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Solution Manual For Introduction To Electrodynamics, 4th Edition

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    Solution Manual Introduction to Electrodynamics Fourth Edition David J. Griffiths 2012

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    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 261 c

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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 θ 1 ︸ ︷︷ ︸ | C | cos θ 2 } | B | sin θ 1 } | C | sin θ 2 θ 1 θ 2 θ 3 (a) From the diagram, | B + C | cos θ 3 = | B | cos θ 1 + | C | cos θ 2 . | 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 ; A · B = +1 + 1 − 1 = 1 = AB cos θ = √ 3 √ 3 cos θ ⇒ cos θ . 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 ): c © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. (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 CHAPTER 1. VECTOR ANALYSIS 3 Chapter 1 Vector Analysis Problem 1.1 ! " # $ A B C B + C ︸ ︷︷ ︸ | B | cos θ 1 ︸ ︷︷ ︸ | C | cos θ 2 } | B | sin θ 1 } | C | sin θ 2 θ 1 θ 2 θ 3 (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 y + 0 z ; = − 1 x + 0 y + 3 z . c © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. 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 θ 1 ︸ ︷︷ ︸ | C | cos θ 2 } | B | sin } | C | sin θ 1 θ 2 θ 3 (a) From the diagram, | B + C | cos θ 3 = | B | cos θ 1 + | C | cos θ 2 . | 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) 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 ; A · B = +1 + 1 − 1 = 1 = AB cos θ = √ 3 √ 3 cos θ ⇒ cos θ . 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 exam 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. 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 ˆ ˆ ˆ ˆ ˆ

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    CHAPTER 1. VECTOR ANALYSIS 5 A × B = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z − 1 2 0 − 1 0 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 6 ˆ x + 3 ˆ y + 2 ˆ z . This has the right direction , but the wrong magnitude . To make a unit vector out of it, simply divide by its length: | A × B | = √ 36 + 9 + 4 = 7. ˆ n = A × B | A × B | = 6 7 ˆ x + 3 7 ˆ y + 2 7 ˆ z . Problem 1.5 A × ( B × C ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z A x A y A z ( B y C z − B z C y ) ( B z C x − B x C z ) ( B x C y − B y C x ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x [ A y ( B x C y − B y C x ) − A z ( B z C x − B x C z )] + ˆ y () + ˆ z () (I’ll just check the x-component; the others go the same way) = ˆ x ( A y B x C y − A y B y C x − A z B z C x + A z B x C z ) + ˆ y () + ˆ z (). B ( A · C ) − C ( A · B ) = [ B x ( A x C x + A y C y + A z C z ) − C x ( A x B x + A y B y + A z B z )] ˆ x + () ˆ y + () ˆ z = ˆ x ( A y B x C y + A z B x C z − A y B y C x − A z B z C x ) + ˆ y () + ˆ z (). They agree. Problem 1.6 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 = (4 ˆ x + 6 ˆ y + 8 ˆ z ) − (2 ˆ x + 8 ˆ y + 7 ˆ z ) = 2 ˆ x − 2 ˆ y + ˆ z r = √ 4 + 4 + 1 = 3 ˆ r = r r = 2 3 ˆ x − 2 3 ˆ y + 1 3 ˆ z Problem 1.8 (a) ¯ A y ¯ B y + ¯ A z ¯ B z = (cos φA y + sin φA z )(cos φB y + sin φB z ) + ( − sin φA y + cos φA z )( − sin φB y + cos φB z ) = cos 2 φA y B y + sin φ cos φ ( A y B z + A z B y ) + sin 2 φA z B z + sin 2 φA y B y − sin φ cos φ ( A y B z + A z B y ) + cos 2 φA z B z = (cos 2 φ + sin 2 φ ) A y B y + (sin 2 φ + cos 2 φ ) A z B z = A y B y + A z B z . X (b) ( A x ) 2 + ( A y ) 2 + ( A z ) 2 = Σ 3 i =1 A i A i = Σ 3 i =1 ( Σ 3 j =1 R ij A j ) ( Σ 3 k =1 R ik A k ) = Σ j,k (Σ i R ij R ik ) A j A k . This equals A 2 x + A 2 y + A 2 z provided Σ 3 i =1 R ij R ik = { 1 if j = k 0 if j 6 = k } Moreover, if R is to preserve lengths for all vectors A , then this condition is not only sufficient but also necessary . For suppose A = (1 , 0 , 0). Then Σ j,k (Σ i R ij R ik ) A j A k = Σ i R i 1 R i 1 , and this must equal 1 (since we want A 2 x + A 2 y + A 2 z = 1). Likewise, Σ 3 i =1 R i 2 R i 2 = Σ 3 i =1 R i 3 R i 3 = 1. To check the case j 6 = k , choose A = (1 , 1 , 0). Then we want 2 = Σ j,k (Σ i R ij R ik ) A j A k = Σ i R i 1 R i 1 + Σ i R i 2 R i 2 + Σ i R i 1 R i 2 + Σ i R i 2 R i 1 . But we already know that the first two sums are both 1; the third and fourth are equal , so Σ i R i 1 R i 2 = Σ i R i 2 R i 1 = 0, and so on for other unequal combinations of j , k . X In matrix notation: ˜ RR = 1, where ˜ R is the transpose of R . c

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    6 CHAPTER 1. VECTOR ANALYSIS Problem 1.9 CHAPTER 1. VECTOR ANALYSIS 5 ! x " y # z $ % Looking down the axis: " y & x ' z " z ′ & y ′ ' x ′ ( ) * c © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights res protected un Looking down the axis: CHAPTER 1. VECTOR ANALYSIS 5 ! x " y # z $ % Looking down the axis: " y & x ' z " z ′ & y ′ ' x ′ ( ) * c © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. A 120 ◦ rotation carries the z axis into the y (= z ) axis, y into x (= y ), and x into z (= x ). So A x = A z , A y = A x , A z = A y . R =   0 0 1 1 0 0 0 1 0   Problem 1.10 (a) No change. ( A x = A x , A y = A y , A z = A z ) (b) A −→ − A , in the sense ( A x = − A x , A y = − A y , A z = − A z ) (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 ) ∇ f = 2 x ˆ x + 3 y 2 ˆ y + 4 z 3 ˆ z ( b ) ∇ f = 2 xy 3 z 4 ˆ x + 3 x 2 y 2 z 4 ˆ y + 4 x 2 y 3 z 3 ˆ z ( c ) ∇ f = e x sin y ln z ˆ x + e x cos y ln z ˆ y + e x sin y (1 /z ) ˆ z Problem 1.12 (a) ∇ h = 10[(2 y − 6 x − 18) ˆ x + (2 x − 8 y + 28) ˆ y ]. ∇ h = 0 at summit, so 2 y − 6 x − 18 = 0 2 x − 8 y + 28 = 0 = ⇒ 6 x − 24 y + 84 = 0 } 2 y − 18 − 24 y + 84 = 0. 22 y = 66 = ⇒ y = 3 = ⇒ 2 x − 24 + 28 = 0 = ⇒ x = − 2. Top is 3 miles north, 2 miles west, of South Hadley. (b) Putting in x = − 2, y = 3: h = 10( − 12 − 12 − 36 + 36 + 84 + 12) = 720 ft. (c) Putting in x = 1, y = 1: ∇ h = 10[(2 − 6 − 18) ˆ x + (2 − 8 + 28) ˆ y ] = 10( − 22 ˆ x + 22 ˆ y ) = 220( − ˆ x + ˆ y ). | ∇ h | = 220 √ 2 ≈ 311 ft/mile; direction: northwest. c

