Solid generated by revolving region, find the diameter of the hole

[isukatphysics69]

Homework Statement

A solid is generated by revolving region bounded by y=(1/2)x^2 and y=12 about the y axis. A hole centered along the axis of revolution is drilled through this solid so that 1/4 of the volume is removed. find the diameter of the hole.

Homework Equations

y=(1/12)x^2 y = 12
shell method integral

The Attempt at a Solution

i figured out the boundries are x=0 to x=12 and integrated using the shell method and got the answer of the total volume as 864*pi. I divided that by 4 and got 216*pi as the area that needs to be removed but now i am stuck and don't know how to figure out the diameter of the hole! [/B]

 

[tnich]

Homework Statement

A solid is generated by revolving region bounded by y=(1/2)x^2 and y=12 about the y axis. A hole centered along the axis of revolution is drilled through this solid so that 1/4 of the volume is removed. find the diameter of the hole.

Homework Equations

y=(1/12)x^2 y = 12
shell method integral

The Attempt at a Solution

i figured out the boundries are x=0 to x=12 and integrated using the shell method and got the answer of the total volume as 864*pi. I divided that by 4 and got 216*pi as the area that needs to be removed but now i am stuck and don't know how to figure out the diameter of the hole! [/B]

Have you made a sketch of the solid and the hole? That might help you figure it out. Actually, just start with the intersection of the solid and the hole with the x,y-plane, then it should be pretty clear how to integrate the solid of revolution for the hole. Give the hole a radius of r, do the integration, and then figure out what r has to be to give you the right volume for the hole.