Final Exam Solution

Then s oriented positively according to the

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Unformatted text preview: nh2 u du dv 2 du dv = 8 ln 2. 0 − ln 2 Math 3010 Final Exam - Page 9 of 14 Dec. 8, 2012 6. Integral theorems (a) (5 points) Verify Green’s theorem for the line integral ∂ D x2 y dx + y dy when ∂D is the boundary of the region bounded by the curves y = x and y = x3 , x ∈ [0, 1]. Solution: Traverse ∂D from (0, 0) to (0, 1) along y = x3 , and then back along y = x. c1 (t) = t, t3 , t ∈ [0, 1] c2 (t) = (t, t) , t ∈ [0, 1] We need to travel along c1 and then opposite the direction of c2 . x2 y dx + y dy = ∂D c1 1 x2 y dx + y dy − x2 y dx + y dy c2 t5 , t3 · c′1 (t) − t3 , t · c′2 (t) dt = 0 1 t5 , t3 · 1, 3t2 − t3 , t · (1, 1) dt = 0 1 4t5 − t3 − t dt = = 0 211 1 − − =− . 342 12 Compare with D ∂2 ∂ y− x y dA = − ∂x ∂y =− x2 dA D 1 x x2 dy dx x3 0 1 =− 2 xy 0 x x3 1 dx = − 0 x3 − x5 dx = − (b) (5 points) Verify Stokes’ theorem for the upper hemisphere z = z 0 and the radial vector field F (x, y, z ) = (x, y, z ). 1 12 1 − x2 − y 2 , Solution: Let S be the given surface with either orientation. Then ∂S (oriented positively according to the orientation of S ) is a closed curve. Since F = 1 1 ∇ 2 x2 + 2 y 2 + 1 z 2 then 2 ∂S Also, ∇ × F = 0 so D F · ds = 0. ∇ × F · d S = 0. Math 3010 Final Exam - Page 10 of 14 Dec. 8, 2012 7. Let S ⊂ R3 be a closed surface and F be a C 2 vector field defined on S . (a) (5 points) Prove that S ∇ × F · dS = 0 with the help of Stokes’ theorem. Solution: Sp...
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