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.
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 $\displaystyle x = -1$ and that the area bounded on either side of $\displaystyle x = -1$ are mirror images of each other.
Use disc integration, where
$\displaystyle V = \pi \int_a^b f(x)^2 dx$
But since the region in $\displaystyle -2<x<-1$ is a mirror of the one at $\displaystyle -1<x<0$, you can simply integrate over the second region and multiply the volume by two: $\displaystyle V = 2 \pi \int_{-1}^0 (x^2)^2 dx$
Essentially, what you are doing is taking the volume of a cylinder $\displaystyle V = \pi h r^2$ whose radius $\displaystyle r$ varies. In this case, the radius is dependent on the parabola that you are integrating under (do you see it from your drawing)?
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.