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Practice Exam 2 Solutions The answer to evey multiple choice question is a . 9. Under the change of variables x = s 2 - t 2 , y = 2 st the quarter circular region in the st -plane given by s 2 + t 2 1 , s 0 , t 0 is mapped onto a certain region D of the xy -plane. Evaluate ZZ D dx dy p x 2 + y 2 . To evaluate this integral, we note that ( x, y ) ( s, t ) = 2 s - 2 t 2 t 2 s = 4 s 2 - ( - 4 t 2 ) = 4 s 2 + 4 t 2 > 0 and 1 p x 2 + y 2 = 1 p ( s 2 - t 2 ) 2 + (2 st ) 2 = 1 s 4 + 2 s 2 t 2 + t 4 = 1 p ( s 2 + t 2 ) 2 = 1 s 2 + t 2 . Now by the change of variables formula ( § 15.10): ZZ D dx dy p x 2 + y 2 = ZZ { s 2 + t 2 1 , s 0 , t 0 } 4( s 2 + t 2 ) s 2 + t 2 dA = 4 ZZ { s 2 + t 2 1 , s 0 , t 0 } dA = 4 π 4 = π since the area of the quater circle in the st -plane is π 4 . 10. Let F = z 2 i + ze yz j + (2 xz + cos z + ye yz ) k . Find a function f ( x, y, z ) so that f = F . We need f x ( x, y, z ) = z 2 so f ( x, y, x ) = xz 2 + g ( y, z ) where g y ( y, z ) = ze yz implying g ( y, z ) = e yz + h ( z ). Finally, 2 xz + cos z + ye yz = f z ( x, y, z ) = 2 xz + g z ( y, z ) = 2 xz + ye yz + h z ( z ) so h z ( z ) = cos z and h ( z ) = sin z . Therefore f ( x, y, z ) = xz 2 + e yz + sin z .

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