓sin?ˆ y + cos✓ˆ z; @r @✓= rcos✓cos?ˆ x + rcos✓sin?ˆ y ? rsin✓ˆ z; @r @?= ?rsin✓sin?ˆ x + rsin✓cos?ˆ y; ˆ r = sin✓cos?ˆ x + sin✓sin?ˆ y + cos✓ˆ z. ˆ✓ = cos✓cos?ˆ x+cos✓sin?ˆ y?sin✓ˆ z. ˆ? = ?sin?ˆ x + cos?ˆ y. Check: ˆ r·ˆ r = sin2✓(cos2? + sin2?) + cos2✓ = sin2✓ + cos2✓ = 1, X ˆ✓·ˆ? = ?cos✓sin?cos? + cos✓sin?cos? = 0, X sin✓ˆ r = sin2✓cos?ˆ x + sin2✓sin?ˆ y + sin✓cos✓ˆ z. cos✓ˆ✓ = cos2✓cos?ˆ x + cos2✓sin?ˆ y ? sin✓cos✓ˆ z. Add these: (1) sin✓ˆ r + cos✓ˆ✓ = +cos?ˆ x + sin?ˆ y; (2) ˆ? = ?sin?ˆ x + cos?ˆ y. Multiply (1) by cos?, (2) by sin?, and subtract: ??2= r2cos2✓cos2? + r2cos2✓sin2? + r2sin2✓ = r2. @✓ ??@r @? ) etc. ˆ x = sin✓cos?ˆ r + cos✓cos?ˆ✓ ? sin?ˆ?. 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. @solutionmanual1

15. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 15 CHAPTER 1. VECTOR ANALYSIS Multiply (1) by sin?, (2) by cos?, and add: ˆ y = sin✓sin?ˆ r + cos✓sin?ˆ✓ + cos?ˆ?. cos✓ˆ r = sin✓cos✓cos?ˆ x + sin✓cos✓sin?ˆ y + cos2✓ˆ z. sin✓ˆ✓ = sin✓cos✓cos?ˆ x + sin✓cos✓sin?ˆ y ? sin2✓ˆ z. Subtract these: ˆ z = cos✓ˆ r ? sin✓ˆ✓. Problem 1.39 (a) r·v1= R(r·v1)d⌧ =R(4r)(r2sin✓drd✓d?) = (4)RR ?r2 1 They don’t agree! The point is that this divergence is zero except at the origin, where it blows up, so our calculation ofR(r·v2) is incorrect. The right answer is 4⇡. r·v = = = 3cos✓ + 2cos✓ ? sin? = 5cos✓ ? sin? R(r·v)d⌧ =R(5cos✓ ? sin?)r2sin✓drd✓d? =RR = 3 ,!sin2✓ = @ @r(r2r2) = 1 r24r3= 4r 1 r2 ⇣ ⌘ 0sin✓d✓R2⇡ d? = 4⇡R4X (Note: at surface of sphere r = R.) 0r3drR⇡ R4 4 0d? = (4) (2)(2⇡) = 4⇡R4 Rv1·da =R(r2ˆ r)·(r2sin✓d✓d?ˆ r) = r4R⇡ (b) r·v2= Rv2·da =R?1 0sin✓d✓R2⇡ 0 ?= 0 ) R(r·v2)d⌧ = 0 1 r2 @ @r r2 r2ˆ r?(r2sin✓d✓d?ˆ r) =Rsin✓d✓d? = 4⇡. Problem 1.40 @ @r(r2rcos✓) + @ @✓(sin✓rsin✓) + @?(rsin✓cos?) 1 r2 1 r23r2cos✓ + 1 1 @ r sin✓ r sin✓ r sin✓r2sin✓cos✓ + r sin✓rsin✓(?sin?) 1 1 hR2⇡ i 0r2drR ✓ 0(5cos✓ ? sin?)d? ,!2⇡(5cos✓) d✓sin✓ 2 0 ⇣ ⌘ (10⇡)R ⇡ R3 0sin✓cos✓d✓ 2 ??? ⇡ 2 =1 2 2 0 5⇡ 3R3. Two surfaces—one the hemisphere: da = R2sin✓d✓d?ˆ r; r = R; ? : 0 ! 2⇡, ✓ : 0 !⇡ Rv·da =R(rcos✓)R2sin✓d✓d? = R3R ⇡ Rv·da =R(rsin✓)(rdrd?) =RR Problem 1.41 rt = (cos✓ + sin✓cos?)ˆ r + (?sin✓ + cos✓cos?)ˆ✓ + 2. d? = R3?1 ?(2⇡) = ⇡R3. 0sin✓cos✓d✓R2⇡ 0r2drR2⇡ 2 2 2). r : 0 ! R, ? : 0 ! 2⇡. 0 other the flat bottom: da = (dr)(rsin✓d?)(+ˆ✓) = rdrd?