# Calculus

posted by on .

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 - ,

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 - ,

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 - ,

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.