Quiz 4 Solution - Math 3010 Quiz 4 Nov 28 2012 1(4 points...

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Math 3010 Quiz 4 Nov. 28, 2012 1. (4 points) For a surface S and ±xed vector vV R 3 , prove that 2 ii S · ˆ n dS = i ∂S ( × v r ) · d v S. where v r ( x, y, z ) = ( x, y, z ). Solution: Let = ( a, b, c ) and v F = × v r . Then × v r = v v v v v v ˆ ı ˆ ˆ k a b c x y z v v v v v v = ( bz cy, cx az, ay bx ) . v ∇ × ( × v r ) == v v v v v v ˆ ı ˆ ˆ k ∂x ∂y ∂z bz cy cx az ay bx v v v v v v = (2 a, 2 b, 2 c ) = 2 vV. By Stokes’ theorem, II S p v ∇ × v F P · ˆ n dS = I ∂S v F · dvs . Thus, 2 S · ˆ n dS = i ∂S ( × v r ) · d v 2. (a) (2 points) Find f : R 2 R such that ( y 3 + 1 , 3 xy 2 + 1) = v f , if it exists. Solution: This is a conservative vector ±eld since v F is de±ned everywhere on R 2 and v ∇ × v F = 0. To ±nd f , solve the following system of equations: ∂f ∂x = y 3 + 1 , ∂y = 3 xy 2 + 1 Solving for f in the ±rst equation leads to f ( x, y ) = xy 3 + x + g ( y ). Using this in the second equation yields g ( y ) = 1, so g ( y ) = y + any constant. So we can choose f ( x, y ) = xy 3 + x + y .
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Quiz 4 Solution - Math 3010 Quiz 4 Nov 28 2012 1(4 points...

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