ˆ✓ (here ✓ =⇡ d? = 2⇡R3 3. Rv·da = ⇡R3+2 0 3⇡R3. X Total: 3⇡R3=5 / sin?)ˆ? (?sin✓ 1 / sin✓ r2t = r·(rt) = = = = ?r2(cos✓ + sin✓cos?)?+ ⇥(sin2✓ + cos2✓)cos? ? cos?⇤= 0. 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. @ @✓(sin✓(?sin✓ + cos✓cos?)) + r sin✓(?2sin✓cos✓ + cos2✓cos? ? sin2✓cos?) ? @?(?sin?) 1 r sin✓cos? 1 r2 1 r22r(cos✓ + sin✓cos?) + 1 r sin✓[2sin✓cos✓ + 2sin2✓cos? ? 2sin✓cos✓ + cos2✓cos? ? sin2✓cos? ? cos?] 1 r sin✓ @ @r 1 1 @ r sin✓ r sin✓ 1 @solutionmanual1

16. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 16 CHAPTER 1. VECTOR ANALYSIS ) r2t = 0 Check: rcos✓ = z, rsin✓cos? = x ) in Cartesian coordinates t = x + z. Obviously Laplacian is zero. Gradient Theorem: Segment 1: ✓ =⇡ Rrt·dl =R2 Segment 2: ✓ =⇡ 2. rt·dl = (?sin?)(2d?) = ?2sin?d?. Segment 3: r = 2, ? =⇡ dl = rd✓ˆ✓ = 2d✓ˆ✓; rt·dl = (?sin✓ + cos✓cos?)(2d✓) = ?2sin✓d✓. Rrt·dl = ?R0 Total: Rb art·dl = t(b) ? t(a) 2, ? = 0, r : 0 ! 2. dl = drˆ r; rt·dl = (cos✓ + sin✓cos?)dr = (0 + 1)dr = dr. 0dr = 2. dl = rsin✓d?ˆ? = 2d?ˆ?. Rrt·dl = ?R ⇡ 2, r = 2, ? : 0 !⇡ ⇡ 2 0= ?2. 02sin?d? = 2cos?| 2 2; ✓ :⇡ 2! 0. 22sin✓d✓ = 2cos✓|0 2= 2. ⇡ ⇡ Rb art·dl = 2 ? 2 + 2 = 2 . Meanwhile, t(b) ? t(a) = [2(1 + 0)] ? [0( )] = 2. X Problem 1.42 From Fig. 1.42, ˆ s = cos?ˆ x + sin?ˆ y;ˆ? = ?sin?ˆ x + cos?ˆ y; ˆ z = ˆ z Multiply first by cos?, second by sin?, and subtract: ˆ scos? ?ˆ?sin? = cos2?ˆ x + cos?sin?ˆ y + sin2?ˆ x ? sin?cos?ˆ y = ˆ x(sin2? + cos2?) = ˆ x. So ˆ x = cos?ˆ s ? sin?ˆ?. Multiply first by sin?, second by cos?, and add: ˆ ssin? +ˆ?cos? = sin?cos?ˆ x + sin2?ˆ y ? sin?cos?ˆ x + cos2?ˆ y = ˆ y(sin2? + cos2?) = ˆ y. So ˆ y = sin?ˆ s + cos?ˆ?. ˆ z = ˆ z. Problem 1.43 (a) r·v = ?ss(2 + sin2?)?+1 @?(ssin?cos?) + ss(cos2? ? sin2?) + 3 @ @z(3z) 1 s 1 s2s(2 + sin2?) +1 = 4 + 2sin2? + cos2? ? sin2? + 3 = 4 + sin2? + cos2? + 3 = 8. (b)R(r·v)d⌧ =R(8)sdsd?dz = 8R2 Meanwhile, the surface integral has five parts: top: z = 5, da = sdsd?ˆ z; v·da = 3z sdsd? = 15sdsd?. bottom: z = 0, da = ?sdsd?ˆ z; v·da = ?3z sdsd? = 0. back: ? =⇡ left: ? = 0, da = ?dsdzˆ?; v·da = ?ssin?cos?dsdz = 0. front: s = 2, da = sd?dzˆ s; v·da = s(2 + sin2?)sd?dz = 4(2 + sin2?)d?dz. Rv·da = 4R ⇡ ⇣ +1 s @? = @ @s @ s = 0dz = 8(2)?⇡ ?(5) = 40⇡. Rv·da = 15R2 Rv·da = 0. 0sdsR ⇡ 0d?R5 2 2 0sdsR ⇡ 0d? = 15⇡. 