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Unformatted text preview: rhodes (ajr2283) – 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 3 dxdy with A = braceleftBig ( x, y ) : 3 ≤ x ≤ 6 , 2 ≤ y ≤ 6 bracerightBig by first identifying it as the volume of a solid. 1. I = 40 2. I = 36 correct 3. I = 38 4. I = 42 5. I = 44 Explanation: The value of I is the volume of the solid below the graph of z = f ( x, y ) = 3 and above the region A = braceleftBig ( x, y ) : 3 ≤ x ≤ 6 , 2 ≤ y ≤ 6 bracerightBig . Since A is a rectangle, this solid is a box with base A and height 3. Its volume, therefore, is given by length × width × height = (6 3) × (6 2) × 3 . Consequently, I = 36 . 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 ≤ 6 } in the xyplane by first identifying it as the volume of a solid. 1. I = 25 2. I = 26 3. I = 27 correct 4. I = 28 5. I = 29 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 ≤ 6 } as its base. Thus the solid is the wedge z 5 2 x y (5 , 6) having a triangular face of height 3 and base 3. Since the wedge has length 6, the solid thus has volume = 27 . keywords: rhodes (ajr2283) – HW15 – Gilbert – (56380) 2 003 10.0 points Determine the value of the iterated integral I = integraldisplay 3 braceleftBig integraldisplay 2 1 (5 + 2 xy ) dx bracerightBig dy . 1. I = 49 2 2. I = 61 2 3. I = 57 2 correct 4. I = 53 2 5. I = 65 2 Explanation: Integrating with respect to x and holding y fixed, we see that integraldisplay 2 1 (5 + 2 xy ) dx = bracketleftBig 5 x + x 2 y bracketrightBig x =2 x =1 . Thus I = integraldisplay 3 braceleftBig 5 + 3 y bracerightBig dy = bracketleftBig 5 y + 3 2 y 2 bracketrightBig 3 . Consequently, I = braceleftBig 15 + 27 2 bracerightBig = 57 2 . keywords: 004 10.0 points Evaluate the iterated integral I = integraldisplay ln 4 parenleftBigg integraldisplay ln 5 e 2 x y dx parenrightBigg dy . 1. I = 7 2. I = 6 3. I = 5 4. I = 8 5. I = 9 correct Explanation: Integrating with respect to x with y fixed, we see that integraldisplay ln 5 e 2 x y dx = 1 2 bracketleftBig e 2 x y bracketrightBig ln 5 = 1 2 parenleftBig e 2 ln 5 y e y parenrightBig = parenleftBig 5 2 1 2 parenrightBig e y . Thus I = 12 integraldisplay ln 4 e y dy = 12 bracketleftBig e y bracketrightBig ln 4 = 12 parenleftBig e ln 4 1 parenrightBig ....
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This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas.
 Spring '07
 TextbookAnswers
 Multivariable Calculus

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