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Unformatted text preview: Version 084 Homework 13 Gilbert (59825) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 10.0 points Evaluate the iterated integral I = integraldisplay 3 / 2 integraldisplay cos 3 e sin drd . 1. I = 3 parenleftBig 1 e 1 parenrightBig correct 2. I = 0 3. I = 3( e 1) 4. I = e 3 5. I = 3 e 6. I = 1 e 3 Explanation: After simple integration integraldisplay cos 3 e sin dr = bracketleftBig 3 r e sin bracketrightBig cos = 3 cos e sin . In this case, I = integraldisplay 3 / 2 3 cos e sin d = bracketleftBig 3 e sin bracketrightBig 3 / 2 . Consequently, I = 3 parenleftBig 1 e 1 parenrightBig . 002 10.0 points Find the value of the integral I = integraldisplay integraldisplay A (6 x 2 2 y 2 ) dxdy when A = braceleftBig ( x, y ) : 0 y 2 x, x 2 bracerightBig . 1. I = 163 6 2. I = 80 3 correct 3. I = 27 4. I = 161 6 5. I = 82 3 Explanation: As an iterated integral, I = integraldisplay 2 bracketleftbiggintegraldisplay 2 x (6 x 2 2 y 2 ) dy bracketrightbigg dx = integraldisplay 2 bracketleftbigg 6 x 2 y 2 3 y 3 bracketrightbigg 2 x dx = integraldisplay 2 20 3 x 3 dx . Consequently, I = 80 3 . 003 10.0 points The graph of f ( x, y ) = 4 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 Version 084 Homework 13 Gilbert (59825) 2 Find the volume of the solid under this graph over the region A . 1. Volume = 81 8 cu. units 2. Volume = 81 2 cu. units correct 3. Volume = 27 cu. units 4. Volume = 81 4 cu. units 5. Volume = 81 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 4 xy dy parenrightBig dx. Now the inner integral is equal to bracketleftBig 2 xy 2 bracketrightBig 9 x 2 = 2 x (9 x 2 ) . Thus V = 2 integraldisplay 3 x (9 x 2 ) dx = bracketleftBig 1 2 (9 x 2 ) 2 bracketrightBig 3 . Consequently, Volume = 81 2 cu. units. 004 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A (2 x + 3 y ) dxdy when A is the region enclosed by the graphs of x = 1 , x y = 1 , y = 1 . 1. I = 4 3 2. I = 5 3 3. I = 2 4. I = 1 5. I = 7 3 correct Explanation: The graphs of x = 1 , x y = 1 , y = 1 are straight lines intersecting at the points (1 , 0) , (1 , 1) , (2 , 1) . Thus the region of integration is the shaded triangular region (1 , 0) (2 , 1) (1 , 1) ( x, x 1) ( x, 1) ( x, 0) Version 084 Homework 13 Gilbert (59825) 3 so I can be written as the repeated integral I = integraldisplay 2 1 parenleftbiggintegraldisplay 1 x 1 (2 x + 3 y ) dy parenrightbigg dx , integrating first with respect to y from y = x 1 to y = 1. Now the inner integral is equal= 1....
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This note was uploaded on 05/08/2008 for the course M 408 M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

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