Posted by **Becca** on Sunday, January 29, 2012 at 4:33pm.

Find the volume of the solid formed when the region bounded by y=3x^2 and y=36-x^2 is revolved about the x-axis.

- Calculus -
**Reiny**, Sunday, January 29, 2012 at 6:04pm
I see several posts made by you dealing mainly with these integrations problems, without any indication by you showing any steps or attempts.

I will do this one. Try to follow these steps to solve the others

Make a sketchl

First you need the intersection to see the region that you are revolving.

y = 3x^2 and y = 36-x^2

3x^2 = 36-x^2

4x^2 = 36

x = ± 3

Because of the symmetry, I will boundaries from 0 to 3, then double our volume

in general,

Volume = π∫y^2 dx from left boundary to right boundary

the y , the height or radius of our rotation, is

36-x^2 - 3x^2 = 36-4x^2

V = 2π∫ (36-4x^2)^2 dx from 0 to 3

= 2π∫ (1296 - 288x^2 + 16x^4) dx from 0 to 3

= 2π [ 1296x - 96x^3 + (16/5)x^5] from 0 to 3

= 2π (3888 - 2592 + 3888/5 - 0 - 0 - 0)

= (20736/5)π = appr. 13028.8

- Calculus -
**Becca**, Sunday, January 29, 2012 at 6:44pm
I use this to check my answers. I apologize for not realizing that I should have proven to you that I was doing them by myself at first.

- Calculus -
**Reiny**, Sunday, January 29, 2012 at 6:57pm
Fair enough.

In that case, you should post the answer you got, that way I can just do some fast scribbling on a piece of scrap paper and either confirm or reject your answer.

It would save us a lot of unnecessary typing of the whole solution.

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