integrate $2\int _0^2\:r^2\sqrt{(1-(r-1)^2)} dr$

Question

I have to solve the integral $\int _0^2\:\int _{-arccos\left(r-1\right)}^{arccos\left(r-1\right)}\: r(rcos(\theta)+rsin(\theta)) d\theta dr$. I have to do it in this order specifically, so I can't change it to $drd\theta$ .

I got to $2\int _0^2\:r^2\sqrt{(1-(r-1)^2)} dr$, but I don't know how to continute from here. I have tried to substitute $r-1$ with $sin(p)$, but that didn't seem to work either.

(I'd prefer a hint or suggestion for what kind of technique or subsitution I'm supposed to use here over a complete answer)

Answer 1

If $r-1 = \sin(p)$ then the integral becomes

$$ 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin(p)+1)^2 \cos^2(p) dp = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[ \sin^2(p)\cos^2(p) + 2\sin(p)\cos^2(p) + \cos^2(p)\right] $$

Integrating the second and third terms should be fairly straight forward (right?). For the first term, we can use $$ \sin^2(p)\cos^2(p) = \frac{1}{4} \sin^2(2p) = \frac{1- \cos(4p)}{8}$$