1. ## Find the volume

$x=1-y^4$, $x=0$; about the $x=9$

I drew a picture and I turned it as if it was $y=1-x^4$ being rotated about $y=9$ and I made into 2 sections.

So I'd have:
$2\int_{0}^{1}\left( 2\pi (9-x)\left(1-x^4\right)\right)dx=\frac{412}{15}\pi$

Can someone tell if that's correct. I'm using webassign and I'm on my last try lol

2. ## Re: Find the volume

Hey amthomasjr.

You said in your post two different things: rotating about x = 9 and then rotating about y = 9. Which one is it?

Also what rotation method are you using? (Shell method? Other method?)

The rotation should be something like integrating pi*y^2*dx or pi*x^2*dy with a shift in the function if you are not rotating around the x or y axes.

3. ## Re: Find the volume

Rotating about the x = 9 is the problem. The other stuff is from me turning the picture I drew.
I am using the shell method. Should I use something different?

4. ## Re: Find the volume

Hint: Try shifting the function to the origin and then use the normal formula for rotating about the x (or y depending on the problem) axis.

5. ## Re: Find the volume

I spent some more time on the problem this morning and I finally figured it out. I should have been using the washer method.

$2\int_{0}^{1} \pi\left[9^2-(9-(1-y^4))^2\right]dy=2\pi \int_{0}^{1}\left[81-(8+y^4)^2\right]dy=2\pi \int_{0}^{1}\left[17-16y^4-y^8\right]dy=\frac{1232}{45}\pi$