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# vbcbcvbc - MATH 241H - Exam 4 - May 7, 2007 - Solutions 1....

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Unformatted text preview: MATH 241H - Exam 4 - May 7, 2007 - Solutions 1. The force ﬁeld is conservative with potential function f (x, y, z ) = xy sin z + y 2 . The curve has initial point r(0) = (1, 1, 0) and terminal point r(1) = (−1, e, π/2). Thus the work integral is equal to f (−1, e, π/2) − f (1, 1, 0) = −e + e2 − 1. 2. r (t) = √ 14 so (xy + z ) ds = C 1 (t(3t) 0 √ √ + 2t + 3) 14 dt = 5 14. 3. Using Green’s theorem the work integral is equal to 1 −1 0 1−x2 −2 dydx = −8/3. 4. Using Divergence Theorem (with a minus sign since normals are inward instead of outward) and setting up the triple integral in cylindrical coordinates, the ﬂux integral is equal to 2π 3 0 r 3 − 0 4zr dzdrdθ = −81π. 5a. Σ √ the graph of the function z = 5 − x − 3y with fx = −1 and fy = −3. Thus is √ dS = 1 + 1 + 9 dA = 11 dA. Σ is the part of the graph lying over the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, so 2 1 0 z dS = Σ 0 √ √ (5 − x − 3y ) 11 dxdy = 3 11. 5b. Use Stokes Theorem. The upward pointing normal vector is N = i + 3j + k, and curl F = i + y j +(1 − z )k = i + y j +(1 − (5 − x − 3y ))k. Thus curl F · N = 1+3y − 4+ x +3y = x + 6y − 3. Thus the work integral is equal to 2 0 0 1 (x + 6y − 3) dxdy = 7. ...
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## This note was uploaded on 04/04/2010 for the course MATH 113 taught by Professor Staff during the Spring '08 term at Maryland.

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