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    CHAPTER 1. VECTOR ANALYSIS 7 Problem 1.13 r = ( x − x ′ ) ˆ x + ( y − y ′ ) ˆ y + ( z − z ′ ) ˆ z ; r = √ ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 . (a) ∇ ( r 2 ) = ∂ ∂x [( x − x ′ ) 2 +( y − y ′ ) 2 +( z − z ′ ) 2 ] ˆ x + ∂ ∂y () ˆ y + ∂ ∂z () ˆ z = 2( x − x ′ ) ˆ x +2( y − y ′ ) ˆ y +2( z − z ′ ) ˆ z = 2 r . (b) ∇ ( 1 r ) = ∂ ∂x [( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ ) 2 ] − 1 2 ˆ x + ∂ ∂y () − 1 2 ˆ y + ∂ ∂z () − 1 2 ˆ z = − 1 2 () − 3 2 2( x − x ′ ) ˆ x − 1 2 () − 3 2 2( y − y ′ ) ˆ y − 1 2 () − 3 2 2( z − z ′ ) ˆ z = − () − 3 2 [( x − x ′ ) ˆ x + ( y − y ′ ) ˆ y + ( z − z ′ ) ˆ z ] = − (1 / r 3 ) r = − (1 / r 2 ) ˆ r . (c) ∂ ∂x ( r n ) = n r n − 1 ∂ r ∂x = n r n − 1 ( 1 2 1 r 2 r x ) = n r n − 1 ˆ r x , so ∇ ( r n ) = n r n − 1 ˆ r Problem 1.14 y = + y cos φ + z sin φ ; multiply by sin φ : y sin φ = + y sin φ cos φ + z sin 2 φ. z = − y sin φ + z cos φ ; multiply by cos φ : z cos φ = − y sin φ cos φ + z cos 2 φ. Add: y sin φ + z cos φ = z (sin 2 φ + cos 2 φ ) = z . Likewise, y cos φ − z sin φ = y . So ∂y ∂y = cos φ ; ∂y ∂z = − sin φ ; ∂z ∂y = sin φ ; ∂z ∂z = cos φ . Therefore ( ∇ f ) y = ∂f ∂y = ∂f ∂y ∂y ∂y + ∂f ∂z ∂z ∂y = + cos φ ( ∇ f ) y + sin φ ( ∇ f ) z ( ∇ f ) z = ∂f ∂z = ∂f ∂y ∂y ∂z + ∂f ∂z ∂z ∂z = − sin φ ( ∇ f ) y + cos φ ( ∇ f ) z } So ∇ f transforms as a vector. qed Problem 1.15 ( a ) ∇· v a = ∂ ∂x ( x 2 ) + ∂ ∂y (3 xz 2 ) + ∂ ∂z ( − 2 xz ) = 2 x + 0 − 2 x = 0 . ( b ) ∇· v b = ∂ ∂x ( xy ) + ∂ ∂y (2 yz ) + ∂ ∂z (3 xz ) = y + 2 z + 3 x. ( c ) ∇· v c = ∂ ∂x ( y 2 ) + ∂ ∂y (2 xy + z 2 ) + ∂ ∂z (2 yz ) = 0 + (2 x ) + (2 y ) = 2( x + y ) Problem 1.16 ∇· v = ∂ ∂x ( x r 3 ) + ∂ ∂y ( y r 3 ) + ∂ ∂z ( z r 3 ) = ∂ ∂x [ x ( x 2 + y 2 + z 2 ) − 3 2 ] + ∂ ∂y [ y ( x 2 + y 2 + z 2 ) − 3 2 ] + ∂ ∂z [ z ( x 2 + y 2 + z 2 ) − 3 2 ] = () − 3 2 + x ( − 3 / 2)() − 5 2 2 x + () − 3 2 + y ( − 3 / 2)() − 5 2 2 y + () − 3 2 + z ( − 3 / 2)() − 5 2 2 z = 3 r − 3 − 3 r − 5 ( x 2 + y 2 + z 2 ) = 3 r − 3 − 3 r − 3 = 0. This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin. How, then, can ∇· v = 0? The answer is that ∇· 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, ∇· v is infinite at that one point, and zero elsewhere, as we shall see in Sect. 1.5. Problem 1.17 v y = cos φ v y + sin φ v z ; v z = − sin φ v y + cos φ v z . ∂v y ∂y = ∂v y ∂y cos φ + ∂v z ∂y sin φ = ( ∂v y ∂y ∂y ∂y + ∂v y ∂z ∂z ∂y ) cos φ + ( ∂v z ∂y ∂y ∂y + ∂v z ∂z ∂z ∂y ) sin φ . Use result in Prob. 1.14: = ( ∂v y ∂y cos φ + ∂v y ∂z sin φ ) cos φ + ( ∂v z ∂y cos φ + ∂v z ∂z sin φ ) sin φ . ∂v z ∂z = − ∂v y ∂z sin φ + ∂v z ∂z cos φ = − ( ∂v y ∂y ∂y ∂z + ∂v y ∂z ∂z ∂z ) sin φ + ( ∂v z ∂y ∂y ∂z + ∂v z ∂z ∂z ∂z ) cos φ = − ( − ∂v y ∂y sin φ + ∂v y ∂z cos φ ) sin φ + ( − ∂v z ∂y sin φ + ∂v z ∂z cos φ ) cos φ . So c

