Mistake in the proof of the volume of a torus

Question

We define the function $$f(x)=\sqrt{r^2-(x-R)^2}$$ which graphs a semi-circle with radius $r$ with its center being at $(R,0)$ If we were to rotate this function around the $y$-axis, we'd get half a torus, thus, the formula for the volume of a tors with radii $r$ and $R$ would be $$2\pi\int\limits_{R-r}^{R+r}\left(f(x)\right)^2dx=2\pi\int\limits_{R-r}^{R+r}\left(r^2-x^2-R^2+2xR\right)dx=2\pi\left[xr^2-\frac{x^3}{3}-R^2x+Rx^2\right]_{R-r}^{R+r}$$ This simplifies to $$\left(Rr^2+r^3-\frac{(R+r)^3}{3}-R^3-R^2r+R(R+r)^2\right)-\left(Rr^2-r^3-\frac{(R-r)^3}{3}-R^3+R^2r+R(R-r)^2\right)$$ $$2r^3+\frac{(R-r)^3-(R+r)^3}{3}-2R^2r+4R^2r$$ $$2r^3+\frac{-2r^3-6R^2r}{3}+2R^2r$$ $$\frac{4}{3}r^3\Rightarrow\frac{8\pi r^3}{3}$$ which has no dependence on $R$. This cannot be correct. Where is my mistake?