# Rotational volume problem

• Nov 30th 2012, 11:38 PM
Nora314
Rotational volume problem
Hi everyone!

I am having a bit trouble with a question about rotational volume. So, this is the problem:

"An area R in the xy plane is given by:

0 ≤ x ≤ 1

and

0 ≤ y ≤ x3

Find the volume when rotating R around the axis y = -1
"

Here is how I was thinking. First, I drew this figure:

Attachment 25999

My idea was to draw the boundaries, and two radius in red, which I need to find in order to use the washer method. So, I thought the lower boundary must be r(y) = g(y) - c = x3 - (-1) = x3 + 1 (The second red line in the picture)

But I am unsure what I should put as the upper boundary (The first radius in the picture).

I hope someone can help and tell me in what way I am thinking wrong.

Thanks!
• Dec 1st 2012, 01:04 AM
MarkFL
Re: Rotational volume problem
The region $\displaystyle R$ is bounded by $\displaystyle y=0,\,x=1,\,y=x^3$.

An arbitrary slice is the washer:

$\displaystyle dV=\pi(r_o^2-r_i^2)\,dx$

$\displaystyle r_o=x^3+1$

$\displaystyle r_i=1$
The "stack" of washers runs from $\displaystyle x=0$ to $\displaystyle x=1$, so summing the washers in the stack, we find the volume is:
$\displaystyle V=\pi\int_0^1 (x^3+1)^2-1^2\,dx$