2 Rv·da = 0. Rv·da = 0. 2, da = dsdzˆ?; v·da = ssin?cos?dsdz = 0. 0(2 + sin2?)d?R5 0dz = (4)(⇡ +⇡ 4)(5) = 25⇡. 2 SoHv·da = 15⇡ + 25⇡ = 40⇡. X (c) r⇥v = ⇣ ⌘ ˆ s +?@ ?s(2 + sin2?)?? ˆ z @s(3z)?ˆ? @?(3z) ? @ @s(s2sin?cos?) ? s(2ssin?cos? ? s2sin?cos?)ˆ z = 0. @ @z(ssin?cos?) 1 s @ @ @z ?s(2 + sin2?)?⌘ @ 1 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. @solutionmanual1

17. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 17 CHAPTER 1. VECTOR ANALYSIS Problem 1.44 (a) 3(32) ? 2(3) ? 1 = 27 ? 6 ? 1 = 20. (b) cos⇡ = -1. (c) zero. (d) ln(?2 + 3) = ln1 = zero. Problem 1.45 (a)R2 (b) By Eq. 1.94, ?(1 ? x) = ?(x ? 1), so 1 + 3 + 2 = 6. (c)R1 (d) 1 (if a > b), 0 (if a < b). ?2(2x + 3)1 3?(x)dx =1 3(0 + 3) = 1. 3)dx = 9??1 ?21 ?19x2 1 3?(x +1 3= 1 3. 3 Problem 1.46 (a)R1 So the integral is ?R1 So, xd (b)R1 Problem 1.47 ?1f(x)⇥xd dx?(x)⇤dx = xf(x)?(x)|1 ⇣ qed ?1?R1 ?(x)dx = 0 ? f(0) = ?f(0) = ?R1 ?1?R1 dx(xf(x))?(x)dx. d ?1 dx(xf(x)) = xdf dxf = xdf ?1f(x)?(x)dx. The first term is zero, since ?(x) = 0 at ±1; dx+dx dx+ f. d ⌘ xdf dx+ f ?1 dx?(x) = ??(x). ?1f(x)d✓ = f(0) =R1 dx✓(x)dx = f(1) ?R1 dxdx = f(x)✓(x)|1 ?1f(x)?(x)dx. Sod✓ df df dxdx = f(1) ? (f(1) ? f(0)) ?1 0 dx= ?(x). qed R⇢(r)d⌧ = qR?3(r ? r0)d⌧ = q. X (a) ⇢(r) = q?3(r ? r0). Check: (b) ⇢(r) = q?3(r ? a) ? q?3(r). (c) Evidently ⇢(r) = A?(r ? R). To determine the constant A, we require Q =R⇢d⌧ =RA?(r ? R)4⇡r2dr = A4⇡R2. Problem 1.48 Q Q So A = ⇢(r) = 4⇡R2?(r ? R). 4⇡R2. (a) a2+ a·a + a2= 3a2. (b)R(r ? b)2 1 (c) c2= 25 + 9 + 4 = 38 > 36 = 62, so c is outside V, so the integral is zero. (d) (e ? (2ˆ x + 2ˆ y + 2ˆ z))2= (1ˆ x + 0ˆ y + (?1)ˆ z)2= 1 + 1 = 2 < (1.5)2= 2.25, so e is inside V, and hence the integral is e·(d ? e) = (3,2,1)·(?2,0,2) = ?6 + 0 + 2 = -4. Problem 1.49 First method: use Eq. 1.99 to write J =Re?r?4⇡?3(r)?d⌧ = 4⇡e?0= 4⇡. 53?3(r)d⌧ = 125b2= 125(42+ 32) = 1 1 1 5. Second method: integrating by parts (use Eq. 1.59). 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. @solutionmanual1

18. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 18 CHAPTER 1. VECTOR ANALYSIS ✓@ ◆ Z V I S e?rˆ r ˆ r r2· r(e?r)d⌧ + e?rˆ r r?e?r?= Z J = ? But ˆ r = ?e?rˆ r. @re?r r2· da. Z Z Z R 1 r2e?r4⇡r2dr + = r2· r2sin✓d✓d?ˆ r = 4⇡ e?rdr + e?R sin✓d✓d? 0 = 4⇡??e?r???R 0+ 4⇡e?R= 4⇡??e?R+ e?0?+ 4⇡e?R= 4⇡.X ?Here R = 1, so e?R= 0.? ?x2?= 0 ; ?x2?= ?2xˆ y ; ?x3+ y2+ z2? @x @x+@y Problem 1.50 (a) r·F1= @x(0) + @y(0) + ?????? U2=1 =?@Ax r·F2=@x @y+@z ?????? @z= 1 + 1 + 1 = 3 ?????? @ @ @ @z ?????? ˆ x @ @x 0 ˆ y @ @y 0 x2 ˆ z @ @z ˆ x @ @x x ˆ y @ @y y ˆ z @ @z z = ?ˆ y@ r⇥F1= r⇥F2= = 0 @x F2is a gradient; F1is a curl would do (F2= rU2). @x?@Ax 2 ⇣@Ay (F1= r⇥A1). (But these are not unique.) ⌘ ?= 0; Ay=x3 @Ay For A1, we want @y= x2. 3, Ax= Az= 0 would do it. @z?@Az @z?@Az @y A1=1 3x2ˆ y ?????? ?????? ˆ x @ @x yz xz xy ˆ y @ @y ˆ z @ @z (b) r·F3= @x(yz) + @y(xz) + @ @z(xy) = 0; r⇥F3= = ˆ x(x ? x) + ˆ y(y ? y) + ˆ z(z ? z) = 0. @ @ So F3can be written as the gradient of a scalar (F3= rU3) and as the curl of a vector (F3= r⇥A3). In fact, U3= xyz does the job. For the vector potential, we have 8 > 9 > @Az @y?@Ay @Ax @z?@Az @Ay @x?@Ax @z= yz, which suggests Az=1 @x= xz, suggesting @y= xy, so 4y2z + f(x,z); Ay= ?1 Ax=1 Ay=1 4yz2+ g(x,y) 4zx2+ j(y,z) 4xy2+ l(x,z) > : > ; < = 4z2x + h(x,y); Az= ?1 4x2y + k(y,z); Ax= ?1 ?x?z2? y2?ˆ x + y?x2? z2?ˆ y + z?y2? x2?ˆ z (Eq. 1.44 – curl of gradient is always zero). HF · dl =R(r⇥F) · da = 0 (Eq. 1.57–Stokes’ theorem). Zb (b) ) (c): same as (c) ) (b), only in reverse; (c) ) (a): same as (a)) (c). Problem 1.52 (d) ) (a): r·F = r·(r⇥W) = 0 (a) ) (c): Putting this all together: A3=1 (again, not unique). 4 Problem 1.51 (d) ) (a): r⇥F = r⇥(?rU) = 0 (a) ) (c): (c) ) (b): Rb a IF · dl ?Rb a IIF · dl =Rb a IF · dl +Ra b IIF · dl =HF · dl = 0, so F · dl = a II Zb F · dl. a I (Eq 1.46—divergence of curl is always zero). HF · da =R(r·F)d⌧ = 0 (Eq. 1.56—divergence theorem). 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. @solutionmanual1

19. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 19 CHAPTER 1. VECTOR ANALYSIS R IF · da ?R IIF · da =HF · da = 0, so (c) ) (b): Z Z F · da = F · da. I II (Note: sign change because forHF · da, da is outward, whereas for surface II it is inward.) Problem 1.53 In Prob. 1.15 we found that r·va= 0; in Prob. 1.18 we found that r⇥vc= 0. So vccan be written as the gradient of a scalar; vacan be written as the curl of a vector. (b) ) (c): same as (c) ) (b), in reverse; (c)) (a): same as (a)) (c) . (a) To find t: (1) @t @x= y2) t = y2x + f(y,z) @t @y=?2xy + z2? (2) (3) @t @z= 2yz @f @z= 2yz ) f = yz2+ g(y) ) t = y2x + yz2+ g(y), so 2xy + z2(from (2)) )@g (b) To find W: Pick Wx= 0; then @t @y= 2xy + z2+@g From (1) & (3) we get @y= @y= 0. We may as well pick g = 0; then t = xy2+ yz2. @Wz @y?