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Unformatted text preview: romasko (qrr58) Assignment 10 luecke (55035) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The graph of f ( x, y ) = 5 xy over the bounded region A in the first quad rant enclosed by y = radicalbig 9 x 2 and the x, yaxes is the surface Find the volume of the solid under this graph over the region A . 1. Volume = 405 16 cu. units 2. Volume = 405 32 cu. units 3. Volume = 405 8 cu. units correct 4. Volume = 135 4 cu. units 5. Volume = 405 4 cu. units Explanation: The volume of the solid under the graph of f is given by the double integral V = integraldisplay integraldisplay A f ( x, y ) dxdy, which in turn can be written as the repeated integral integraldisplay 3 parenleftBig integraldisplay 9 x 2 5 xy dy parenrightBig dx. Now the inner integral is equal to bracketleftBig 5 2 xy 2 bracketrightBig 9 x 2 = 5 2 x (9 x 2 ) . Thus V = 5 2 integraldisplay 3 x (9 x 2 ) dx = bracketleftBig 5 8 (9 x 2 ) 2 bracketrightBig 3 . Consequently, Volume = 405 8 cu. units . 002 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D (5 x 4 y ) dydx where D is the region bounded by the circle with center at the origin and radius 1. 1. I = 3 2. I = 0 correct 3. I = 1 4. I = 2 5. I = 1 Explanation: If we want to integrate first with respect to y , the double integral becomes I = integraldisplay 1 1 integraldisplay 1 x 2 1 x 2 (5 x 4 y ) dydx . But then I = integraldisplay 1 1 bracketleftBig 5 xy 2 y 2 bracketrightBig 1 x 2 1 x 2 = 10 integraldisplay 1 1 x radicalbig 1 x 2 dx . romasko (qrr58) Assignment 10 luecke (55035) 2 Using substitution we thus see that I = 10 3 bracketleftBig ( 1 x 2 ) 3 / 2 bracketrightBig 1 1 . Consequently, I = 0 . 003 10.0 points Hint: You may need to look up the integral of lnx. The graph of f ( x, y ) = 1 x + y + 3 over the triangular region A enclosed by the graphs of x = 1 , x + y = 2 , y + 3 = 0 is the surface Find the volume V of the solid under this graph and over the region A . 1. V = 4 ln 5 correct 2. V = 5 ln 5 3. V = 5 + ln 5 4. V = 4 + ln 5 5. V = 6 ln 5 Explanation: As the region of integration is given by (1 , 3) (5 , 3) (1 , 1) ( x, 3) ( x, 2 x ) (not drawn to scale) the double integral can be written as the repeated integral I = integraldisplay 5 1 parenleftbiggintegraldisplay 2 x 3 1 x + y + 3 dy parenrightbigg dx , integrating first with respect to y from y = 3 to y = 2 x . Now the inner integral is equal to bracketleftBig ln( x + y + 3) bracketrightBig 2 x 3 = ln 5 ln x. Thus I = integraldisplay 5 1 braceleftBig ln 5 ln x bracerightBig dx = 4 ln5 bracketleftBig x ln x x bracketrightBig 5 1 ....
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This note was uploaded on 06/15/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.
 Fall '07
 Gilbert

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