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Unformatted text preview: Rehman (aar638) – HW10 – sachse – (56620) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph of the function z = f ( x, y ) = 4 x is the plane shown in z 4 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 ≤ 4 , ≤ y ≤ 2 bracerightBig in the xyplane by first identifying it as the volume of a solid below the graph of f . 1. I = 16 cu. units correct 2. I = 18 cu. units 3. I = 17 cu. units 4. I = 14 cu. units 5. I = 15 cu. units Explanation: The double integral I = integraldisplay integraldisplay A f ( x, y ) dxdy is the volume of the solid below the graph of f having the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 4 , ≤ y ≤ 2 bracerightBig for its base. Thus the solid is the wedge z 4 4 x y (4 , 2) and so its volume is the area of triangular face multiplied by the thickness of the wedge. Consequently, I = 16 cu. units . keywords: double integral, linear function, volume under graph, volume, rectangular re gion, prism, triangle 002 10.0 points Determine the value of the iterated integral I = integraldisplay 2 braceleftBig integraldisplay 2 1 (5 + 2 xy ) dx bracerightBig dy . 1. I = 20 2. I = 14 3. I = 22 4. I = 16 correct 5. I = 18 Explanation: Rehman (aar638) – HW10 – sachse – (56620) 2 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 2 braceleftBig 5 + 3 y bracerightBig dy = bracketleftBig 5 y + 3 2 y 2 bracketrightBig 2 . Consequently, I = braceleftBig 10 + 6 bracerightBig = 16 . keywords: 003 10.0 points Evaluate the iterated integral I = integraldisplay 3 1 braceleftBig integraldisplay 3 1 ( x + y ) 2 dx bracerightBig dy . 1. I = 1 2 ln 2 2. I = 2 ln2 3. I = ln 2 correct 4. I = 1 2 ln 6 5 5. I = 2 ln 6 5 6. I = ln 6 5 Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that integraldisplay 3 1 ( x + y ) 2 dx = bracketleftBig 1 x + y bracketrightBig 3 = braceleftBig 1 y 1 3 + y bracerightBig . In this case I = integraldisplay 3 1 braceleftBig 1 y 1 3 + y bracerightBig dy = bracketleftBig ln y ln(3 + y ) bracketrightBig 3 1 . Consequently, I = ln parenleftBig (3)(1 + 3) (3 + 3) parenrightBig = ln 2 . 004 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A ( 3 x 2 y 3 x 4 ) dxdy when A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 2 bracerightBig . 1. I = 18 5 correct 2. I = 13 5 3. I = 3 5 4. I = 23 5 5. I = 8 5 Explanation: Since A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 2 bracerightBig is a rectangle with sides parallel to the coor dinate axes, the value of I can be found by interpreting the double integral as the iter ated integral I = integraldisplay 2 bracketleftbiggintegraldisplay...
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV
 Calculus

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