plugin-FExRevProbSolutions2-1

plugin-FExRevProbSolutions2-1 - Math 255 Winter 2010 Review...

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Math 255 Winter 2010 Review problems for Final Exam 1. The electrostatic potential at (0 , 0 , - a ) of a charge of constant density σ on the hemisphere S : x 2 + y 2 + z 2 = a 2 , z 0 is U = Z Z S σ p x 2 + y 2 + ( z + a ) 2 dS. Show that U = 2 πσa (2 - 2). Solution The surface is a graph z = g ( x,y ) = p a 2 - x 2 - y 2 . We therefore project the surface onto the xy plane and obtain the circular disk x 2 + y 2 a 2 . Hence a possible parametrization of the surface is in terms of the parameters x and y x = x, y = y z = p a 2 - x 2 - y 2 for ( x,y ) satisfying x 2 + y 2 a 2 . (1) Hence you can compute U = Z a - a Z a 2 - y 2 - a 2 - y 2 σ p x 2 + y 2 + ( g ( x,y ) + a ) 2 q 1 + ( g x ) 2 + ( g y ) 2 dxdy Since the region of integration is a circular disk, it could be convenient for us to convert to polar coordinates to find that we need to calculate, hence the parametrization 1 becomes x = r cos θ, y = r sin θ z = a 2 - r 2 for r [0 ,a ] θ [0 , 2 π ] . (2) Then the formula for U becomes U = Z a 0 Z 2 π 0 σ p 2 a 2 + 2 a a 2 - r 2 a a 2 - r 2 r d θ d r. After integrating with respect to θ U = 2 π Z a 0 σ p 2 a 2 + 2 a a 2 - r 2 a a 2 - r 2 r d r. we make the substitution w = a 2 - r 2 to find that U = - πσa Z 0 a 2 1 p 2 a 2 + 2 a w 1 w d w. Make another substitution u = 2 a 2 + 2 a w with du = a w dw and convert the integral into U = πσ Z 4 a 2 2 a 2 2 1 u d u = 2 πσ u | 4 a 2 2 a 2 = 2 πσa (2 - 2) .
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2. Find the value of the line integral
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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-FExRevProbSolutions2-1 - Math 255 Winter 2010 Review...

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