Find the volume of the solid generated by revolving the region enclosed by $y=x^3$ and $y=x$ about $x$-axis. I need help with this exercise.
Find the volume of the solid generated by revolving the region enclosed by $y=x^3$ and $y=x$ about $x$-axis.
When I graph I get that these are the enclosed regions:

If I rotate this region about $x$-axis I get the following:

So then the integral will be this one?
$$V=\pi\int_{-1}^{0} [(x^3)^2-(x)^2]\,dx+\pi\int_{0}^{1} [(x)^2-(x^3)^2]\,dx$$
 A: Solving the equation $x=x^{3}$ we get $x=-1,0,1$ then the curves $y=x$ and $y=x^{3}$ has intersection $(-1-1),(0,0)$ and $(1,1)$ as you noticed.
Using the washer method:
Over $[0,1]$ we have,
$$A_{1}(x)=\pi(\text{outer radius})^{2}-\pi(\text{inner radius})^{2}=\pi (x)^{2}-\pi (x^{3})^{2},$$ then
$$V_{1}=\int_{0}^{1}A_{1}(x)\, {\rm d}x=\pi\int_{0}^{1}(x^{2}-x^{6})\, {\rm d}x=\boxed{\frac{4}{21}\pi}$$
Over $[-1,0]$ we have,
$$A_{2}(x)=\pi(\text{outer radius})^{2}-\pi(\text{inner radius})^{2}=\pi(x)^{2}-\pi(x^{3})^{2},$$
then
$$V_{2}(x)=\int_{-1}^{0}A_{2}(x)\, {\rm d}x=\pi\int_{-1}^{0}(x^{2}-x^{6})\, {\rm d}x=\boxed{\frac{4}{21}\pi}$$
Therefore,
$$V=V_{1}+V_{2}=\frac{4}{21}\pi+\frac{4}{21}\pi=\boxed{\frac{ 2\cdot 2^{2}}{21}\pi}$$
Just small remark: When we're considering the washer method we are talking about "radius" and this has notion about the magnitude, so that is true thar over $[-1,0]$ we have $x<x^{3}$ however we need to see in the magnitude and that is $|x|>|x^{3}|$ over $[-1,0]$ so the outer radius will be $x$ and the inner radius will be $x^{3}$ as also happened in the interval $[0,1]$. Moreover using the symmetry of the problem, you could just calculate the volume over $[0,1]$ and then multiply that result by $2$, which is in fact what happened here. In fact it is not even necessary to draw always, it is enough to do algebra to see what radius will be the outer and what radius will be the inside considering their magnitudes. However, your graph is very illustrative, good work.
