Calculating the Diameter of a Drilled Hole in a Revolved Solid

[icesalmon]

Homework Statement

A solid is generated by revolving the region bounded by y = x²/2 and y = 2 around the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-fourth of the volume is removed. Find the diameter of the hole.

The Attempt at a Solution

I'm going with cylindrical shells this time around. I'm integrating from x = 0 to x = 2. I think my height is 2-x and my radii are all going to be generated by the function x²/2. After integration, I get the overall volume is to be 4pi/3. If one quarter of that is taken out after drilling this hole, I have the volume of this figure to be pi. I believe it's a right cylindrical shaped whole, so the volume of a cylinder is pir²h. I need the radius of one of these cross sections so I need some way to relate the volume. Or maybe I don't, this is where I need assistance. Hopefully I'm not over-thinking this. Thanks.

 

[icesalmon]

I believe I've made a mistake on the radius / height. Radius = x and Height = x²/2, after integration I get v_t = 4pi so the volume of the cylinder is 3pi.

 

[haruspex]

I believe I've made a mistake on the radius / height. Radius = x and Height = x²/2, after integration I get v_t = 4pi so the volume of the cylinder is 3pi.

Yes, except that the removed core is not exactly a cylinder. What do you get for the diameter of the hole?

 

[icesalmon]

Yes, except that the removed core is not exactly a cylinder. What do you get for the diameter of the hole?

I don't know how to find it, especially since I don't know what the figure of the hole is.

Edit: the diameter seems like it would be 2(y/2)^1/2 but that's sort of a shot in the dark.
Edit: that's wrong, nvm.

 

[SammyS]

I believe I've made a mistake on the radius / height. Radius = x and Height = x²/2, after integration I get v_t = 4pi so the volume of the cylinder is 3pi.

The height of each shell should be 2 - x²/2 .

Do the same integral but have x go from 0 to a. set that volume to 1/4 the volume without the hole & solve for a .

 

[icesalmon]

I'm getting two answers..2 and 2(2^1/2)
and neither are correct.

Should have included this, sorry. My integrand here has been changed to (x)(2-x²/2). When my bounds are [0,a] I get 2pi(a² - a⁴/8). When my bounds are [0,2] I get the volume as 4pi. If I set the first part equal to 1/4 of the total volume I get 2pi(a² -a⁴/8) = pi and I solve it from there to get that a = 2^1/2

 

[haruspex]

2pi(a² -a⁴/8) = pi

Right

a = 2^1/2

Wrong.

 

[icesalmon]

thanks, got it now.