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Gilbert_Hmwk16sol

# Gilbert_Hmwk16sol - moseley(cmm3869 HW16 Gilbert(56380 This...

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moseley (cmm3869) – HW16 – Gilbert – (56380) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points By changing to polar coordinates evaluate the integral I = integraldisplay integraldisplay R radicalbig x 2 + y 2 dxdy when R is the region braceleftBig ( x, y ) : 9 x 2 + y 2 25 , y 0 bracerightBig in the xy -plane. 1. I = 86 3 π 2. I = 95 3 π 3. I = 92 3 π 4. I = 89 3 π 5. I = 98 3 π correct Explanation: In polar cooordinates, R = braceleftBig ( r, θ ) : 3 r 5 , 0 θ π bracerightBig , while I = integraldisplay integraldisplay R r ( rdrdθ ) = integraldisplay integraldisplay R r 2 drdθ , since radicalbig x 2 + y 2 = r . But then I = integraldisplay 5 3 integraldisplay π 0 r 2 drdθ = π integraldisplay 5 3 r 2 dr . Consequently, I = 1 3 bracketleftBig r 3 bracketrightBig 5 3 = 98 3 π . 002 10.0 points By changing to polar coordinates evaluate the integral I = integraldisplay integraldisplay R 4 e - x 2 - y 2 dxdy when R is the region in the xy -plane bounded by the graph of x = radicalbig 4 - y 2 and the y -axis. 1. I = 2 π (1 - e - 4 ) correct 2. I = 4 π (1 - e - 4 ) 3. I = 2 π (1 - e - 2 ) 4. I = π (1 - e - 4 ) 5. I = π (1 - e - 2 ) 6. I = 4 π (1 - e - 2 ) Explanation: In polar cooordinates, R is the set braceleftBig ( r, θ ) : 0 r 2 , - π 2 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 2 0 integraldisplay π/ 2 - π/ 2 re - r 2 drdθ = 4 π integraldisplay 2 0 re - r 2 dr .

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moseley (cmm3869) – HW16 – Gilbert – (56380) 2 The presence of the term r now allows this last integral to be evaluated by the subsitution u = r 2 . For then I = 2 π bracketleftBig - e - u bracketrightBig 4 0 = 2 π (1 - e - 4 ) . 003 10.0 points The solid shown in lies inside the sphere x 2 + y 2 + z 2 = 9 and outside the cylinder x 2 + y 2 = 4 . Find the volume of the part of this solid lying above the xy -plane. 1. volume = 5 5 2. volume = 5 5 3 π 3. volume = 5 5 3 4. volume = 5 5 π 5. volume = 10 5 3 6. volume = 10 5 3 π correct Explanation: From directly overhead the solid is similar to x y 3 2 θ r In Cartesian coordinates this is the annulus R = braceleftBig ( x, y ) : 4 x 2 + y 2 9 bracerightBig . Thus the volume of the solid above the xy - plane is given by the integral V = integraldisplay integraldisplay R (9 - x 2 - y 2 ) 1 / 2 dxdy . To evaluate V we change to polar coordi- nates. Now R = braceleftBig ( r, θ ) : 2 r 3 , 0 θ 2 π bracerightBig , so that after changing coordinates the integral becomes V = integraldisplay 3 2 integraldisplay 2 π 0 radicalbig 9 - r 2 rdrdθ = 2 π integraldisplay 3 2 r radicalbig 9 - r 2 dr = π bracketleftBig - 2 3 (9 - u ) 3 / 2 bracketrightBig 9 4 , using the substitution u = r 2 . Consequently, volume = V = 10 5 3 π . 004 10.0 points The plane z = 3 and the paraboloid z = 8 - 5 x 2 - 5 y 2 enclose a solid as shown in
moseley (cmm3869) – HW16 – Gilbert – (56380) 3 z y x Use polar coordinates to determine the vol- ume of this solid.

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Gilbert_Hmwk16sol - moseley(cmm3869 HW16 Gilbert(56380 This...

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