EXAM03-solutions-1

# EXAM03-solutions-1 - Version 012 – EXAM03 – gilbert –...

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Unformatted text preview: Version 012 – EXAM03 – gilbert – (55485) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The solid E shown in x y z is bounded by the circular cylinders x 2 + y 2 = 1 , y 2 + z 2 = 1 , and the coordinate planes. Write the triple integral I = integraldisplay integraldisplay integraldisplay E f ( x, y, z ) dV as a repeated integral, integrating first with respect to z , then x , and finally y . 1. integraldisplay 1 integraldisplay 1 − y 2 integraldisplay 1 − x 2 f ( x, y, z ) dz dx dy 2. integraldisplay 1 integraldisplay √ 1 − y 2 integraldisplay √ 1 − y 2 f ( x, y, z ) dz dx dy correct 3. integraldisplay 1 integraldisplay √ 1 − y 2 integraldisplay √ 1 − x 2 f ( x, y, z ) dz dx dy 4. integraldisplay 1 integraldisplay 1 − y 2 integraldisplay 1 − y 2 f ( x, y, z ) dz dx dy Explanation: Because E is bounded by the coordinate planes x, y, z = 0, the graph of the solid shows that E consists of all points ( x, y, z ) in 3-space satisfying ≤ z ≤ radicalbig 1 − y 2 , ≤ x ≤ radicalbig 1 − y 2 , and 0 ≤ y ≤ 1 . Consequently, as a repeated integral, I be- comes integraldisplay 1 parenleftBig integraldisplay √ 1 − y 2 parenleftBig integraldisplay √ 1 − y 2 f ( x, y, z ) dz parenrightBig dx parenrightBig dy . keywords: clicker 002 10.0 points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay E 6 xy 2 dxdydz when E is the set of points ( x, y, z ) such that ≤ z ≤ y ≤ (1 − x 2 ) 1 / 4 , and 0 ≤ x ≤ 1. 1. I = 1 8 2. I = 9 8 3. I = 3 8 correct 4. I = 5 8 5. I = 7 8 Explanation: As a repeated integral I = integraldisplay 1 parenleftBig integraldisplay (1 − x 2 ) 1 / 4 parenleftBig integraldisplay y 6 xy 2 dz parenrightBig dy parenrightBig dx . Version 012 – EXAM03 – gilbert – (55485) 2 Now integraldisplay y 6 xy 2 dz = bracketleftBig 6 xy 2 z bracketrightBig y = 6 xy 3 , while integraldisplay (1 − x 2 ) 1 / 4 6 xy 3 dy = 3 2 bracketleftBig xy 4 bracketrightBig (1 − x 2 ) 1 / 4 = 3 2 x ( − x 2 ) . Thus I = 3 2 integraldisplay 1 x ( − x 2 ) dx = bracketleftBig 3 4 x 2 − 3 8 x 4 bracketrightBig 1 . Consequently, I = 3 8 . keywords: integral, triple integral, repeated integral, linear function, limits of integration, general region, polynomial integrand, mono- mial integrand, evaluation of triple integral 003 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A 4 xye y dxdy when A = braceleftBig ( x, y ) : 0 ≤ x ≤ 2 , ≤ y ≤ 1 bracerightBig . 1. I = 9 2. I = 10 3. I = 6 4. I = 8 correct 5. I = 7 Explanation: The double integral can be rewritten as the repeated integral I = integraldisplay 1 parenleftBig integraldisplay 2 xye y dx parenrightBig dy , integrating first with respect to x . Now integraldisplay 2 4 xye y dx = 2 bracketleftBig x 2 ye y bracketrightBig 2 = 8 ye y ....
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## This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas.

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EXAM03-solutions-1 - Version 012 – EXAM03 – gilbert –...

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