# Thread: [SOLVED] Volume between two curves

1. ## [SOLVED] Volume between two curves

Hi, I'm trying to figure out how to this question:

The area bounded by the curves y = x^2, y = (x+2)^2 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

Any help would be greatly appreciated. Thanks.

2. Originally Posted by T_C
Hi, I'm trying to figure out how to this question:

The area bounded by the curves y = x^2, y = (x+2)^2 and the x-axis is rotated about the x-axis. Find the volume of the solid of revolution formed.

Any help would be greatly appreciated. Thanks.
Draw it out.

You will find the intersection of the two parabola is at $x = -1$ and that the area bounded on either side of $x = -1$ are mirror images of each other.

Use disc integration, where
$V = \pi \int_a^b f(x)^2 dx$

But since the region in $-2 is a mirror of the one at $-1, you can simply integrate over the second region and multiply the volume by two: $V = 2 \pi \int_{-1}^0 (x^2)^2 dx$

Essentially, what you are doing is taking the volume of a cylinder $V = \pi h r^2$ whose radius $r$ varies. In this case, the radius is dependent on the parabola that you are integrating under (do you see it from your drawing)?

3. Thank you very much.

Wow...the question was pretty simple. I already drew the graph and solved the equations simultaneously to get the intersection of the two parabolas before I asked the question, but the wording of the question confused me.

Yeah I understood everything you said but I don't know what you mean by the cylinder thing.

Thanks once again.