Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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318 0 CHAPTER 16 VECTOR CALCULUS (b) curlF = ({)R - {)Q) i + ({)P - {)R) j + ({)Q - {)p) k = (0- 0) i -t (0 - O) j + (o- {)P) k =-{)P k 8y .{)z {)z 8x ax 8y 8y 8y : fJP 0 8Pk . . . . th . d" . S uy vy • · mce -;:;- > , - .q,, IS a vector porntmg m e negative z- 1rect1on. 13. curiF = "il x F = 8f8x 8f8y j k 8f8z = (6xyz 2 - 6xyz 2 ) i - (~y 2 z 2 - 3y 2 z 2 )j + (2yz 3 - 2yz 3 ) k = 0 and F is defined on all oflR 3 with component functions which have continuous partial derivatives, so by Theorem 4, F is conservativ~. Thus, there exists a· function f such that F = "il f. Then f., ( x, y , z) = y 2 z 3 implies f(x,y,z) = xy 2 z 3 + g(y,z) and f 11 (x,y,z) = 2xyz 3 + g 11 (y,z). But f 11 (x,y,z) = 2xyz 3 ,sog(y,z) = h(z) and f(x, y,z) = xy 2 z 3 + h(z). Thus fz(x, y, z) = 3xy 2 z 2 + h'(z) but f:(x, y, z) = 3xy 2 z 2 so· h(z) = K, a constant. Hence a potential function for F is f (x, y, z) = xy 2 z 3 + K. 15. curl F = "il X F = 8f8x 8f8y 8f8z j 3xy 2 z 2 2x 2 yz 3 3x 2 y 2 z 2 = (6x 2 yz 2 - 6x 2 yz 2 ) i - (6xy 2 z 2 - 6xy 2 z) j .+ (4xyz 3 - 6xyz 2 ) k = 6xy 2 z(I- z) j + 2xyz 2 (2z- 3) k =1- 0 so F is not conservative. 17. cur!F = "il x F = 8f8x 8f8y 8f8z j k k F is defined on all oflll 3 , and the partial derivatives of the component functi~ns are continuous,'so F is conservative. Thus there exists a function f such that "il f =F. Then f,(x, y, z) = e 11 = implies f(x, y, z) = xe 11 = + g(y, z) => fu(x, y, z) = xzev= + g 11 (y, z). But / 11 {x, y, z) = xze 11 =, so g(y, z) = h(z) and f(x, y, z) = xe 11 z + h(z). Thus f:(x, y, z) = xye 11 z + h'(z) but f:(x, y, z) = xye 11 = so h(z) =Kanda potential function for F is f(x, y, z) = xe 11 = + K . . 19. No. Assume there is such a G. Then div(curl G ) ='! (x sin y) + ~ (cosy) + ~ (z - xy) =sin y- siny + 1 f 0, uX uy uz which contradicts Theorem 11 . j k 21 . curlF = 8f8x 8f8y 8f8z = {0- 0) i + {0- O)j + {0 - 0) k = 0. Hence F = f(x) i + g(y) j + h(z) k is irrotational. f(x) g(y) h(z) ® 2012 Ccnl!llgC Lcnming. All Rights Reserved. Mny not be sclllUlcd. copied, or duplicated. or posted ro a publicly accessible websire, in whole or in par1.
SECTION 16.5 CURL AND DIVERGENCE 0 319 ForExercises23-29, 1etF(x,y,z) = P1 i+ Qd + R1 ~and G(x,y,z) = P2 i + Q2j + R2 k. 23. div(F +G)= div(Pt + P2 , Q1 + Qz,R1 + R 2 ) = o(P 1 0 + P2 ) + o(Q 1 0 + Qz )" + o(R1 / R 2 ) • X y Z = oP1 + oPz + oQ1 + oQ2 + oR1 + oR2 = (oP1 + oQ1 + oR1) + ( oP2 + oQ2 + oR2) ox ox oy oy o z o z ox oy o z ox oy o z = div(P1 ,,Q1, R1) + div(Pz , Qz , Rz ) = div F + div G 27. div(F X G) ='iJ · (F x G)= = [Q 1 oR2 + R 2 oQ1 _ Q 2 oR1 _ R 1 oQ2] _ [p 1 oRz + Rz oP1 _ Pz oR1 _ R 1 oPz] ox ox ox ox .oy oy ay ay + [Pt oQ2 + Q 2 oP1 _ p 2 oQ1. _ Q 1 oP2 ] oz oz . oz oz = [p 2 (oR1 _ oQ1) +Q 2 (oP1 _ oR1) +Rz(oQ1 _ 8P1 )] oy oz oz ox OX ay _ [p 1 (oRz _ oQz) -f_ Q 1 (oP2 _ oRz) + R 1 (oQz ~ aP2 )] oy az az ox ox 8y = G · curl F - F · curl G j 29. curl{ curl F) = 'iJ x ('iJ x F) = ajax a;ay k a;az Now let's consider grad(div F ) - 'iJ 2 F and compare with the above. (Note that 'iJ 2 F is defined 'on page J J 19 [ET I 095).) [continued] ® 2012 Cengage Lcnrning. All Rights Reserved. May not bo scanned, copied, or duplicated. or posted to a publicly accessible website. in whole or in part.
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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CHAPTER 10 REVIEW 0 35 the length o
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CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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0 PROBLEMS PLUS l lt sin u dx cost
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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