Final Exam Solution

# Then s oriented positively according to the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online