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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 ≤ x^{3}

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 = x^{3} - (-1) = x^{3} + 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!

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:

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

The outer radius is:

$\displaystyle r_o=x^3+1$

The inner radius is:

$\displaystyle r_i=1$

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:

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

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

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 :p