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Unformatted text preview: momin (rrm497) – Homework 09 – cheng – (58520) 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 Evaluate the double integral I = integraldisplay 3 2 integraldisplay 2 e x y dxdy . 1. I = e 3 e 2 + e 1 + 1 2. I = e 3 e 2 e 1 1 3. I = e 3 + e 2 e 1 + 1 4. I = e 3 e 2 e 1 + 1 correct Explanation: After integration with respect to x , I = integraldisplay 3 2 bracketleftbig e x y bracketrightbig 2 dy = integraldisplay 3 2 ( e 2 y e y ) dy . But then, after integrating next with respect to y we see that I = bracketleftbig e 2 y + e y bracketrightbig 3 2 = e 1 + e 3 ( 1 + e 2 ) . Consequently, I = e 3 e 2 e 1 + 1 . 002 10.0 points Determine the value, I , of the integral of the function f ( x, y ) = 5 x + y 1 + 20 y + y 2 over the rectangle A = braceleftBig ( x, y ) : 1 ≤ x ≤ 3 , ≤ y ≤ 1 bracerightBig . 1. I = ln 20 2. I = 20 3. I = 22 4. I = 2 ln22 5. I = ln 22 correct 6. I = 2 ln20 Explanation: The integral of f over A can be written as the iterated integral I = integraldisplay 1 parenleftbiggintegraldisplay 3 1 5 x + y 1 + 20 y + y 2 dx parenrightbigg dy, integrating first with respect to x . Now integraldisplay 3 1 5 x + y 1 + 20 y + y 2 dx = bracketleftBig 5 2 x 2 + xy 1 + 20 y + y 2 bracketrightBig 3 1 = 20 + 2 y 1 + 20 y + y 2 . Thus I = integraldisplay 1 20 + 2 y 1 + 20 y + y 2 dy . To evaluate this last integral we use the sub stitution u = 1 + 20 y + y 2 . For then du dx = 20 + 2 y , while y = 0 = ⇒ u = 1 y = 1 = ⇒ u = 22 . Consequently, I = integraldisplay 22 1 1 u du = bracketleftBig ln u bracketrightBig 22 1 = ln 22 . momin (rrm497) – Homework 09 – cheng – (58520) 2 003 10.0 points Evaluate the double integral I = integraldisplay integraldisplay A 5 + x 2 1 + y 2 dxdy when A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 1 bracerightBig . 1. I = 4 3 π correct 2. I = 7 6 π 3. I = 2 3 π 4. I = 5 6 π 5. I = π Explanation: Since A = braceleftBig ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ 1 bracerightBig is a rectangle with sides parallel to the coor dinate axes, the double integral can be repre sented as the iterated integral I = integraldisplay 1 parenleftbiggintegraldisplay 1 5 + x 2 1 + y 2 dx parenrightbigg dy . Now integraldisplay 1 5 + x 2 1 + y 2 dx = 1 1 + y 2 bracketleftBig 5 x + 1 3 x 3 bracketrightBig 1 . Thus I = 16 3 integraldisplay 1 1 1 + y 2 dy = 16 3 bracketleftBig tan 1 y bracketrightBig 1 . Consequently, I = 4 3 π . 004 10.0 points Calculate the value of the double integral I = integraldisplay integraldisplay A 4 x cos( x + y ) dxdy when A is the rectangle braceleftBig ( x, y ) : 0 ≤ x ≤ π 4 , ≤ y ≤ π 4 bracerightBig ....
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This note was uploaded on 12/11/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin
 Calculus

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