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HW09-solutions

# HW09-solutions - wei(jw35975 – HW09 – kalahurka...

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Unformatted text preview: wei (jw35975) – HW09 – kalahurka – (55230) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The solid shown in lies below the graph of z = f ( x, y ) = 1 + x 2- y 2 above the rectangle- 1 ≤ x ≤ 1 ,- 1 ≤ y ≤ 1 in the xy-plane. Determine the volume of this solid. 1. Volume = 4 3 2. Volume = 1 3 3. Volume = 8 3 4. Volume = 2 3 5. Volume = 1 6. Volume = 4 correct Explanation: The volume of the solid below the graph of z = f ( x, y ) and above the region D in the xy-plane is given by the double integral I = integraldisplay integraldisplay D f ( x, y ) dxdy . So when z = f ( x, y ) = 1 + x 2- y 2 and D is the rectangle- 1 ≤ x ≤ 1 ,- 1 ≤ y ≤ 1 , the volume is given by I = integraldisplay integraldisplay D (1 + x 2- y 2 ) dxdy , which as a repeated integral becomes I = integraldisplay 1- 1 parenleftbiggintegraldisplay 1- 1 (1 + x 2- y 2 ) dx parenrightbigg dy = integraldisplay 1- 1 bracketleftbigg x + 1 3 x 3- y 2 x bracketrightbigg 1- 1 dy . Consequently, the solid has volume = integraldisplay 1- 1 parenleftBig 8 3- 2 y 2 parenrightBig dy = 4 . 002 10.0 points Evaluate the integral I = integraldisplay 1 integraldisplay 2 1 (4 x + 3 x 2 y ) dydx . 1. I = 3 2. I = 4 3. I = 9 2 4. I = 5 5. I = 7 2 correct wei (jw35975) – HW09 – kalahurka – (55230) 2 Explanation: The integral can be written in iterated form I = integraldisplay 1 parenleftBig integraldisplay 2 1 (4 x + 3 x 2 y ) dy parenrightBig dx . Now integraldisplay 2 1 (4 x + 3 x 2 y ) dy = bracketleftBig 4 xy + 3 2 x 2 y 2 bracketrightBig 2 1 = 4 x + 9 2 x 2 . But then I = integraldisplay 1 (4 x + 9 2 x 2 ) dx = bracketleftBig 2 x 2 + 3 2 x 3 bracketrightBig 1 . Consequently, I = 7 2 . keywords: definite integral, iterated integral, polynomial function, 003 10.0 points Evaluate the iterated integral I = integraldisplay 2 1 braceleftBig integraldisplay 2 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . 1. I = 3 2 ln 3 2. I = 2 ln 3 2 3. I = 3 ln 3 2 4. I = 2 ln3 5. I = 3 2 ln 2 6. I = 3 ln2 correct Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 2 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 2 1 = (ln 2) x + 3 2 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 2 1 bracketleftbigg (ln2) x + 3 2 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbiggparenleftbigg x 2 2 parenrightbigg ln2 + 3 2 ln x bracketrightbigg 2 1 . Consequently, I = 3 ln 2 . 004 10.0 points Evaluate the iterated integral I = integraldisplay ln 4 parenleftBigg integraldisplay ln 5 e 2 x- y dx parenrightBigg dy ....
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HW09-solutions - wei(jw35975 – HW09 – kalahurka...

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