Spring 2006 - Nordgren's Class - Exam 2 (Version A)

Spring 2006 - Nordgren's Class - Exam 2 (Version A) -...

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Unformatted text preview: Solutions to Version A of Midterm 2 By H akan Nordgren R EMARK : Please note that there were two versions of the midterm. Problem 1 (25 points): Evaluate Z C ( x + y + z ) ds, where the curve C : [0 , ] R 3 is defined by C ( t ) = (sin( t ) , cos( t ) , 4 t ). Solution: Along the curve C we have x ( t ) = sin( t ) , y ( t ) = cos( t ) , and z ( t ) = 4 t, so dx ( t ) = cos( t ) dt, dy ( t ) =- sin( t ) dt, and dz ( t ) = 4 dt Thus ds = ( cos 2 ( t ) + sin 2 ( t ) + 16 ) 1 2 dt = 17 dt and Z C ( x + y + z ) ds = 17 Z (sin( t ) + cos( t ) + 4 t ) dt =- 17[cos( t )] + 17[sin( t )] + 17 2 t 2 / = 17 ( 2 + 2 2 ) . Problem 2 (25 points): Evaluate the integral Z Z Z E (8 x 2 + 8 y 2 + 18 z 2 ) dxdydz, where E = { ( x,y,z ) R 3 : x 2 9 + y 2 9 + z 2 4 1 } . Solution: Make the change of variables u = x 3 , v = y 3 , and w = z 2 . Then dudvdw = fl fl fl fl ( u,v,w ) ( x,y,z ) fl fl fl fl dxdydz = fl fl fl fl fl fl det 1 3 1...
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Spring 2006 - Nordgren's Class - Exam 2 (Version A) -...

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