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Unformatted text preview: kim (tk5895) HW10 Henry (54974) 1 This printout should have 16 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 ) = 8 x is the plane shown in z 8 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 3 , y 2 bracerightBig in the xyplane by first identifying it as the volume of a solid below the graph of f . 1. I = 39 cu. units correct 2. I = 37 cu. units 3. I = 40 cu. units 4. I = 38 cu. units 5. I = 41 cu. units Explanation: The double integral I = integraldisplay integraldisplay A f ( x, y ) dxxdy is the volume of the solid below the graph of f having the rectangle A = braceleftBig ( x, y ) : 0 x 3 , y 2 bracerightBig for its base. Thus the solid is the wedge z 3 8 x y (3 , 2) and so its volume is the area of trapezoidal face multiplied by the thickness of the wedge. Consequently, I = 39 cu. units . keywords: 002 10.0 points Evaluate the integral I = integraldisplay 1 integraldisplay 2 1 (3 x x 2 y ) dydx . 1. I = 1 2 2. I = 3 2 3. I = 1 correct 4. I = 0 5. I = 1 2 Explanation: kim (tk5895) HW10 Henry (54974) 2 The integral can be written in iterated form I = integraldisplay 1 parenleftBig integraldisplay 2 1 (3 x x 2 y ) dy parenrightBig dx . Now integraldisplay 2 1 (3 x x 2 y ) dy = bracketleftBig 3 xy 1 2 x 2 y 2 bracketrightBig 2 1 = 3 x 3 2 x 2 . But then I = integraldisplay 1 (3 x 3 2 x 2 ) dx = bracketleftBig 3 2 x 2 1 2 x 3 bracketrightBig 1 . Consequently, I = 1 . keywords: definite integral, iterated integral, polynomial function, 003 10.0 points Evaluate the iterated integral I = integraldisplay 3 1 braceleftBig integraldisplay 3 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . 1. I = 3 ln4 2. I = 8 ln3 correct 3. I = 3 ln8 4. I = 4 ln3 5. I = 4 ln8 6. I = 8 ln4 Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 3 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 3 1 = (ln 3) x + 4 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 3 1 bracketleftbigg (ln3) x + 4 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbiggparenleftbigg x 2 2 parenrightbigg ln3 + 4 ln x bracketrightbigg 3 1 . Consequently, I = 8 ln 3 . 004 10.0 points Evaluate the iterated integral I = integraldisplay ln 5 parenleftBigg integraldisplay ln 4 e 2 x y dx parenrightBigg dy . 1. I = 5 2. I = 7 3. I = 8 4. I = 4 5. I = 6 correct 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 ....
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 Fall '08
 Cepparo
 Calculus

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