# Chapter 16 - SECTION 16.1(PAGE 858 R A ADAMS CALCULUS...

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Unformatted text preview: SECTION 16.1 (PAGE 858) R. A. ADAMS: CALCULUS CHAPTER 16. VECTOR CALCULUS Section 16.1 Gradient, Divergence, and Curl (page 858) 1. F = x i + y j divF = ∂ ∂ x ( x ) + ∂ ∂ y ( y ) + ∂ ∂ z ( ) = 1 + 1 = 2 curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z x y = 2. F = y i + x j divF = ∂ ∂ x ( y ) + ∂ ∂ y ( x ) + ∂ ∂ z ( ) = + = curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z y x = ( 1 − 1 ) k = 3. F = y i + z j + x k divF = ∂ ∂ x ( y ) + ∂ ∂ y ( z ) + ∂ ∂ z ( x ) = curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z y z x = − i − j − k 4. F = yz i + xz j + xy k divF = ∂ ∂ x ( yz ) + ∂ ∂ y ( xz ) + ∂ ∂ z ( xy ) = curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z yz xz xy = ( x − x ) i + ( y − y ) j + ( z − z ) k = 5. F = x i + x k divF = ∂ ∂ x ( x ) + ∂ ∂ y ( ) + ∂ ∂ z ( x ) = 1 curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z x x = − j 6. F = xy 2 i − yz 2 j + zx 2 k divF = ∂ ∂ x xy 2 + ∂ ∂ y − yz 2 + ∂ ∂ z zx 2 = y 2 − z 2 + x 2 curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z xy 2 − yz 2 zx 2 = 2 yz i − 2 xz j − 2 xy k 7. F = f ( x ) i + g ( y ) j + h ( z ) k divF = ∂ ∂ x f ( x ) + ∂ ∂ y g ( y ) + ∂ ∂ z h ( z ) = f ( x ) + g ( y ) + h ( z ) curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z f ( x ) g ( y ) h ( z ) = 8. F = f ( z ) i − f ( z ) j divF = ∂ ∂ x f ( z ) + ∂ ∂ y − f ( z ) = curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z f ( z ) − f ( z ) = f ( z )( i + j ) 9. Since x = r cos θ , and y = r sin θ , we have r 2 = x 2 + y 2 , and so ∂ r ∂ x = x r = cos θ ∂ r ∂ y = y r = sin θ ∂ ∂ x sin θ = ∂ ∂ x y r = − xy r 3 = − cos θ sin θ r ∂ ∂ y sin θ = ∂ ∂ y y r = 1 r − y 2 r 3 = x 2 r 3 = cos 2 θ r ∂ ∂ x cos θ = ∂ ∂ x x r = 1 r − x 2 r 3 = y 2 r 3 = sin 2 θ r ∂ ∂ y cos θ = ∂ ∂ y x r = − xy r 3 = − cos θ sin θ r . (The last two derivatives are not needed for this exercise, but will be useful for the next two exercises.) For F = r i + sin θ j , we have div F = ∂ r ∂ x + ∂ ∂ y sin θ = cos θ + cos 2 θ r curl F = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z r sin θ = − sin θ cos θ r − sin θ k . 600 INSTRUCTOR’S SOLUTIONS MANUAL SECTION 16.1 (PAGE 858) 10. F = ˆ r = cos θ i + sin θ j divF = sin 2 θ r + cos 2 θ r = 1 r = 1 x 2 + y 2 curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z cos θ sin θ = − cos θ sin θ r − cos θ sin θ r k = 11. F = ˆ θ = − sin θ i + cos θ j divF = cos θ sin θ r − cos θ sin θ r = curlF = i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z − sin θ cos θ = sin 2 θ r + cos 2 θ r k = 1 r k = 1 x 2 + y 2 k 12. We use the Maclaurin expansion of F , as presented in the proof of Theorem 1: F = F + F 1 x + F 2 y + F 3 z + · · · , where F = F ( , , ) F 1 = ∂ ∂ x F ( x , y , z ) ( , , ) = ∂ F 1 ∂ x i + ∂ F 2 ∂ x j + ∂ F 3 ∂ x k ( , , ) F 2 = ∂ ∂ y F ( x , y , z ) ( , , ) = ∂ F 1 ∂ y i + ∂ F 2 ∂ y j + ∂ F 3 ∂ y k ( , , ) F 3 = ∂ ∂ z F ( x ,...
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## This note was uploaded on 12/16/2009 for the course FEW, FEWEB 400567 taught by Professor Moerdersen during the Fall '09 term at Vrije Universiteit Amsterdam.

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Chapter 16 - SECTION 16.1(PAGE 858 R A ADAMS CALCULUS...

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