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    8 CHAPTER 1. VECTOR ANALYSIS ∂v y ∂y + ∂v z ∂z = ∂v y ∂y cos 2 φ + ∂v y ∂z sin φ cos φ + ∂v z ∂y sin φ cos φ + ∂v z ∂z sin 2 φ + ∂v y ∂y sin 2 φ − ∂v y ∂z sin φ cos φ − ∂v z ∂y sin φ cos φ + ∂v z ∂z cos 2 φ = ∂v y ∂y ( cos 2 φ + sin 2 φ ) + ∂v z ∂z ( sin 2 φ + cos 2 φ ) = ∂v y ∂y + ∂v z ∂z . X Problem 1.18 (a) ∇× v a = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z x 2 3 xz 2 − 2 xz ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x (0 − 6 xz ) + ˆ y (0 + 2 z ) + ˆ z (3 z 2 − 0) = − 6 xz ˆ x + 2 z ˆ y + 3 z 2 ˆ z . (b) ∇× v b = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z xy 2 yz 3 xz ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x (0 − 2 y ) + ˆ y (0 − 3 z ) + ˆ z (0 − x ) = − 2 y ˆ x − 3 z ˆ y − x ˆ z . (c) ∇× v c = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z y 2 (2 xy + z 2 ) 2 yz ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x (2 z − 2 z ) + ˆ y (0 − 0) + ˆ z (2 y − 2 y ) = 0 . Problem 1.19 A x y z v v v v B As we go from point A to point B (9 o’clock to 10 o’clock), x increases, y increases, v x increases, and v y decreases, so ∂v x /∂y > 0, while ∂v y /∂y < 0. On the circle, v z = 0, and there is no dependence on z , so Eq. 1.41 says ∇ × v = ˆ z ( ∂v y ∂x − ∂v x ∂y ) 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.] Problem 1.20 v = y ˆ x + x ˆ y ; or v = yz ˆ x + xz ˆ y + xy ˆ z ; or v = (3 x 2 z − z 3 ) ˆ x + 3 ˆ y + ( x 3 − 3 xz 2 ) ˆ z ; or v = (sin x )(cosh y ) ˆ x − (cos x )(sinh y ) ˆ y ; etc. Problem 1.21 (i) ∇ ( f g ) = ∂ ( f g ) ∂x ˆ x + ∂ ( f g ) ∂y ˆ y + ∂ ( f g ) ∂z ˆ z = ( f ∂g ∂x + g ∂f ∂x ) ˆ x + ( f ∂g ∂y + g ∂f ∂y ) ˆ y + ( f ∂g ∂z + g ∂f ∂z ) ˆ z = f ( ∂g ∂x ˆ x + ∂g ∂y ˆ y + ∂g ∂z ˆ z ) + g ( ∂f ∂x ˆ x + ∂f ∂y ˆ y + ∂f ∂z ˆ z ) = f ( ∇ g ) + g ( ∇ f ). qed (iv) ∇· ( A × B ) = ∂ ∂x ( A y B z − A z B y ) + ∂ ∂y ( A z B x − A x B z ) + ∂ ∂z ( A x B y − A y B x ) = A y ∂B z ∂x + B z ∂A y ∂x − A z ∂B y ∂x − B y ∂A z ∂x + A z ∂B x ∂y + B x ∂A z ∂y − A x ∂B z ∂y − B z ∂A x ∂y + A x ∂B y ∂z + B y ∂A x ∂z − A y ∂B x ∂z − B x ∂A y ∂z = B x ( ∂A z ∂y − ∂A y ∂z ) + B y ( ∂A x ∂z − ∂A z ∂x ) + B z ( ∂A y ∂x − ∂A x ∂y ) − A x ( ∂B z ∂y − ∂B y ∂z ) − A y ( ∂B x ∂z − ∂B z ∂x ) − A z ( ∂B y ∂x − ∂B x ∂y ) = B · ( ∇× A ) − A · ( ∇× B ). qed (v) ∇× ( f A ) = ( ∂ ( f A z ) ∂y − ∂ ( f A y ) ∂z ) ˆ x + ( ∂ ( f A x ) ∂z − ∂ ( f A z ) ∂x ) ˆ y + ( ∂ ( f A y ) ∂x − ∂ ( f A x ) ∂y ) ˆ z c

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    CHAPTER 1. VECTOR ANALYSIS 9 = ( f ∂A z ∂y + A z ∂f ∂y − f ∂A y ∂z − A y ∂f ∂z ) ˆ x + ( f ∂A x ∂z + A x ∂f ∂z − f ∂A z ∂x − A z ∂f ∂x ) ˆ y + ( f ∂A y ∂x + A y ∂f ∂x − f ∂A x ∂y − A x ∂f ∂y ) ˆ z = f [( ∂A z ∂y − ∂A y ∂z ) ˆ x + ( ∂A x ∂z − ∂A z ∂x ) ˆ y + ( ∂A y ∂x − ∂A x ∂y ) ˆ z ] − [( A y ∂f ∂z − A z ∂f ∂y ) ˆ x + ( A z ∂f ∂x − A x ∂f ∂z ) ˆ y + ( A x ∂f ∂y − A y ∂f ∂x ) ˆ z ] = f ( ∇× A ) − A × ( ∇ f ). qed Problem 1.22 (a) ( A · ∇ ) B = ( A x ∂B x ∂x + A y ∂B x ∂y + A z ∂B x ∂z ) ˆ x + ( A x ∂B y ∂x + A y ∂B y ∂y + A z ∂B y ∂z ) ˆ y + ( A x ∂B z ∂x + A y ∂B z ∂y + A z ∂B z ∂z ) ˆ z . (b) ˆ r = r r = x ˆ x + y ˆ y + z ˆ z √ x 2 + y 2 + z 2 . Let’s just do the x component. [( ˆ r · ∇ ) ˆ r ] x = 1 √ ( x ∂ ∂x + y ∂ ∂y + z ∂ ∂z ) x √ x 2 + y 2 + z 2 = 1 r { x [ 1 √ + x ( − 1 2 ) 1 ( √ ) 3 2 x ] + yx [ − 1 2 1 ( √ ) 3 2 y ] + zx [ − 1 2 1 ( √ ) 3 2 z ]} = 1 r { x r − 1 r 3 ( x 3 + xy 2 + xz 2 )} = 1 r { x r − x r 3 ( x 2 + y 2 + z 2 )} = 1 r ( x r − x r ) = 0. Same goes for the other components. Hence: ( ˆ r · ∇ ) ˆ r = 0 . (c) ( v a · ∇ ) v b = ( x 2 ∂ ∂x + 3 xz 2 ∂ ∂y − 2 xz ∂ ∂z ) ( xy ˆ x + 2 yz ˆ y + 3 xz ˆ z ) = x 2 ( y ˆ x + 0 ˆ y + 3 z ˆ z ) + 3 xz 2 ( x ˆ x + 2 z ˆ y + 0 ˆ z ) − 2 xz (0 ˆ x + 2 y ˆ y + 3 x ˆ z ) = ( x 2 y + 3 x 2 z 2 ) ˆ x + ( 6 xz 3 − 4 xyz ) ˆ y + ( 3 x 2 z − 6 x 2 z ) ˆ z = x 2 ( y + 3 z 2 ) ˆ x + 2 xz ( 3 z 2 − 2 y ) ˆ y − 3 x 2 z ˆ z Problem 1.23 (ii) [ ∇ ( A · B )] x = ∂ ∂x ( A x B x + A y B y + A z B z ) = ∂A x ∂x B x + A x ∂B x ∂x + ∂A y ∂x B y + A y ∂B y ∂x + ∂A z ∂x B z + A z ∂B z ∂x [ A × ( ∇× B )] x = A y ( ∇× B ) z − A z ( ∇× B ) y = A y ( ∂B y ∂x − ∂B x ∂y ) − A z ( ∂B x ∂z − ∂B z ∂x ) [ B × ( ∇× A )] x = B y ( ∂A y ∂x − ∂A x ∂y ) − B z ( ∂A x ∂z − ∂A z ∂x ) [( A ·∇ ) B ] x = ( A x ∂ ∂x + A y ∂ ∂y + A z ∂ ∂z ) B x = A x ∂B x ∂x + A y ∂B x ∂y + A z ∂B x ∂z [( B ·∇ ) A ] x = B x ∂A x ∂x + B y ∂A x ∂y + B z ∂A x ∂z So [ A × ( ∇× B ) + B × ( ∇× A ) + ( A ·∇ ) B + ( B ·∇ ) A ] x = A y ∂B y ∂x − A y ∂B x ∂y − A z ∂B x ∂z + A z ∂B z ∂x + B y ∂A y ∂x − B y ∂A x ∂y − B z ∂A x ∂z + B z ∂A z ∂x + A x ∂B x ∂x + A y ∂B x ∂y + A z ∂B x ∂z + B x ∂A x ∂x + B y ∂A x ∂y + B z ∂A x ∂z = B x ∂A x ∂x + A x ∂B x ∂x + B y ( ∂A y ∂x − ∂A x ∂y / + ∂A x ∂y / ) + A y ( ∂B y ∂x − ∂B x ∂y / + ∂B x ∂y / ) + B z ( − ∂A x ∂z / + ∂A z ∂x + ∂A x ∂z / ) + A z ( − ∂B x ∂z / + ∂B z ∂x + ∂B x ∂z / ) = [ ∇ ( A · B )] x (same for y and z ) (vi) [ ∇× ( A × B )] x = ∂ ∂y ( A × B ) z − ∂ ∂z ( A × B ) y = ∂ ∂y ( A x B y − A y B x ) − ∂ ∂z ( A z B x − A x B z ) = ∂A x ∂y B y + A x ∂B y ∂y − ∂A y ∂y B x − A y ∂B x ∂y − ∂A z ∂z B x − A z ∂B x ∂z + ∂A x ∂z B z + A x ∂B z ∂z [( B ·∇ ) A − ( A ·∇ ) B + A ( ∇· B ) − B ( ∇· A )] x = B x ∂A x ∂x + B y ∂A x ∂y + B z ∂A x ∂z − A x ∂B x ∂x − A y ∂B x ∂y − A z ∂B x ∂z + A x ( ∂B x ∂x + ∂B y ∂y + ∂B z ∂z ) − B x ( ∂A x ∂x + ∂A y ∂y + ∂A z ∂z ) c

