# Thread: Volume between two curves

1. ## Volume between two curves

Hello, here is my question:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
So, I tired to solve it and I'm getting the wrong answer. Here is a graph:
I did the integral of (1-x^6)-(x^6-5)dx from 0 to 1 and it wasn't right.

Thanks,
Matt

2. I didn't see "pi" in the integral, did you leave it out of the answer as well?

3. No, I did put it in for the answer, I forgot to put it in for the post...

4. Okay, I think I almost got it, $pi/int_0,1 (5-x^6)^2-16 dx$ I meant from 0 to 1 but I couldn't get the limits to work good in latex. I solved it and it still isn't the right answer.

5. Is is not $\pi\int_0^{1}\left[(5-x^6)^2-16\right]dx=\frac{696}{91}\pi$?

6. Yep, that's it, but I found out that I had to multiple the answer by two because it wanted the volume of the whole shape, so it includes -1 to 0 as well. I have a similar problem here:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

$\pi\int_2^{6}\left[\frac{1}{x^3}\right]^2dx=$