014 100points by changing to polar coordinates

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014 10.0points 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 ) : 16 x 2 + y 2 25 , y 0 bracerightBig in the xy -plane. 1. I = 64 3 π 2. I = 67 3 π 3. I = 70 3 π 4. I = 61 3 π correct 5. I = 73 3 π Explanation: In polar cooordinates, R = braceleftBig ( r, θ ) : 4 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 4 parenleftbiggintegraldisplay π 0 r 2 parenrightbigg dr = π integraldisplay 5 4 r 2 dr . Consequently, I = 1 3 bracketleftBig r 3 bracketrightBig 5 4 = 61 3 π . 015 10.0points Evaluate the triple integral I = integraldisplay 1 0 integraldisplay x 0 integraldisplay x + y 0 ( x - 3 y ) dzdydx . 1. I = 0
jiang (xj842) – HW13 – allen – (54060) 11 2. I = - 1 6 3. I = - 1 12 4. I = 1 12 5. I = - 1 4 correct Explanation: As a repeated integral, I = integraldisplay 1 0 parenleftBig integraldisplay x 0 parenleftBig integraldisplay x + y 0 ( x - 3 y ) dz parenrightBig dy parenrightBig dx . Now integraldisplay x + y 0 ( x - 3 y ) dz = bracketleftBig ( x - 3 y ) z bracketrightBig x + y 0 = ( x - 3 y )( x + y ) = x 2 - 2 xy - 3 y 2 , while integraldisplay x 0 ( x 2 - 2 xy - 3 y 2 ) dy = bracketleftBig x 2 y - xy 2 - y 3 bracketrightBig x 0 = - 1 x 3 . Consequently, I = integraldisplay 1 0 - 1 x 3 dx = - 1 4 . keywords: integral, triple integral, re- peated integral, linear function,polynomial integrand, binomial integrand, evaluation of triple integral 016 10.0points Evaluate the triple integral I = integraldisplay integraldisplay integraldisplay B x e - 2 y +3 z dV where B is the rectangular box in 3-space shown in x y z having one corner at the origin and three adjacent faces in the coordinate planes. 1. I = 3 4 ( e 6 + e - 6 - 2) 2. I = 4 3 ( e 6 + e - 6 ) 3. I = 3 4 ( e 6 + e - 6 ) 4. I = 4 3 ( e 6 + e - 6 - 1) 5. I = 4 3 ( e 6 + e - 6 - 2) correct 6. I = 3 4 ( e 6 + e - 6 - 1) Explanation: Since B consists of all points ( x, y, z ) in 3-space such that 0 x 4 , 0 y 3 , 0 z 2 , we see that I can be written as a repeated integral I = integraldisplay 2 0 parenleftBig integraldisplay 3 0 parenleftBig integraldisplay 4 0 x e - 2 y +3 z dx parenrightBig dy parenrightBig dz . Now integraldisplay 4 0 x e - 2 y +3 z dx = 1 2 bracketleftBig x 2 e - 2 y +3 z bracketrightBig 4 0 = 8 e - 2 y +3 z ,
jiang (xj842) – HW13 – allen – (54060) 12 while integraldisplay 3 0 8 e - 2 y +3 z dy = bracketleftBig - 4 e - 2 y +3 z bracketrightBig 3 0 = - 4( e - 6+3 z - e 3 z ) . Thus I = - 4 integraldisplay 2 0 ( e - 6+3 z - e 3 z ) dz = - 4 3 bracketleftBig e - 6+3 z - e 3 z bracketrightBig 2 0 . Consequently, I = 4 3 ( e 6 + e - 6 - 2) . keywords: integral, triple integral, repeated integral, linear function,integral over a speci- fied volume, limits of integration, polynomial integrand, exponential integrand, evaluation of triple integral

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