plugin-FExRevProbSolutions

plugin-FExRevProbSolutions - Math 255 Winter 2010 Review...

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Unformatted text preview: Math 255 Winter 2010 Review problems for Final Exam 1. The electrostatic potential at (0 , ,- a ) of a charge of constant density on the hemisphere S : x 2 + y 2 + z 2 = a 2 , z 0 is U = S x 2 + y 2 + ( z + a ) 2 dS. Show that U = 2 a (2- 2). Solution The surface is a graph z = a 2- x 2- y 2 . We therefore project the integral onto the xy plane and convert to polar coordinates to find that we need to calculate 2 a 2 a 2 + 2 a a 2- r 2 a a 2- r 2 r d r d . After integrating with respect to we make the substitution w = a 2- r 2 to find that 2 a a 2 d( a + w ) a + w = 2 a (2- 2) . 2. Find the value of the line integral C yzdx + xzdy + xydz where C is the curve with initial point P = (1 , , 0) given by. a) x = cos t , y = sin t , z = t 2 with t [0 , ]. b) The line segment between P and (- 1 , ,- ). c) The ellipse x 2 + 4 y 2 = 1 and z = 0 parametrized clockwise. d) The circle x 2 + y 2 = 9 and z = 0 parametrized counterclockwise as viewed from above. Use the fact that the path C is the boundary of the region x 2 + y 2 9 with z = 0. Use Greens Theorem to compute value of the line integral. e) The same path as in part d ), but now using Stokes Theorem. Use the fact that the path C in this case is the boundary of the hemisphere x 2 + y 2 + z 2 = 9 with z 0....
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This note was uploaded on 12/31/2011 for the course MATH 255 taught by Professor Jackwaddell during the Fall '08 term at University of Michigan.

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plugin-FExRevProbSolutions - Math 255 Winter 2010 Review...

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