solution9_pdf9 - Horstman (mdh995) – HW09 – Radin –...

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Unformatted text preview: Horstman (mdh995) – HW09 – Radin – (58505) 1 This print-out should have 15 questions. Multiple-choice 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 correct 3. I = e- 3- e- 2 + e- 1 + 1 4. I = e- 3 + e- 2- e- 1 + 1 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 = ln20 2. I = 2 ln22 3. I = 20 4. I = 2 ln20 5. I = 22 6. I = ln22 correct 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 = ln22 . Horstman (mdh995) – HW09 – Radin – (58505) 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 = 5 6 π 2. I = 3 2 π 3. I = π 4. I = 7 6 π 5. I = 4 3 π correct 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 x sin( 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 11/30/2008 for the course M 58510 taught by Professor Radin during the Fall '08 term at University of Texas.

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solution9_pdf9 - Horstman (mdh995) – HW09 – Radin –...

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