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    10 CHAPTER 1. VECTOR ANALYSIS = B y ∂A x ∂y + A x ( − ∂B x ∂x / + ∂B x ∂x / + ∂B y ∂y + ∂B z ∂z ) + B x ( ∂A x ∂x / − ∂A x ∂x / − ∂A y ∂y − ∂A z ∂z ) + A y ( − ∂B x ∂y ) + A z ( − ∂B x ∂z ) + B z ( ∂A x ∂z ) = [ ∇× ( A × B )] x (same for y and z ) Problem 1.24 ∇ ( f /g ) = ∂ ∂x ( f /g ) ˆ x + ∂ ∂y ( f /g ) ˆ y + ∂ ∂z ( f /g ) ˆ z = g ∂f ∂x − f ∂g ∂x g 2 ˆ x + g ∂f ∂y − f ∂g ∂y g 2 ˆ y + g ∂f ∂z − f ∂g ∂z g 2 ˆ z = 1 g 2 [ g ( ∂f ∂x ˆ x + ∂f ∂y ˆ y + ∂f ∂z ˆ z ) − f ( ∂g ∂x ˆ x + ∂g ∂y ˆ y + ∂g ∂z ˆ z )] = g ∇ f − f ∇ g g 2 . qed ∇· ( A /g ) = ∂ ∂x ( A x /g ) + ∂ ∂y ( A y /g ) + ∂ ∂z ( A z /g ) = g ∂Ax ∂x − A x ∂g ∂x g 2 + g ∂Ay ∂y − A y ∂g ∂y g 2 + g ∂Az ∂z − A z ∂g ∂x g 2 = 1 g 2 [ g ( ∂A x ∂x + ∂A y ∂y + ∂A z ∂z ) − ( A x ∂g ∂x + A y ∂g ∂y + A z ∂g ∂z )] = g ∇ · A − A · ∇ g g 2 . qed [ ∇× ( A /g )] x = ∂ ∂y ( A z /g ) − ∂ ∂z ( A y /g ) = g ∂Az ∂y − A z ∂g ∂y g 2 − g ∂Ay ∂z − A y ∂g ∂z g 2 = 1 g 2 [ g ( ∂A z ∂y − ∂A y ∂z ) − ( A z ∂g ∂y − A y ∂g ∂z )] = g ( ∇ × A ) x +( A × ∇ g ) x g 2 (same for y and z ) . qed Problem 1.25 (a) A × B = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z x 2 y 3 z 3 y − 2 x 0 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x (6 xz ) + ˆ y (9 zy ) + ˆ z ( − 2 x 2 − 6 y 2 ) ∇· ( A × B ) = ∂ ∂x (6 xz ) + ∂ ∂y (9 zy ) + ∂ ∂z ( − 2 x 2 − 6 y 2 ) = 6 z + 9 z + 0 = 15 z ∇× A = ˆ x ( ∂ ∂y (3 z ) − ∂ ∂z (2 y ) ) + ˆ y ( ∂ ∂z ( x ) − ∂ ∂x (3 z ) ) + ˆ z ( ∂ ∂x (2 y ) − ∂ ∂y ( x ) ) = 0; B · ( ∇× A ) = 0 ∇× B = ˆ x ( ∂ ∂y (0) − ∂ ∂z ( − 2 x ) ) + ˆ y ( ∂ ∂z (3 y ) − ∂ ∂x (0) ) + ˆ z ( ∂ ∂x ( − 2 x ) − ∂ ∂y (3 y ) ) = − 5 ˆ z ; A · ( ∇× B ) = − 15 z ∇· ( A × B ) ? = B · ( ∇× A ) − A · ( ∇× B ) = 0 − ( − 15 z ) = 15 z. X (b) A · B = 3 xy − 4 xy = − xy ; ∇ ( A · B ) = ∇ ( − xy ) = ˆ x ∂ ∂x ( − xy ) + ˆ y ∂ ∂y ( − xy ) = − y ˆ x − x ˆ y A × ( ∇× B ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z x 2 y 3 z 0 0 − 5 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x ( − 10 y ) + ˆ y (5 x ); B × ( ∇× A ) = 0 ( A ·∇ ) B = ( x ∂ ∂x + 2 y ∂ ∂y + 3 z ∂ ∂z ) (3 y ˆ x − 2 x ˆ y ) = ˆ x (6 y ) + ˆ y ( − 2 x ) ( B ·∇ ) A = ( 3 y ∂ ∂x − 2 x ∂ ∂y ) ( x ˆ x + 2 y ˆ y + 3 z ˆ z ) = ˆ x (3 y ) + ˆ y ( − 4 x ) A × ( ∇× B ) + B × ( ∇× A ) + ( A ·∇ ) B + ( B ·∇ ) A = − 10 y ˆ x + 5 x ˆ y + 6 y ˆ x − 2 x ˆ y + 3 y ˆ x − 4 x ˆ y = − y ˆ x − x ˆ y = ∇· ( A · B ) . X (c) ∇× ( A × B ) = ˆ x ( ∂ ∂y ( − 2 x 2 − 6 y 2 ) − ∂ ∂z (9 zy ) ) + ˆ y ( ∂ ∂z (6 xz ) − ∂ ∂x ( − 2 x 2 − 6 y 2 ) ) + ˆ z ( ∂ ∂x (9 zy ) − ∂ ∂y (6 xz ) ) = ˆ x ( − 12 y − 9 y ) + ˆ y (6 x + 4 x ) + ˆ z (0) = − 21 y ˆ x + 10 x ˆ y ∇· A = ∂ ∂x ( x ) + ∂ ∂y (2 y ) + ∂ ∂z (3 z ) = 1 + 2 + 3 = 6; ∇· B = ∂ ∂x (3 y ) + ∂ ∂y ( − 2 x ) = 0 c