@Wy @Wy @x?@Wx @z= x2; @x= 3z2x; @y= ?2xz. @Wx @z?@Wz = ?3xz2) Wz= ?3 @Wz @x @Wy @x 2x2z2+ f(y,z) = ?2xz ) Wy= ?x2z + g(y,z). @Wz @y?@Wy W = ?x2z ˆ y ?3 @z=@f @y+ x2?@g 2x2z2ˆ z. ?????? @z= x2)@f @y?@g @z= 0. May as well pick f = g = 0. ?????? ˆ x @ @x 0 ?x2z ?3 ˆ y @ @y ˆ z @ @z 2x2z2 = ˆ x?x2?+ ˆ y?3xz2?+ˆ z(?2xz).X Check: r⇥W = You can add any gradient (rt) to W without changing its curl, so this answer is far from unique. Some other solutions: W = xz3ˆ x ? x2z ˆ y; W =?2xyz + xz3?ˆ x + x2yˆ z; W = xyz ˆ x ?1 2x2?y ? 3z2?ˆ z. 2x2z ˆ y +1 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. @solutionmanual1

20. https://gioumeh.com/product/introduction-to-electrodynamics-solutions/https://gioumeh.com/product/introduction-to-electrodynamics-solutions/ cl i ck h ere t o d ow nl oad 20 CHAPTER 1. VECTOR ANALYSIS Problem 1.54 1 r2 1 r24r3cos✓ + =rcos✓ sin✓ 1 1 ?r2r2cos✓?+ ?sin✓r2cos??+ 1 rsin✓ ??r2cos✓sin?? @ @r @ @✓ @ r·v = rsin✓ rsin✓ @? 1 ??r2cos✓cos?? = rsin✓cos✓r2cos? + [4sin✓ + cos? ? cos?] = 4rcos✓. ⇡/2 Z 0 ⇡/2 Z 0 Z Z ?R4?✓1 Z 0 R (r·v)d⌧ = (4rcos✓)r2sin✓drd✓d? = 4 cos✓sin✓d✓ r3dr d? ◆⇣⇡ ⌘ ⇡R4 4. = = 2 2 Surface consists of four parts: (1) Curved: da = R2sin✓d✓d?ˆ r; r = R. v · da =?R2cos✓??R2sin✓d✓d??. ⇡/2 Z 0 ✓1 ◆⇣⇡ ⇡/2 Z 0 Z ⌘ =⇡R4 v · da = R4 cos✓sin✓d✓ d? = R4 . 2 2 4 v · da =?r2cos✓sin??(rdrd✓) = 0. Rv · da = 0. (2) Left: da = ?rdrd✓ˆ?; ? = 0. (3) Back: da = rdrd✓ˆ?; ? = ⇡/2. v · da =??r2cos✓sin??(rdrd✓) = ?r3cos✓drd✓. R Z 0 0 ✓1 ◆ ⇡/2 Z Z (+1) = ?1 v · da = cos✓d✓ = ? r3dr 4R4 4R4. v · da =?r2cos??(rdrd?). R Z 0 0 (4) Bottom: da = rsin✓drd?ˆ✓; ✓ = ⇡/2. ⇡/2 Z Z cos?d? =1 v · da = r3dr 4R4. Hv · da = ⇡R4/4 + 0 ?1 Problem 1.55 ?????? so dy = ?(x/y)dx. So 4R4=⇡R4 X Total: 4R4+1 4. ?????? ˆ x @ @x ay bx 0 ˆ y @ @y ˆ z @ @z R(r⇥v) · da = (b ? a)⇡R2. r⇥v = = ˆ z(b ? a). So v · dl = (ay ˆ x + bxˆ y) · (dxˆ x + dy ˆ y + dzˆ z) = ay dx + bxdy; x2+ y2= R2) 2xdx + 2y dy = 0, v · dl = ay dx + bx(?x/y)dx =1 ?ay2? bx2?dx. y 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. @solutionmanual1