How do I solve this integral (volume of a torus)? I figured the volume of a torus is 
$4\pi R \int_{-r}^{r}\sqrt{r^2-y^2}dy$
But I don't know how to solve an integral like this.  How is it done?

Use the trig substitution: $y=r\sin(t)$
Then $dy = r\cos(t)\,dt$ and then limits change to $\pm 1 =\sin(t)$ so that $t=\pm \pi/2$. Thus 
$$ 4\pi R\int_{-r}^r \sqrt{r^2-y^2}\,dy = 4\pi R \int_{-\pi/2}^{\pi/2} \sqrt{r^2-r^2\sin^2(t)}\cdot r\cos(t)\,dt $$ $$= 4\pi R \int_{-\pi/2}^{\pi/2} \sqrt{r^2\cos^2(t)}\cdot r\cos(t)\,dt = 4\pi R \int_{-\pi/2}^{\pi/2} r^2\cos^2(t)\,dt$$ $$= 4\pi R \int_{-\pi/2}^{\pi/2} \frac{r^2}{2}(1+\cos(2t))\,dt = 4\pi R \frac{r^2}{2}\pi$$
Alternatively, you could just note that the integral is computing the area of the upper-half of a circle of radius $r$ (thus $0.5\pi r^2$).
