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Unformatted text preview: moseley (cmm3869) HW15 Gilbert (56380) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A 2 dxdy with A = braceleftBig ( x, y ) : 5 x 8 , 6 y 8 bracerightBig by first identifying it as the volume of a solid. 1. I = 4 2. I = 12 correct 3. I = 8 4. I = 10 5. I = 6 Explanation: The value of I is the volume of the solid below the graph of z = f ( x, y ) = 2 and above the region A = braceleftBig ( x, y ) : 5 x 8 , 6 y 8 bracerightBig . Since A is a rectangle, this solid is a box with base A and height 2. Its volume, therefore, is given by length width height = (8 5) (8 6) 2 . Consequently, I = 12 . keywords: volume, double integral, rectangu lar region, rectangular solid 002 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay R (5 x ) dxdy over the region R = { ( x, y ) : 2 x 5 , y 4 } in the xyplane by first identifying it as the volume of a solid. 1. I = 15 2. I = 18 correct 3. I = 16 4. I = 17 5. I = 14 Explanation: The double integral I is the volume of the solid below the graph of z = 5 x having the rectangle R = { ( x, y ) : 2 x 5 , y 4 } as its base. Thus the solid is the wedge z 5 2 x y (5 , 4) having a triangular face of height 3 and base 3. Since the wedge has length 4, the solid thus has volume = 18 . keywords: moseley (cmm3869) HW15 Gilbert (56380) 2 003 10.0 points Determine the value of the iterated integral I = integraldisplay 3 braceleftBig integraldisplay 3 1 (7 + 2 xy ) dx bracerightBig dy . 1. I = 84 2. I = 76 3. I = 82 4. I = 80 5. I = 78 correct Explanation: Integrating with respect to x and holding y fixed, we see that integraldisplay 3 1 (7 + 2 xy ) dx = bracketleftBig 7 x + x 2 y bracketrightBig x =3 x =1 . Thus I = integraldisplay 3 braceleftBig 14 + 8 y bracerightBig dy = bracketleftBig 14 y + 4 y 2 bracketrightBig 3 . Consequently, I = braceleftBig 42 + 36 bracerightBig = 78 . keywords: 004 10.0 points Evaluate the iterated integral I = integraldisplay ln 3 parenleftBigg integraldisplay ln 4 e 2 x y dx parenrightBigg dy . 1. I = 5 correct 2. I = 7 3. I = 3 4. I = 4 5. I = 6 Explanation: Integrating with respect to x with y fixed, we see that integraldisplay ln 4 e 2 x y dx = 1 2 bracketleftBig e 2 x y bracketrightBig ln 4 = 1 2 parenleftBig e 2 ln 4 y e y parenrightBig = parenleftBig 4 2 1 2 parenrightBig e y . Thus I = 15 2 integraldisplay ln 3 e y dy = 15 2 bracketleftBig e y bracketrightBig ln 3 = 15 2 parenleftBig e ln 3 1 parenrightBig ....
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This note was uploaded on 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler
 Multivariable Calculus

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