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    CHAPTER 1. VECTOR ANALYSIS 11 ( B ·∇ ) A − ( A ·∇ ) B + A ( ∇· B ) − B ( ∇· A ) = 3 y ˆ x − 4 x ˆ y − 6 y ˆ x + 2 x ˆ y − 18 y ˆ x + 12 x ˆ y = − 21 y ˆ x + 10 x ˆ y = ∇× ( A × B ) . X Problem 1.26 (a) ∂ 2 T a ∂x 2 = 2; ∂ 2 T a ∂y 2 = ∂ 2 T a ∂z 2 = 0 ⇒ ∇ 2 T a = 2 . (b) ∂ 2 T b ∂x 2 = ∂ 2 T b ∂y 2 = ∂ 2 T b ∂z 2 = − T b ⇒ ∇ 2 T b = − 3 T b = − 3 sin x sin y sin z. (c) ∂ 2 T c ∂x 2 = 25 T c ; ∂ 2 T c ∂y 2 = − 16 T c ; ∂ 2 T c ∂z 2 = − 9 T c ⇒ ∇ 2 T c = 0 . (d) ∂ 2 v x ∂x 2 = 2 ; ∂ 2 v x ∂y 2 = ∂ 2 v x ∂z 2 = 0 ⇒ ∇ 2 v x = 2 ∂ 2 v y ∂x 2 = ∂ 2 v y ∂y 2 = 0 ; ∂ 2 v y ∂z 2 = 6 x ⇒ ∇ 2 v y = 6 x ∂ 2 v z ∂x 2 = ∂ 2 v z ∂y 2 = ∂ 2 v z ∂z 2 = 0 ⇒ ∇ 2 v z = 0      ∇ 2 v = 2 ˆ x + 6 x ˆ y . Problem 1.27 ∇· ( ∇× v ) = ∂ ∂x ( ∂v z ∂y − ∂v y ∂z ) + ∂ ∂y ( ∂v x ∂z − ∂v z ∂x ) + ∂ ∂z ( ∂v y ∂x − ∂v x ∂y ) = ( ∂ 2 v z ∂x ∂y − ∂ 2 v z ∂y ∂x ) + ( ∂ 2 v x ∂y ∂z − ∂ 2 v x ∂z ∂y ) + ( ∂ 2 v y ∂z ∂x − ∂ 2 v y ∂x ∂z ) = 0, by equality of cross-derivatives. From Prob. 1.18: ∇× v a = − 6 xz ˆ x +2 z ˆ y +3 z 2 ˆ z ⇒ ∇· ( ∇× v a ) = ∂ ∂x ( − 6 xz )+ ∂ ∂y (2 z )+ ∂ ∂z (3 z 2 ) = − 6 z +6 z = 0 . Problem 1.28 ∇× ( ∇ t ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z ∂t ∂x ∂t ∂y ∂t ∂z ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x ( ∂ 2 t ∂y ∂z − ∂ 2 t ∂z ∂y ) + ˆ y ( ∂ 2 t ∂z ∂x − ∂ 2 t ∂x ∂z ) + ˆ z ( ∂ 2 t ∂x ∂y − ∂ 2 t ∂y ∂x ) = 0, by equality of cross-derivatives. In Prob. 1.11(b), ∇ f = 2 xy 3 z 4 ˆ x + 3 x 2 y 2 z 4 ˆ y + 4 x 2 y 3 z 3 ˆ z , so ∇× ( ∇ f ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 xy 3 z 4 3 x 2 y 2 z 4 4 x 2 y 3 z 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ˆ x (3 · 4 x 2 y 2 z 3 − 4 · 3 x 2 y 2 z 3 ) + ˆ y (4 · 2 xy 3 z 3 − 2 · 4 xy 3 z 3 ) + ˆ z (2 · 3 xy 2 z 4 − 3 · 2 xy 2 z 4 ) = 0 . X Problem 1.29 (a) (0 , 0 , 0) −→ (1 , 0 , 0) . x : 0 → 1 , y = z = 0; d l = dx ˆ x ; v · d l = x 2 dx ; ∫ v · d l = ∫ 1 0 x 2 dx = ( x 3 / 3) | 1 0 = 1 / 3 . (1 , 0 , 0) −→ (1 , 1 , 0) . x = 1 , y : 0 → 1 , z = 0; d l = dy ˆ y ; v · d l = 2 yz dy = 0; ∫ v · d l = 0 . (1 , 1 , 0) −→ (1 , 1 , 1) . x = y = 1 , z : 0 → 1; d l = dz ˆ z ; v · d l = y 2 dz = dz ; ∫ v · d l = ∫ 1 0 dz = z | 1 0 = 1 . Total: ∫ v · d l = (1 / 3) + 0 + 1 = 4/3. (b) (0 , 0 , 0) −→ (0 , 0 , 1) . x = y = 0 , z : 0 → 1; d l = dz ˆ z ; v · d l = y 2 dz = 0; ∫ v · d l = 0 . (0 , 0 , 1) −→ (0 , 1 , 1) . x = 0 , y : 0 → 1 , z = 1; d l = dy ˆ y ; v · d l = 2 yz dy = 2 y dy ; ∫ v · d l = ∫ 1 0 2 y dy = y 2 | 1 0 = 1 . (0 , 1 , 1) −→ (1 , 1 , 1) . x : 0 → 1 , y = z = 1; d l = dx ˆ x ; v · d l = x 2 dx ; ∫ v · d l = ∫ 1 0 x 2 dx = ( x 3 / 3) | 1 0 = 1 / 3 . Total: ∫ v · d l = 0 + 1 + (1 / 3) = 4/3. (c) x = y = z : 0 → 1; dx = dy = dz ; v · d l = x 2 dx + 2 yz dy + y 2 dz = x 2 dx + 2 x 2 dx + x 2 dx = 4 x 2 dx ; ∫ v · d l = ∫ 1 0 4 x 2 dx = (4 x 3 / 3) | 1 0 = 4/3. (d) ∮ v · d l = (4 / 3) − (4 / 3) = 0. c

