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HW15-solutions-1 - zakaria(mmz255 – HW15 – gilbert...

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Unformatted text preview: zakaria (mmz255) – HW15 – gilbert – (55485) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the triple integral I = integraldisplay 1 integraldisplay x integraldisplay x- y (3 x + 2 y ) dzdydx . 1. I = 5 8 2. I = 3 8 3. I = 7 24 4. I = 11 24 correct 5. I = 13 24 Explanation: As a repeated integral, I = integraldisplay 1 parenleftBig integraldisplay x parenleftBig integraldisplay x- y (3 x + 2 y ) dz parenrightBig dy parenrightBig dx . Now integraldisplay x- y (3 x + 2 y ) dz = bracketleftBig (3 x + 2 y ) z bracketrightBig x- y = (3 x + 2 y )( x- y ) = 3 x 2- xy- 2 y 2 , while integraldisplay x (3 x 2- xy- 2 y 2 ) dy = bracketleftBig 3 x 2 y- 1 2 xy 2- 2 3 y 3 bracketrightBig x = 11 6 x 3 . Consequently, I = integraldisplay 1 11 6 x 3 dx = 11 24 . keywords: integral, triple integral, re- peated integral, linear function,polynomial integrand, binomial integrand, evaluation of triple integral 002 10.0 points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay E 2 xy 2 dxdydz when E is the set of points ( x, y, z ) such that ≤ z ≤ y ≤ (4- x 2 ) 1 / 4 , and 0 ≤ x ≤ 2. 1. I = 3 2 2. I = 9 4 3. I = 2 correct 4. I = 7 4 5. I = 5 2 Explanation: As a repeated integral I = integraldisplay 2 parenleftBig integraldisplay (4- x 2 ) 1 / 4 parenleftBig integraldisplay y 2 xy 2 dz parenrightBig dy parenrightBig dx . Now integraldisplay y 2 xy 2 dz = bracketleftBig 2 xy 2 z bracketrightBig y = 2 xy 3 , while integraldisplay (4- x 2 ) 1 / 4 2 xy 3 dy = 1 2 bracketleftBig xy 4 bracketrightBig (4- x 2 ) 1 / 4 = 1 2 x (4- x 2 ) . Thus I = 1 2 integraldisplay 2 x (4- x 2 ) dx = bracketleftBig x 2- 1 8 x 4 bracketrightBig 2 . zakaria (mmz255) – HW15 – gilbert – (55485) 2 Consequently, I = 2 ....
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