HW 11 solution

# HW 11 solution - gutierrez(ig3472 – HW11 – Neitzke...

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Unformatted text preview: gutierrez (ig3472) – HW11 – Neitzke – (56585) 1 This print-out should have 18 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 1- 1 integraldisplay y y 2 (3 x − 2 y ) dxdy . 1. I = − 17 15 2. I = − 11 15 3. I = − 14 15 correct 4. I = − 8 15 5. I = − 1 3 Explanation: Treating I as an iterated integral, integrat- ing first with respect to x with y fixed, we see that I = integraldisplay 1- 1 braceleftBig integraldisplay y y 2 (3 x − 2 y ) dx bracerightBig dy = integraldisplay 1- 1 bracketleftBig 3 2 x 2 − 2 xy bracketrightBig y y 2 dy . Thus I = integraldisplay 1- 1 3 2 ( y 2 − y 4 ) dy − integraldisplay 1- 1 2( y 2 − y 3 ) dy = bracketleftBig 3 2 parenleftBig y 3 3 − y 5 5 parenrightBig − 2 parenleftBig y 3 3 − y 4 4 parenrightBigbracketrightBig 1- 1 . Consequently, I = − 14 15 . 002 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D 6 x 2 y 3 dxdy when D = braceleftBig ( x, y ) : 0 ≤ y ≤ 1 , − y ≤ x ≤ y bracerightBig . 1. I = 0 2. I = 3 7 3. I = 1 2 4. I = 2 3 5. I = 4 7 correct Explanation: The double integral can be rewritten as the repeated integral I = integraldisplay 1 parenleftBig integraldisplay y- y 6 x 2 y 3 dx parenrightBig dy , integrating first with respect to x . Now integraldisplay y- y 6 x 2 y 3 dy = bracketleftBig 2 x 3 y 3 bracketrightBig y- y = 4 y 6 . Consequently, I = 4 integraldisplay 1 y 6 dy = 4 7 . 003 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D 8 y x 2 + 1 dxdy when D is the region braceleftBig ( x, y ) : 2 ≤ x ≤ 6 , ≤ y ≤ √ x bracerightBig in the xy-plane. gutierrez (ig3472) – HW11 – Neitzke – (56585) 2 1. I = 8 ln 7 2. I = 2 ln 37 5 correct 3. I = 4 ln 37 5 4. I = 2 ln 7 5. I = 4 ln 7 6. I = 8 ln 37 5 Explanation: As an iterated integral, integrating first with respect to y , we see that I = integraldisplay 6 2 braceleftBig integraldisplay √ x 8 y x 2 + 1 dy bracerightBig dx . Now integraldisplay √ x y x 2 + 1 dy = 1 2 bracketleftBig y 2 x 2 + 1 bracketrightBig √ x = x 2( x 2 + 1) . In this case, I = 4 integraldisplay 6 2 x x 2 + 1 dx = 2 bracketleftBig ln( x 2 + 1) bracketrightBig 6 2 . Consequently, I = 2 ln 37 5 . 004 10.0 points The graph of f ( x, y ) = 7 xy over the bounded region A in the first quad- rant enclosed by y = radicalbig 16 − x 2 and the x, y-axes is the surface Find the volume of the solid under this graph over the region A . 1. Volume = 448 3 cu. units 2. Volume = 112 cu. units 3. Volume = 224 cu. units correct 4. Volume = 448 cu. units 5. Volume = 56 cu. units Explanation: The volume of the solid under the graph of f is given by the double integral V = integraldisplay integraldisplay A f ( x, y ) dxdy, which in turn can be written as the repeated integral integraldisplay 4 parenleftBig integraldisplay √ 16- x 2 7 xy dy parenrightBig dx....
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HW 11 solution - gutierrez(ig3472 – HW11 – Neitzke...

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