prac3bsol

prac3bsol - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.02 Practice Exam 3 B Solutions (1,2) y = 2x x=1 1. a) (1,1) y = x b) 0 1 y dxdy + y/2 1 2 1 dxdy. y/2 (the first integral corresponds to the bottom half 0 y 1, the second integral to the top half 1 y 2.) 2. a) dA = r sin rdrd = sin drd. r2 3 M= dA = sin drd = R 0 1 0 2 sin d = - 2 cos 0 = 4. 1 b) x = M 1 xdA = 4 R 0 3 r cos sin drd 1 The reason why one knows that x = 0 without computation is that the region and the density are symmetric with respect to the y-axis ((x, y) = (-x, y)). 3. a) Nx = -12y = My , hence F is conservative. b) fx = 3x2 - 6y 2 f = x3 - 6y 2 x + c(y) fy = -12xy + c (y) = -12xy + 4y. So c (y) = 4y, thus c(y) = 2y 2 (+ constant). In conclusion f = x3 - 6xy 2 + 2y 2 (+ constant). c) The curve C starts at (1, 0) and ends at (1, 1), therefore F dr = f (1, 1) - f (1, 0) = (1 - 6 + 2) - 1 = -4. C 4. a) The parametrization of the circle C is x = cos t, y = sin t, for 0 t < 2; then dx = - sin tdt, dy = cos tdt and 2 W = (5 cos t + 3 sin t)(- sin t)dt + (1 + cos(sin t)) cos tdt. 0 b) Let R be the unit disc inside C; F dr = (Nx - My )dA = (0 - 3)dA = -3 area(R) = -3. C R R C3 (1, 4) 5. a) (0, 4) C4 C2 ^ F n ds = = (0, 0) C1 (1, 0) (y + cos x cos y - cos x cos y)dxdy = R 4 1 4 = ydxdy = ydy = [y 2 /2]4 = 8. 0 0 0 0 C div F dxdy R ydxdy R ^ ^ b) On C4 , x = 0, so F = - sin y ^, whereas n = -^. Hence F n = 0. Therefore the flux of F i through C4 equals 0. Thus ^ ^ ^ ^ F n ds = F n ds - F n ds = F n ds ; C1 +C2 +C3 C C4 C and the total flux through C1 + C2 + C3 is equal to the flux through C. ux uy 2 -1 = = 3. 6. Let u = 2x - y and v = x + y - 1. The Jacobian vx vy 1 1 1 Hence dudv = 3dxdy and dxdy = dudv, so that 3 V = (4 - (2x - y)2 - (x + y - 1)2 ) dxdy (2x-y)2 +(x+y-1)2 <4 = 1 (4 - u2 - v 2 ) dudv 3 2 0 u2 +v 2 <4 = = 2 0 2 0 1 (4 - r ) rdrd = 3 2 2 0 2 2 1 r - r4 3 12 2 d 0 4 8 d = . 3 3 ...
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This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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