I\'m given the question, find an iterated intégral for the volume of a solid abo

Question

I'm given the question, find an iterated intégral for the volume of a solid above the equation z = x^2 +y^2, and below z =2y. Normally we look for an expression where the two equations intersect projected onto the xy plane for the domain of the iterated integral. Here the intersection is an ellipse, but it is projected as a circle on the xy plane. Is there any way to get the equation of this ellipse? Unless the projection is supposed to be a circle. Please show me how to approach this problem. I'm writing in 2 hours so this would be very helpful

Explanation / Answer

given  z = x2 +y2, z =2y

when both intersect  ,x2 +y2 =2y

=>x2 +y2 - 2y +1 =1

=>x2+(y-1)2=1

use x=u , y-1 =v ,z=z

=>x=u , y =1+v

jacobian =(1*1)-(0*0)=1

z = x2 +y2, z =2y

=>z = u2 +(1+v)2, z =2(1+v)

=>z = u2 +v2+2v+1, z =2+2v

x2+(y-1)2=1

=>u2+v2=1

in cylindrical coordinates :u=rcos, v=rsin,u2+v2=r2,z=z

u2+v2=1

z = u2 +v2+2v+1, z =2+2v

=>z = r2+2rsin+1, z =2+2rsin

02,0r1 , r2+2rsin+1 z2+2rsin

dV =r dz dr d

volume of the solid =[0 to 2] [0 to 1] [r2+2rsin+1 to 2+2rsin] r dz dr d

volume of the solid =[0 to 2] [0 to 1] |[r2+2rsin+1 to 2+2rsin] r z dr d

volume of the solid =[0 to 2] [0 to 1] r(2+2rsin-r2-2rsin-1) dr d

volume of the solid =[0 to 2] [0 to 1] (r-r3) dr d

volume of the solid =[0 to 2] |[0 to 1] ((1/2)r2-(1/4)r4) d

volume of the solid =[0 to 2] [((1/2)12-(1/4)14)-((1/2)02-(1/4)04)] d

volume of the solid =[0 to 2] (1/4)d

volume of the solid =|[0 to 2] (1/4)

volume of the solid =(1/4)(2-0)

volume of the solid = (/2)

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