M408m-Luecke_HW11

# M408m-Luecke_HW11 - romasko(qrr58 – Assignment 11 –...

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Unformatted text preview: romasko (qrr58) – Assignment 11 – luecke – (55035) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Hint: The volume under the graph of f(x,y) and over a region D is the integral of f(x,y) over D. The solid shown in is bounded by the paraboloid z = 4- 1 4 ( x 2 + y 2 ) and the xy-plane. Find the volume of this solid. 1. volume = 16 π 2. volume = 8 π 3. volume = 32 π correct 4. volume = 8 5. volume = 16 6. volume = 32 Explanation: The paraboloid intersects the xy-plane when z = 0, i.e. , when x 2 + y 2- 16 = 0 . Thus the solid lies below the graph of z = 4- 1 4 ( x 2 + y 2 ) and above the disk D = braceleftBig ( x, y ) : x 2 + y 2 ≤ 16 bracerightBig , so its volume is given by the integral V = integraldisplay integraldisplay D parenleftBig 4- 1 4 ( x 2 + y 2 ) parenrightBig dxdy . In polar coordinates this becomes V = 1 4 integraldisplay 4 integraldisplay 2 π r (16- r 2 ) dθdr = 1 2 π integraldisplay 4 (16 r- r 3 ) dr = 1 2 π bracketleftBig 8 r 2- r 4 4 bracketrightBig 4 . Consequently, volume = V = 32 π . 002 10.0 points Evaluate the integral I = integraldisplay integraldisplay R 4 e- x 2- y 2 dxdy when R is the region in the first quadrant of the xy-plane inside the graph of x = radicalbig 16- y 2 . 1. I = 2 π (1- e- 4 ) 2. I = 2 π (1- e- 16 ) 3. I = 4 π (1- e- 4 ) 4. I = 4 π (1- e- 16 ) romasko (qrr58) – Assignment 11 – luecke – (55035) 2 5. I = π (1- e- 4 ) 6. I = π (1- e- 16 ) correct Explanation: In polar cooordinates, R is the set braceleftBig ( r, θ ) : 0 ≤ r ≤ 4 , ≤ θ ≤ π 2 bracerightBig , while I = integraldisplay integraldisplay R 4 e- r 2 ( rdrdθ ) = integraldisplay integraldisplay R 4 re- r 2 drdθ , since x 2 + y 2 = r 2 . But then I = 4 integraldisplay 4 integraldisplay π/ 2 re- r 2 dθdr = 2 π integraldisplay 4 re- r 2 dr . The presence of the term r now allows this last integral to be evaluated by the subsitution u = r 2 . For then I = π bracketleftBig- e- u bracketrightBig 16 = π (1- e- 16 ) . 003 10.0 points Which, if any, of the following are correct?...
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M408m-Luecke_HW11 - romasko(qrr58 – Assignment 11 –...

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