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    12 CHAPTER 1. VECTOR ANALYSIS Problem 1.30 x, y : 0 → 1 , z = 0; d a = dx dy ˆ z ; v · d a = y ( z 2 − 3) dx dy = − 3 y dx dy ; ∫ v · d a = − 3 ∫ 2 0 dx ∫ 2 0 y dy = − 3( x | 2 0 )( y 2 2 | 2 0 ) = − 3(2)(2) = − 12. In Ex. 1.7 we got 20, for the same boundary line (the square in the 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 ∫ T dτ = ∫ z 2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last: ∫ z 2 [∫ (∫ dx ) dy ] dz. The sloping surface is x + y + z = 1, so the x integral is ∫ (1 − y − z ) 0 dx = 1 − y − z. For a given z , y ranges from 0 to 1 − z , so the y integral is ∫ (1 − z ) 0 (1 − y − z ) dy = [(1 − z ) y − ( y 2 / 2)] | (1 − z ) 0 = (1 − z ) 2 − [(1 − z ) 2 / 2] = (1 − z ) 2 / 2 = (1 / 2) − z + ( z 2 / 2) . Finally, the z integral is ∫ 1 0 z 2 ( 1 2 − z + z 2 2 ) dz = ∫ 1 0 ( z 2 2 − z 3 + z 4 2 ) dz = ( z 3 6 − z 4 4 + z 5 10 ) | 1 0 = 1 6 − 1 4 + 1 10 = 1/60. ˆ Problem 1.32 T ( b ) = 1 + 4 + 2 = 7; T ( a ) = 0 . ⇒ T ( b ) − T ( a ) = 7 . ∇ T = (2 x + 4 y ) ˆ x + (4 x + 2 z 3 ) y + (6 yz 2 ) ˆ z ; ∇ T · d l = (2 x + 4 y ) dx + (4 x + 2 z 3 ) dy + (6 yz 2 ) dz (a) Segment 1: x : 0 → 1 , y = z = dy = dz = 0 . ∫ ∇ T · d l = ∫ 1 0 (2 x ) dx = x 2 ∣ ∣ ∣ 1 0 = 1 . Segment 2: y : 0 → 1 , x = 1 , z = 0 , dx = dz = 0 . ∫ ∇ T · d l = ∫ 1 0 (4) dy = 4 y | 1 0 = 4 . Segment 3: z : 0 → 1 , x = y = 1 , dx = dy = 0 . ∫ ∇ T · d l = ∫ 1 0 (6 z 2 ) dz = 2 z 3 ∣ ∣ ∣ 1 0 = 2 .      ∫ b a ∇ T · d l = 7 . X (b) Segment 1: z : 0 → 1 , x = y = dx = dy = 0 . ∫ ∇ T · d l = ∫ 1 0 (0) dz = 0 . Segment 2: y : 0 → 1 , x = 0 , z = 1 , dx = dz = 0 . ∫ ∇ T · d l = ∫ 1 0 (2) dy = 2 y | 1 0 = 2 . Segment 3: x : 0 → 1 , y = z = 1 , dy = dz = 0 . ∫ ∇ T · d l = ∫ 1 0 (2 x + 4) dx = ( x 2 + 4 x ) ∣ ∣ ∣ 1 0 = 1 + 4 = 5 .          ∫ b a ∇ T · d l = 7 . X (c) x : 0 → 1 , y = x, z = x 2 , dy = dx, dz = 2 x dx. ∇ T · d l = (2 x + 4 x ) dx + (4 x + 2 x 6 ) dx + (6 xx 4 )2 x dx = (10 x + 14 x 6 ) dx. ∫ b a ∇ T · d l = ∫ 1 0 (10 x + 14 x 6 ) dx = (5 x 2 + 2 x 7 ) ∣ ∣ ∣ 1 0 = 5 + 2 = 7 . X Problem 1.33 ∇· v = y + 2 z + 3 x ∫ ( ∇· v ) dτ = ∫ ( y + 2 z + 3 x ) dx dy dz = ∫∫ { ∫ 2 0 ( y + 2 z + 3 x ) dx } dy dz ↪ → [ ( y + 2 z ) x + 3 2 x 2 ] 2 0 = 2( y + 2 z ) + 6 = ∫ { ∫ 2 0 (2 y + 4 z + 6) dy } dz ↪ → [ y 2 + (4 z + 6) y ] 2 0 = 4 + 2(4 z + 6) = 8 z + 16 = ∫ 2 0 (8 z + 16) dz = (4 z 2 + 16 z ) ∣ ∣ ∣ 2 0 = 16 + 32 = 48. Numbering the surfaces as in Fig. 1.29: c

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    CHAPTER 1. VECTOR ANALYSIS 13 (i) d a = dy dz ˆ x , x = 2 . v · d a = 2 y dy dz. ∫ v · d a = ∫∫ 2 y dy dz = 2 y 2 ∣ ∣ ∣ 2 0 = 8 . (ii) d a = − dy dz ˆ x , x = 0 . v · d a = 0 . ∫ v · d a = 0 . (iii) d a = dx dz ˆ y , y = 2 . v · d a = 4 z dx dz. ∫ v · d a = ∫∫ 4 z dx dz = 16 . (iv) d a = − dx dz ˆ y , y = 0 . v · d a = 0 . ∫ v · d a = 0 . (v) d a = dx dy ˆ z , z = 2 . v · d a = 6 x dx dy. ∫ v · d a = 24 . (vi) d a = − dx dy ˆ z , z = 0 . v · d a = 0 . ∫ v · d a = 0 . ⇒ ∫ v · d a = 8 + 16 + 24 = 48 X Problem 1.34 ∇× v = ˆ x (0 − 2 y ) + ˆ y (0 − 3 z ) + ˆ z (0 − x ) = − 2 y ˆ x − 3 z ˆ y − x ˆ z . d a = dy dz ˆ x , if we agree that the path integral shall run counterclockwise. So ( ∇× v ) · d a = − 2 y dy dz. ∫ ( ∇× v ) · d a = ∫ { ∫ 2 − z 0 ( − 2 y ) dy } dz ↪ → y 2 ∣ ∣ ∣ 2 − z 0 = − (2 − z ) 2 = − ∫ 2 0 (4 − 4 z + z 2 ) dz = − ( 4 z − 2 z 2 + z 3 3 )∣ ∣ ∣ ∣ 2 0 = − ( 8 − 8 + 8 3 ) = − 8 3 - 6 z y @ @ @ @ @ @ y = 2 − z Meanwhile, v · d l = ( xy ) dx + (2 yz ) dy + (3 zx ) dz. There are three segments. - 6 z y @ @ @ @ @ @ - (1) @ @ I (2) ? (3) (1) x = z = 0; dx = dz = 0 . y : 0 → 2 . ∫ v · d l = 0 . (2) x = 0; z = 2 − y ; dx = 0 , dz = − dy, y : 2 → 0 . v · d l = 2 yz dy. ∫ v · d l = ∫ 0 2 2 y (2 − y ) dy = − ∫ 2 0 (4 y − 2 y 2 ) dy = − ( 2 y 2 − 2 3 y 3 )∣ ∣ ∣ 2 0 = − ( 8 − 2 3 · 8 ) = − 8 3 . (3) x = y = 0; dx = dy = 0; z : 2 → 0 . v · d l = 0 . ∫ v · d l = 0 . So ∮ v · d l = − 8 3 . X Problem 1.35 By Corollary 1, ∫ ( ∇× v ) · d a should equal 4 3 . ∇× v = (4 z 2 − 2 x ) ˆ x + 2 z ˆ z . (i) d a = dy dz ˆ x , x = 1; y, z : 0 → 1 . ( ∇× v ) · d a = (4 z 2 − 2) dy dz ; ∫ ( ∇× v ) · d a = ∫ 1 0 (4 z 2 − 2) dz = ( 4 3 z 3 − 2 z ) ∣ ∣ ∣ 1 0 = 4 3 − 2 = − 2 3 . (ii) d a = − dx dy ˆ z , z = 0; x, y : 0 → 1 . ( ∇× v ) · d a = 0; ∫ ( ∇× v ) · d a = 0 . (iii) d a = dx dz ˆ y , y = 1; x, z : 0 → 1 . ( ∇× v ) · d a = 0; ∫ ( ∇× v ) · d a = 0 . (iv) d a = − dx dz ˆ y , y = 0; x, z : 0 → 1 . ( ∇× v ) · d a = 0; ∫ ( ∇× v ) · d a = 0 . (v) d a = dx dy ˆ z , z = 1; x, y : 0 → 1 . ( ∇× v ) · d a = 2 dx dy ; ∫ ( ∇× v ) · d a = 2 . ⇒ ∫ ( ∇× v ) · d a = − 2 3 + 2 = 4 3 . X c

