Calculus
posted by Becca on .
Find the volume of the solid formed when the region bounded by y=3x^2 and y=36x^2 is revolved about the xaxis.

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 = 36x^2
3x^2 = 36x^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
36x^2  3x^2 = 364x^2
V = 2π∫ (364x^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 
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.

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.