HW09-solutions.pdf

# HW09-solutions.pdf - young(toy68 HW09 villafuerte...

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young (toy68) – HW09 – villafuerte altu – (53615) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Evaluate the iterated integral I = integraldisplay 2 1 integraldisplay x - 1 0 (2 y - 1) dy dx . 1. I = - 1 6 correct 2. I = 1 2 3. I = 1 6 4. I = 1 3 5. I = 0 Explanation: After integration with respect to y we see that I = integraldisplay 2 1 bracketleftbig y 2 - y bracketrightbig x - 1 0 dx = integraldisplay 2 1 ( ( x - 1) 2 - ( x - 1) ) dx = bracketleftbigg 1 3 ( x - 1) 3 - 1 2 ( x - 1) 2 bracketrightbigg 2 1 . Consequently, I = 1 3 - 1 2 = - 1 6 . 002 10.0points Which, if any, of the following are correct? A. For all continuous functions f , integraldisplay 1 0 integraldisplay 2 0 f ( x, y ) dx dy = integraldisplay 2 0 integraldisplay 1 0 f ( x, y ) dx dy. B. For all continuous functions g , integraldisplay 1 0 integraldisplay y 0 g ( x, y ) dx dy = integraldisplay 1 0 integraldisplay x 0 g ( x, y ) dy dx. 1. A only 2. both of them 3. neither of them correct 4. B only Explanation: A. False: incorrect reversal of the order of integration when integrating over a rectangle. B. False: incorrect reversal of the order of integration when integrating over the upper triangle in the square [0 , 1] × [0 , 1]. 003 10.0points Evaluate the iterated integral I = integraldisplay 3 π/ 2 0 integraldisplay cos( θ ) 0 2 e sin( θ ) dr dθ . 1. I = 2 parenleftbigg 1 e - 1 parenrightbigg correct 2. I = 0 3. I = 2 e 4. I = e - 2 5. I = 2( e - 1) 6. I = 1 e - 2 Explanation: After simple integration integraldisplay cos( θ ) 0 2 e sin( θ ) dr = bracketleftBig 2 r e sin( θ ) bracketrightBig cos( θ ) 0 = 2 cos( θ ) e sin( θ ) .

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young (toy68) – HW09 – villafuerte altu – (53615) 2 In this case, I = integraldisplay 3 π/ 2 0 2 cos( θ ) e sin( θ ) = bracketleftBig 2 e sin( θ ) bracketrightBig 3 π/ 2 0 . Consequently, I = 2 parenleftbigg 1 e - 1 parenrightbigg . 004 10.0points Evaluate the double integral I = integraldisplay integraldisplay D 4 y x 2 + 1 dxdy when D is the region braceleftBig ( x, y ) : 0 x 1 , 0 y x bracerightBig in the xy -plane. 1. I = 4 2. I = 1 3. I = 2 ln(2) 4. I = 4 ln(2) 5. I = ln(2) correct 6. I = 2 Explanation: As an iterated integral, integrating first with respect to y , we see that I = integraldisplay 1 0 parenleftBigg integraldisplay x 0 4 y x 2 + 1 dy parenrightBigg dx . Now integraldisplay x 0 4 y x 2 + 1 dy = 2 bracketleftBig y 2 x 2 + 1 bracketrightBig x 0 = 2 x x 2 + 1 . In this case, I = 2 integraldisplay 1 0 x x 2 + 1 dx = bracketleftBig ln( x 2 + 1) bracketrightBig 1 0 . Consequently, I = ln(2) . keywords: 005 10.0points Evaluate the double integral I = integraldisplay integraldisplay D parenleftBig x 3 y + 3 xy 2 parenrightBig dA when D = braceleftBig ( x, y ) : 0 x 1 , - x y x bracerightBig . 1. I = 1 2. I = 6 5 3. I = 3 5 4. I = 2 5 correct 5. I = 4 5 Explanation: The double integral can be rewritten as the repeated integral I = integraldisplay 1 0 parenleftBig integraldisplay x - x braceleftBig x 3 y + 3 xy 2 bracerightBig dy parenrightBig dx , integrating first with respect to y . Now integraldisplay x - x braceleftBig x 3 y + 3 xy 2 bracerightBig dy = bracketleftBig 1 2 x 3 y 2 + xy 3 bracketrightBig x - x = 2 x 4 ,
young (toy68) – HW09 – villafuerte altu – (53615) 3 (( - x ) 2 = x 2 remember!). Consequently,

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