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

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

This note was uploaded on 04/04/2010 for the course MATH 113 taught by Professor Staff during the Spring '08 term at Maryland.

Ask a homework question - tutors are online