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YellowFinalSolutions

YellowFinalSolutions - yellow MATH 32B Final Exam LAST NAME...

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yellow MATH 32B Final Exam March 20, 2008 LAST NAME FIRST NAME ID NO. Your TA: To receive credit, you must write your answer in the space provided . DO NOT WRITE BELOW THIS LINE 1 (20 pts) 5 (20 pts) 2 (20 pts) 6 (20 pts) 3 (20 pts) 7 (20 pts) 4 (20 pts) TOTAL
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2 PROBLEM 1 (20 Points) State the Divergence Theorem. Then verify the Divergence Theorem for the vector field F = h x,y,z i and the region W defined by x 2 + y 2 + z 2 16. Answer : Solution: We have div( F ) = 3, so Z Z Z W div( F ) dV = 3vol( W ) = 3( 4 3 π 4 3 ) = 4 4 π = 256 π The boundary of W is the sphere S of radius 3, so we must show that Z Z S F · d S = 256 π We use spherical coordinates. The vector field, when re- stricted to the sphere of radius 4 is equal to F = 4 e r where e r is the unit radial vector and n = 4 2 sin φ e r Z Z S F · d S = Z 2 π θ =0 Z π φ =0 (4 e r ) · (16 sin φ e r ) dφdθ = 2 π Z π φ =0 (4)(16) sin φdφ = 2 π (4)(16)(2) = 256 π This verifies the Divergence Theorem.
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3 PROBLEM 2 (20 Points) Evaluate Z Z D ( x 2 - y 2 ) dA where D is the shaded domain in the figure below. Answer : Figure 1. Domain Solution: Use the change of variables formula with u = xy, v = x - y Then ± ± ± ± ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v ± ± ± ± = ± ± ± ± ± ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y ± ± ± ± ± - 1 = ± ± ± ± y x 1 - 1 ± ± ± ± - 1 = 1 | x + y |
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4 We obtain Z Z D ( x 2 - y 2 ) dA = Z 4 u =2 Z 3 v =0 ( x 2 - y 2 ) ± 1 | x + y | ² dv du = Z 4 u =2 Z 3 v =0 | x - y | dv dv = Z 4 u =2 Z 3 v =0 v dv du = 2 Z 3 v =0 v dv = 9
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PROBLEM 3 (20 Points) Let S be the surface defined by x 2 + y 2 - z 2 = 4 , 1 z 2 Let C 1 be the upper boundary curve of S and let C 2 be lower boundary curve, both oriented counterclockwise when viewed from above. Suppose that F is a vector field such that I C 1 F · d s = 5 , curl( F ) = h yz,xz, 2 xy i Compute I C 2 F · d s . Answer : Solution: With the given orientations, I S F · d s = I C 1 F · d s - I C 2 F · d s = 5 - I C 2 F · d s By Stokes’ Theorem, 5 - I C 2 F · d s = Z Z S curl( F ) · d S = Z Z S h yz,xz, 2 xy i · d S We parametrize S as follows: Φ( x,y ) = h x,y, p x 2 + y 2 - 4 i n = h - x p x 2 + y 2 - 4 , - y p x 2 + y 2 - 4 , 1 i = h - x z , - y z , 1 i Then h yz,xz, 2 xy i · n = h yz,xz, 2 xy i · h - x z , - y z , 1 i = 0. Therefore
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YellowFinalSolutions - yellow MATH 32B Final Exam LAST NAME...

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