12.07 - F around the closed path C equals the surface...

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Questions about group work from last time? Note: Friday’s exam will have four questions rather than three. (Not sure about Monday’s yet.) If there are more questions, they’ll be easier individually. Concept Check for chapter 12: questions 3, 7, 9 [Hand out answer sheets] [Hand out new group-work sheets: 13.8.2, 13.9.1] Section 13.8: [Group Work 2] ..?. . 1. Use the parametrization x = x , y = y , z = – ( d + ax + by )/ c to obtain ∫∫ D sqrt(1 + (– a / c ) 2 + (– b / c ) 2 ) dA . 2. See Section 12.5. 3. (a) Curl F = ( / y ) (–4 x + y ) – ( / z ) (2 x ), ( / z ) ( x 3 + z +2 y ) – ( / x ) (–4 x + y ), ( / x ) (2 x ) – ( / y ) ( x 3 + z +2 y ) = 1 – 0, 1 + 4, 2 – 2 = 1,5,0 . (b) The Instructor’s Guide says: “Since the curl is a
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constant vector field, the net flux through the closed surface (formed by D , S , and the sides created between them by dropping perpendicular to D ) is 0. Now Stokes’ Theorem gives us that the line integral is zero.” But I say: Stokes’ Theorem tells us that the line integral of
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Unformatted text preview: F around the closed path C equals the surface integral of curl F over S . Write S (curl F ) d S = S (curl F ) n dS = S 1,5,0 a , b , c /sqrt( a 2 + b 2 + c 2 ) dS = S ( a +5 b )/sqrt( a 2 + b 2 + c 2 ) dS , which equals ( a +5 b )/sqrt( a 2 + b 2 + c 2 ) times the area A S ; combining this with our formula A S = ( A D /| c |) sqrt( a 2 + b 2 + c 2 ) we see that the original line integral equals ( a +5 b )/| c | times A D . Section 13.9: [Group Work: Group Work 1] ..?. . 1. Since div F > 0, the Divergence Theorem tells us that the net flux is positive. 2. 1 1 1 1 sqrt(1 yy ) sqrt(1 yy ) ( x 2 + y 2 ) dx dy dz = 1 1 2 1 ( r 2 ) r dr d dz = ( 1 1 dz ) ( 2 d ) ( 1 r 3 dr ) = (2)(2 )(1/4) = ....
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12.07 - F around the closed path C equals the surface...

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