# Finding volume of region rotated around the x-axis: $y=x^3\;y=x$

I'm trying to find the volume of the region enclosed by the curves above, rotated along the x-axis. I know how to do this if $x\ge0$ but with this there are two regions, and I'm not sure how to handle that.

For example, if I find the intersections to find x: $$x^3=x\rightarrow x^3-x=0 \rightarrow x(x^2-x)=0\rightarrow x=\pm1 \;and\;x=0$$ So, I have three x's and I do not know how to set up the integration part of the problem.

Thanks for any help!

• You can use some symmetry here. The volume of the region bounded between the curves from $x=0$ to $x=1$ is the same as the volume of the region bounded between the curves from $x=-1$ to $x=0$. – Twenty-six colours Oct 1 '17 at 6:46
• @Twenty-sixcolours So once I find the volume of either one, I could just multiply it by two to get the total volume of both? – JustHeavy Oct 1 '17 at 6:48
• Yes. By the way, if you had a different question, and you got $3$ or more x-values as a result of solving the intersections, you can always just add the volumes. So pretend you didn't know that it was symmetric for this question, you could have just found the volume bounded in $[-1,0]$ and added the volume bounded in $[0,1]$. – Twenty-six colours Oct 1 '17 at 6:49
• Well, $f(x) = (x)^2 - (x^3)^2$ is even. So you could just solve one side and multiply by two. – G_D Oct 1 '17 at 6:50

Just calculate $$2\pi\int\limits_{0}^1(x^2-x^6)dx$$
• Thank you, I get it now. The function is even, so finding the volume of one interval will result in the same volume of the other side, allowing me to multiply by 2. Also, the original setup you had would have been $$\int_0^1 2\pi r^2 dx$$ correct? – JustHeavy Oct 1 '17 at 6:52
• would you be able to quickly check my work? $$=2\pi({x^3\over 3}-{x^7\over 7}) \; ]_0^1$$ Next, $${2\pi \over3}-{2\pi \over 7}-0$$ And now my final answer is, $${8\pi \over21}$$ Does that area make sense to you? – JustHeavy Oct 1 '17 at 7:06