Volume of the region bounded by $y=x^2$ and $y=x^3$ when it's rotated about the line $x=1$

Question

I have to find the volume of the region bounded by $y=x^2$ and $y=x^3$ when it's rotated about the line $x=1$.

Since $y=x^2\Leftrightarrow x=\sqrt{y}$ and $y=x^2\Leftrightarrow x=\sqrt[3]{y}$, so the outer radius would be $$1-\sqrt{y}$$ while the inner radius is equal to $$1-\sqrt[3]{y}$$ Hence, the volume is given by $$\pi\int_{0}^{1}\left((1-\sqrt{y})^{2}-(1-\sqrt[3]{y})^{2}\right)dy$$ I just want to know if it is correct.

Answer 1

This plot confirms, that when rotating around the line x=1, the outer radius comes from $y=x^2$ and the inner radius comes from $y=x^3$. This makes the outer radius $1-y^{\frac12}$ and the inner radius $1-y^{\frac13}$. Also, that this area is closed, this is the only close area bounded by these two y's, and the points of intersection.

Note: Technically, only algebra can conclusively demonstrate these facts, the intersection points, etc., and the graphs merely but strongly suggests these facts.

Answer 2

One can do without sowtfare plots or algebraically in the following way: $$1-\sqrt y\stackrel{?}{>}1-\sqrt[3]y$$ $$\sqrt[3]y\stackrel{?}{>}\sqrt y$$ Taking sixth power $$y^2\stackrel{?}{>}y^3$$ $$1>y$$ It checks since $y\in (0,1)$.