Math252Sols4F07

Math252Sols4F07 - QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252...

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QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252 – FALL 2007 – KUNIYUKI SCORED OUT OF 125 POINTS MULTIPLIED BY 0.84 105% POSSIBLE 1) Reverse the order of integration, and evaluate the resulting double integral: cos 1 + y 4 ( ) dy dx x 3 3 0 27 . Give a simplified exact answer; do not approximate. Sketch the region of integration. (20 points) Sketch the region of integration, R . It may help to rewrite the double integral with labels: cos 1 + y 4 ( ) dy dx y = x 3 y = 3 x = 0 x = 27 We aim / fix x , shoot y , and slide x . Reverse the order of integration. Solve y = x 3 for x : x = y 3
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We aim y , shoot x , and slide y . cos 1 + y 4 ( ) dxdy x = 0 x = y 3 y = 0 y = 3 = x cos 1 + y 4 ( ) x = 0 x = y 3 dy y = 0 y = 3 = y 3 cos 1 + y 4 ( ) 0 ( ) dy 0 3 = y 3 cos 1 + y 4 ( ) dy 0 3 We now perform a standard u -substitution. Let u = 1 + y 4 du = 4 y 3 dy y 3 dy = 1 4 du You can also “Compensate”: y 3 cos 1 + y 4 ( ) dy 0 3 = 1 4 4 y 3 cos 1 + y 4 ( ) dy 0 3 Change the limits of integration: y = 0 u = 1 + 0 ( ) 4 u = 1 y = 3 u = 1 + 3 ( ) 4 u = 82 = cos u 1 4 du u = 1 u = 82 = 1 4 sin u 1 82 = 1 4 sin82 sin1 ( ) Note: This is about 0.13206. ( )
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2) Let R be the region in the xy -plane that is bounded by the rectangle with vertices 1,3 ( ) , 7,3 ( ) , 7,5 ( ) , and 1,5 ( ) . Set up a double integral for the surface area of the portion of the graph of 9 x 2 + 4 y 2 + z 2 = 1000 z > 0 ( ) that lies over R . Make sure your double integral is as detailed as possible; do not leave in generic notation like R or f . Also, do not leave dA in your final answer; break it down into d variable ( ) d variable ( ) . Do not evaluate. (14 points) Observe that the graph of 9 x 2 + 4 y 2 + z 2 = 1000 is an ellipsoid. With the restriction z > 0 ( ) , we only consider [part of] the upper half. Solve 9 x 2 + 4 y 2 + z 2 = 1000 for z : 9 x 2 + 4 y 2 + z 2 = 1000 z > 0 ( ) z 2 = 1000 9 x 2 4 y 2 z > 0 ( ) z = 1000 9 x 2 4 y 2 Let f x , y ( ) = 1000 9 x 2 4 y 2 . Its domain contains R . S = + f x x , y ( ) [ ] 2 + f y x , y ( ) [ ] 2 dA R ∫∫ f x , y ( ) = 1000 9 x 2 4 y 2 ( ) 1 2 f x x , y ( ) = 1 2 1000 9 x 2 4 y 2 ( ) 1 2 18 x ( ) = 9 x 1000 9 x 2 4 y 2 f y x , y ( ) = 1 2 1000 9 x 2 4 y 2 ( ) 1 2 8 y ( ) = 4 y 1000 9 x 2 4 y 2 S = 1 + 9 x 1000 9 x 2 4 y 2 2 + 4 y 1000 9 x 2 4 y 2 2 dydx 3 5 1 7 or S = 1 + 81 x 2 1000 9 x 2 4 y 2 + 16 y 2 1000 9 x 2 4 y 2 3 5 1 7 or S = 1 + 81 x 2 + 16 y 2 1000 9 x 2 4 y 2 3 5 1 7
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Math252Sols4F07 - QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 252...

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