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    14 CHAPTER 1. VECTOR ANALYSIS Problem 1.36 (a) Use the product rule ∇× ( f A ) = f ( ∇× A ) − A × ( ∇ f ) : ∫ S f ( ∇× A ) · d a = ∫ S ∇× ( f A ) · d a + ∫ S [ A × ( ∇ f )] · d a = ∮ P f A · d l + ∫ S [ A × ( ∇ f )] · d a . 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 ) · d a + ∫ V A · ( ∇× B ) dτ. qed (I used the divergence theorem in the last step.) Problem 1.37 r = √ x 2 + y 2 + z 2 ; θ = cos − 1 ( z √ x 2 + y 2 + z 2 ) ; φ = tan − 1 ( y x ) . 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 = x ˆ x + y ˆ y + z ˆ z = r sin θ cos φ ˆ x + r sin θ sin φ ˆ y + r cos θ ˆ z . If I only vary r slightly, then d r = ∂ ∂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: ˆ r = ∂ r ∂r ∣ ∣ ∣ ∂ r ∂r ∣ ∣ ∣ ; ˆ θ = ∂ r ∂θ ∣ ∣ ∣ ∂ r ∂θ ∣ ∣ ∣ ; ˆ φ = ∂ r ∂φ ∣ ∣ ∣ ∂ r ∂φ ∣ ∣ ∣ . ∂ r ∂r = sin θ cos φ ˆ x + sin θ sin φ ˆ y + cos θ ˆ z ; ∣ ∣ ∣ ∂ r ∂r ∣ ∣ ∣ 2 = sin 2 θ cos 2 φ + sin 2 θ sin 2 φ + cos 2 θ = 1 . ∂ r ∂θ = r cos θ cos φ ˆ x + r cos θ sin φ ˆ y − r sin θ ˆ z ; ∣ ∣ ∣ ∂ r ∂θ ∣ ∣ ∣ 2 = r 2 cos 2 θ cos 2 φ + r 2 cos 2 θ sin 2 φ + r 2 sin 2 θ = r 2 . ∂ r ∂φ = − r sin θ sin φ ˆ x + r sin θ cos φ ˆ y ; ∣ ∣ ∣ ∂ r ∂φ ∣ ∣ ∣ 2 = r 2 sin 2 θ sin 2 φ + r 2 sin 2 θ cos 2 φ = r 2 sin 2 θ. ⇒ ˆ r = sin θ cos φ ˆ x + sin θ sin φ ˆ y + cos θ ˆ z . ˆ θ = cos θ cos φ ˆ x + cos θ sin φ ˆ y − sin θ ˆ z . ˆ φ = − sin φ ˆ x + cos φ ˆ y . Check: ˆ r · ˆ r = sin 2 θ (cos 2 φ + sin 2 φ ) + cos 2 θ = sin 2 θ + cos 2 θ = 1 , X ˆ θ · ˆ φ = − cos θ sin φ cos φ + cos θ sin φ cos φ = 0 , X etc. sin θ ˆ r = sin 2 θ cos φ ˆ x + sin 2 θ sin φ ˆ y + sin θ cos θ ˆ z . cos θ ˆ θ = cos 2 θ cos φ ˆ x + cos 2 θ sin φ ˆ y − sin θ cos θ ˆ z . Add these: (1) sin θ ˆ r + cos θ ˆ θ = + cos φ ˆ x + sin φ ˆ y ; (2) ˆ φ = − sin φ ˆ x + cos φ ˆ y . Multiply (1) by cos φ , (2) by sin φ , and subtract: ˆ x = sin θ cos φ ˆ r + cos θ cos φ ˆ θ − sin φ ˆ φ . c

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    CHAPTER 1. VECTOR ANALYSIS 15 ˆ Multiply (1) by sin φ , (2) by cos φ , and add: y = sin θ sin φ ˆ r + cos θ sin φ ˆ θ + cos φ ˆ φ . cos θ ˆ r = sin θ cos θ cos φ ˆ x + sin θ cos θ sin φ ˆ y + cos 2 θ ˆ z . sin θ ˆ θ = sin θ cos θ cos φ ˆ x + sin θ cos θ sin φ ˆ y − sin 2 θ ˆ z . Subtract these: ˆ z = cos θ ˆ r − sin θ ˆ θ . Problem 1.39 (a) ∇· v 1 = 1 r 2 ∂ ∂r ( r 2 r 2 ) = 1 r 2 4 r 3 = 4 r ∫ ( ∇· v 1 ) dτ = ∫ (4 r )( r 2 sin θ dr dθ dφ ) = (4) ∫ R 0 r 3 dr ∫ π 0 sin θ dθ ∫ 2 π 0 dφ = (4) ( R 4 4 ) (2)(2 π ) = 4 πR 4 ∫ v 1 · d a = ∫ ( r 2 ˆ r ) · ( r 2 sin θ dθ dφ ˆ r ) = r 4 ∫ π 0 sin θ dθ ∫ 2 π 0 dφ = 4 πR 4 X ( Note: at surface of sphere r = R .) (b) ∇· v 2 = 1 r 2 ∂ ∂r ( r 2 1 r 2 ) = 0 ⇒ ∫ ( ∇· v 2 ) dτ = 0 ∫ v 2 · d a = ∫ ( 1 r 2 ˆ r ) ( r 2 sin θ dθ dφ ˆ r ) = ∫ sin θ dθ dφ = 4 π . They don’t agree! The point is that this divergence is zero except at the origin , where it blows up, so our calculation of ∫ ( ∇· v 2 ) is incorrect . The right answer is 4 π . Problem 1.40 ∇· v = 1 r 2 ∂ ∂r ( r 2 r cos θ ) + 1 r sin θ ∂ ∂θ (sin θ r sin θ ) + 1 r sin θ ∂ ∂φ ( r sin θ cos φ ) = 1 r 2 3 r 2 cos θ + 1 r sin θ r 2 sin θ cos θ + 1 r sin θ r sin θ ( − sin φ ) = 3 cos θ + 2 cos θ − sin φ = 5 cos θ − sin φ ∫ ( ∇· v ) dτ = ∫ (5 cos θ − sin φ ) r 2 sin θ dr dθ dφ = ∫ R 0 r 2 dr ∫ θ 2 0 [ ∫ 2 π 0 (5 cos θ − sin φ ) dφ ] dθ sin θ ↪ → 2 π (5 cos θ ) = ( R 3 3 ) (10 π ) ∫ π 2 0 sin θ cos θ dθ ↪ → sin 2 θ 2 ∣ ∣ ∣ ∣ π 2 0 = 1 2 = 5 π 3 R 3 . Two surfaces—one the hemisphere: d a = R 2 sin θ dθ dφ ˆ r ; r = R ; φ : 0 → 2 π, θ : 0 → π 2 . ∫ v · d a = ∫ ( r cos θ ) R 2 sin θ dθ dφ = R 3 ∫ π 2 0 sin θ cos θ dθ ∫ 2 π 0 dφ = R 3 ( 1 2 ) (2 π ) = πR 3 . other the flat bottom: d a = ( dr )( r sin θ dφ )(+ ˆ θ ) = r dr dφ ˆ θ (here θ = π 2 ). r : 0 → R, φ : 0 → 2 π. ∫ v · d a = ∫ ( r sin θ )( r dr dφ ) = ∫ R 0 r 2 dr ∫ 2 π 0 dφ = 2 π R 3 3 . Total: ∫ v · d a = πR 3 + 2 3 πR 3 = 5 3 πR 3 . X Problem 1.41 ∇ t = (cos θ + sin θ cos φ ) ˆ r + ( − sin θ + cos θ cos φ ) ˆ θ + 1 sin θ / ( − sin θ / sin φ ) ˆ φ ∇ 2 t = ∇· ( ∇ t ) = 1 r 2 ∂ ∂r ( r 2 (cos θ + sin θ cos φ ) ) + 1 r sin θ ∂ ∂θ (sin θ ( − sin θ + cos θ cos φ )) + 1 r sin θ ∂ ∂φ ( − sin φ ) = 1 r 2 2 r (cos θ + sin θ cos φ ) + 1 r sin θ ( − 2 sin θ cos θ + cos 2 θ cos φ − sin 2 θ cos φ ) − 1 r sin θ cos φ = 1 r sin θ [2 sin θ cos θ + 2 sin 2 θ cos φ − 2 sin θ cos θ + cos 2 θ cos φ − sin 2 θ cos φ − cos φ ] = 1 r sin θ [ (sin 2 θ + cos 2 θ ) cos φ − cos φ ] = 0 . c

