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Unformatted text preview: suleimenov (bs26835) – HW 12 – rusin – (55565) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This will be the last homework. It is due Sunday, May 8. I did not include any ques tions from section 16.9 (Change of Variables) because I am not sure we will have time to dis cuss that very use topic :( . ***Remember, the final exam is May 17 from 2 till 5 pm.*** 001 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A 3 dxdy with A = braceleftBig ( x, y ) : 2 ≤ x ≤ 8 , 6 ≤ y ≤ 8 bracerightBig by first identifying it as the volume of a solid. 1. I = 36 correct 2. I = 40 3. I = 42 4. I = 34 5. I = 38 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 ) : 2 ≤ x ≤ 8 , 6 ≤ y ≤ 8 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 = (8 2) × (8 6) × 3 . Consequently, I = 36 . keywords: volume, double integral, rectangu lar region, rectangular solid 002 10.0 points The graph of the function z = f ( x, y ) = 7 x is the plane shown in z 7 x y Determine the value of the double integral I = integraldisplay integraldisplay A f ( x, y ) dxdy over the region A = braceleftBig ( x, y ) : 0 ≤ x ≤ 5 , ≤ y ≤ 4 bracerightBig in the xyplane by first identifying it as the volume of a solid below the graph of f . 1. I = 87 cu. units 2. I = 90 cu. units correct 3. I = 91 cu. units 4. I = 88 cu. units 5. I = 89 cu. units Explanation: The double integral I = integraldisplay integraldisplay A f ( x, y ) dxxdy suleimenov (bs26835) – HW 12 – rusin – (55565) 2 is the volume of the solid below the graph of f having the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 5 , ≤ y ≤ 4 bracerightBig for its base. Thus the solid is the wedge z 5 7 x y (5 , 4) and so its volume is the area of trapezoidal face multiplied by the thickness of the wedge. Consequently, I = 90 cu. units . keywords: 003 10.0 points Determine the value of the iterated integral I = integraldisplay 1 braceleftBig integraldisplay 2 1 (3 + 2 xy ) dx bracerightBig dy . 1. I = 17 2 2. I = 1 2 3. I = 5 2 4. I = 13 2 5. I = 9 2 correct Explanation: Integrating with respect to x and holding y fixed, we see that integraldisplay 2 1 (3 + 2 xy ) dx = bracketleftBig 3 x + x 2 y bracketrightBig x =2 x =1 . Thus I = integraldisplay 1 braceleftBig 3 + 3 y bracerightBig dy = bracketleftBig 3 y + 3 2 y 2 bracketrightBig 1 . Consequently, I = braceleftBig 3 + 3 2 bracerightBig = 9 2 . keywords: 004 10.0 points Evaluate the iterated integral I = integraldisplay 3 1 braceleftBig integraldisplay 2 2 ( x + y ) 2 dx bracerightBig dy ....
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This note was uploaded on 09/15/2011 for the course M 55565 taught by Professor Rusin during the Spring '11 term at University of Texas.
 Spring '11
 RUSIN

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