# Thread: Rotational volume problem

1. ## 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:

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!

2. ## Re: Rotational volume problem

The region $R$ is bounded by $y=0,\,x=1,\,y=x^3$.

An arbitrary slice is the washer:

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

The outer radius is:

$r_o=x^3+1$

The inner radius is:

$r_i=1$

The "stack" of washers runs from $x=0$ to $x=1$, so summing the washers in the stack, we find the volume is:

$V=\pi\int_0^1 (x^3+1)^2-1^2\,dx$

Expand and simplify the integrand, then use the anti-derivative form of the FTOC to compute the volume.

3. ## Re: Rotational volume problem

Thank you so much for the help, I understood it much better now, I also noticed my figure is completely wrong.... How embarrassing