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    16 CHAPTER 1. VECTOR ANALYSIS ⇒ ∇ 2 t = 0 Check: r cos θ = z, r sin θ cos φ = x ⇒ in Cartesian coordinates t = x + z . Obviously Laplacian is zero. Gradient Theorem: ∫ b a ∇ t · d l = t ( b ) − t ( a ) Segment 1: θ = π 2 , φ = 0 , r : 0 → 2 . d l = dr ˆ r ; ∇ t · d l = (cos θ + sin θ cos φ ) dr = (0 + 1) dr = dr. ∫ ∇ t · d l = ∫ 2 0 dr = 2 . Segment 2: θ = π 2 , r = 2 , φ : 0 → π 2 . d l = r sin θ dφ ˆ φ = 2 dφ ˆ φ . ∇ t · d l = ( − sin φ )(2 dφ ) = − 2 sin φ dφ. ∫ ∇ t · d l = − ∫ π 2 0 2 sin φ dφ = 2 cos φ | π 2 0 = − 2 . Segment 3: r = 2 , φ = π 2 ; θ : π 2 → 0 . d l = r dθ ˆ θ = 2 dθ ˆ θ ; ∇ t · d l = ( − sin θ + cos θ cos φ )(2 dθ ) = − 2 sin θ dθ. ∫ ∇ t · d l = − ∫ 0 π 2 2 sin θ dθ = 2 cos θ | 0 π 2 = 2 . Total: ∫ b a ∇ t · d l = 2 − 2 + 2 = 2 . Meanwhile, t ( b ) − t ( a ) = [2(1 + 0)] − [0( )] = 2 . X ˆ ˆ ˆ ˆ Problem 1.42 From Fig. 1.42, ˆ s = cos φ ˆ x + sin φ ˆ y ; ˆ φ = − sin φ ˆ x + cos φ ˆ y ; ˆ z = ˆ z Multiply first by cos φ , second by sin φ , and subtract: s cos φ − ˆ φ sin φ = cos 2 φ ˆ x + cos φ sin φ ˆ y + sin 2 φ ˆ x − sin φ cos φ ˆ y = ˆ x (sin 2 φ + cos 2 φ ) = ˆ x . So ˆ x = cos φ s − sin φ ˆ φ . Multiply first by sin φ , second by cos φ , and add: s sin φ + ˆ φ cos φ = sin φ cos φ ˆ x + sin 2 φ ˆ y − sin φ cos φ ˆ x + cos 2 φ ˆ y = ˆ y (sin 2 φ + cos 2 φ ) = ˆ y . So ˆ y = sin φ s + cos φ ˆ φ . ˆ z = ˆ z . Problem 1.43 (a) ∇· v = 1 s ∂ ∂s ( s s (2 + sin 2 φ ) ) + 1 s ∂ ∂φ ( s sin φ cos φ ) + ∂ ∂z (3 z ) = 1 s 2 s (2 + sin 2 φ ) + 1 s s (cos 2 φ − sin 2 φ ) + 3 = 4 + 2 sin 2 φ + cos 2 φ − sin 2 φ + 3 = 4 + sin 2 φ + cos 2 φ + 3 = 8. (b) ∫ ( ∇· v ) dτ = ∫ (8) s ds dφ dz = 8 ∫ 2 0 s ds ∫ π 2 0 dφ ∫ 5 0 dz = 8(2) ( π 2 ) (5) = 40 π. Meanwhile, the surface integral has five parts: top: z = 5 , d a = s ds dφ ˆ z ; v · d a = 3 z s ds dφ = 15 s ds dφ. ∫ v · d a = 15 ∫ 2 0 s ds ∫ π 2 0 dφ = 15 π. bottom: z = 0 , d a = − s ds dφ ˆ z ; v · d a = − 3 z s ds dφ = 0 . ∫ v · d a = 0 . back: φ = π 2 , d a = ds dz ˆ φ ; v · d a = s sin φ cos φ ds dz = 0 . ∫ v · d a = 0 . left: φ = 0 , d a = − ds dz ˆ φ ; v · d a = − s sin φ cos φ ds dz = 0 . ∫ v · d a = 0 . front: s = 2 , d a = s dφ dz ˆ s ; v · d a = s (2 + sin 2 φ ) s dφ dz = 4(2 + sin 2 φ ) dφ dz. ∫ v · d a = 4 ∫ π 2 0 (2 + sin 2 φ ) dφ ∫ 5 0 dz = (4)( π + π 4 )(5) = 25 π. So ∮ v · d a = 15 π + 25 π = 40 π. X (c) ∇× v = ( 1 s ∂ ∂φ (3 z ) − ∂ ∂z ( s sin φ cos φ ) ) ˆ s + ( ∂ ∂z ( s (2 + sin 2 φ ) ) − ∂ ∂s (3 z ) ) ˆ φ + 1 s ( ∂ ∂s ( s 2 sin φ cos φ ) − ∂ ∂φ ( s (2 + sin 2 φ ) ) ) ˆ z = 1 s (2 s sin φ cos φ − s 2 sin φ cos φ ) ˆ z = 0 . c

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