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cal11 - choi(kc24547 HW11 BERG(56525 This print-out should...

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choi (kc24547) – HW11 – BERG – (56525) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the double integral I = integraldisplay 1 - 1 integraldisplay y y 2 ( x + 3 y ) dxdy . 1. I = 23 15 2. I = 32 15 correct 3. I = 7 3 4. I = 29 15 5. I = 26 15 Explanation: Treating I as an iterated integral, integrat- ing first with respect to x with y fixed, we see that I = integraldisplay 1 - 1 braceleftBig integraldisplay y y 2 ( x + 3 y ) dx bracerightBig dy = integraldisplay 1 - 1 bracketleftBig 1 2 x 2 + 3 xy bracketrightBig y y 2 dy . Thus I = integraldisplay 1 - 1 1 2 ( y 2 y 4 ) dy + integraldisplay 1 - 1 3( y 2 y 3 ) dy = bracketleftBig 1 2 parenleftBig y 3 3 y 5 5 parenrightBig +3 parenleftBig y 3 3 y 4 4 parenrightBigbracketrightBig 1 - 1 . Consequently, I = 32 15 . 002 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D 8 x 2 y 3 dxdy when D = braceleftBig ( x, y ) : 0 y 1 , y x y bracerightBig . 1. I = 16 21 correct 2. I = 2 3 3. I = 8 9 4. I = 4 7 5. I = 0 Explanation: The double integral can be rewritten as the repeated integral I = integraldisplay 1 0 parenleftBig integraldisplay y - y 8 x 2 y 3 dx parenrightBig dy , integrating first with respect to x . Now integraldisplay y - y 8 x 2 y 3 dy = bracketleftBig 8 3 x 3 y 3 bracketrightBig y - y = 16 3 y 6 . Consequently, I = 16 3 integraldisplay 1 0 y 6 dy = 16 21 . 003 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D 2 y x 2 + 1 dxdy when D is the region braceleftBig ( x, y ) : 1 x 6 , 0 y x bracerightBig in the xy -plane.
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choi (kc24547) – HW11 – BERG – (56525) 2 1. I = ln 37 2 2. I = 1 2 ln 37 2 correct 3. I = 2 ln 37 2 4. I = ln 35 2 5. I = 1 2 ln 35 2 6. I = 2 ln 35 2 Explanation: As an iterated integral, integrating first with respect to y , we see that I = integraldisplay 6 1 braceleftBig integraldisplay x 0 2 y x 2 + 1 dy bracerightBig dx . Now integraldisplay x 0 y x 2 + 1 dy = 1 2 bracketleftBig y 2 x 2 + 1 bracketrightBig x 0 = x 2( x 2 + 1) . In this case, I = integraldisplay 6 1 x x 2 + 1 dx = 1 2 bracketleftBig ln( x 2 + 1) bracketrightBig 6 1 . Consequently, I = 1 2 ln 37 2 . 004 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 1 x 2 and the x, y -axes is the surface Find the volume of the solid under this graph over the region A . 1. Volume = 1 8 cu. units 2. Volume = 1 4 cu. units 3. Volume = 1 cu. units 4. Volume = 1 3 cu. units 5. Volume = 1 2 cu. units correct 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 1 0 parenleftBig integraldisplay 1 - x 2 0 4 xy dy parenrightBig dx. Now the inner integral is equal to bracketleftBig 2 xy 2 bracketrightBig 1 - x 2 0 = 2 x (1 x 2 ) .
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