s10wk12 - solid we have the Theorem that Curl ± F = ±...

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Math 23 B. Dodson Week 12 Homework: [due April 23] 16.5 Curl and Divergence 16.6 Surfaces and Surface Area 16.7 Surface integrals 16.8 Stokes’ Theorem [first half] Week 13 Homework: [due Wednesday, April 28] 16.8 Stokes’ Theorem [second half] 16.9 Divergence Theorem [Syllabus covers just the “first half”] Problem 16.5.5: Find Div( F ) and Curl( F ) when F = < e x sin y, e x cos y, z > . Solution: For F = < P, Q, R >, Div( F ) = P x + Q y + R z , so Div( F ) = e x sin y - e x sin y + 1 = 1 . Using the cross product formula we have Curl( F ) = < R y - Q z , R x - P z , Q x - P y > = < 0 , 0 , 0 > . Notice that if F = < f x , f y , f z > is the gradient of f, we always get Curl = 0 (why?) and on a region like R 3 , a solid ball, or a rectangular
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Unformatted text preview: solid we have the Theorem that Curl( ± F ) = ± exactly when there is f so that ± F = Grad( f ). 2 Problem 16.9.7: Use the Divergence Theorem to calculate the surface integral RR S ± F · d ± S, where ± F ( x, y, z ) = ( e x sin y ) ± i + ( e x sin y ) ± j + yz 2 ± k, and S is the surface of the box B bounded by planes x = 0 , x = 1 , y = 0 , y = 1 , z = 0 , z = 2 . Solution: The divergence div( ± F ) = = e x sin y + e x (-sin y ) + 2 yz = 2 yz. So RR S ± F · d ± S = RRR B 2 yz dV = Z 1 Z 1 Z 2 2 yz dzdydx = 2 ....
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