I'm having difficulty with this integral: $$ \int_{\pi}^{2\pi}\int_{0}^{2}r\sin \theta+\sqrt{4-r^2\cos^2\theta} \,r\,drd\theta $$ I can find a solution using symbolab, however this seems much too complex for the course difficulity.
The integral is derived from finding a 3D-body, defined by $$ z=f(x,y)=y+\sqrt{4-x^2} $$ and the $(x,y)$-plane when $x$ and $y$ are bounded by $x\leq 0$ and circle with $r=2$.
Anyone knows some trick to this (i assume, a trick is used here to reduce complexity).
Rearranging slightly $$\int_{\pi}^{2\pi}\int_{0}^{2}r\sin \theta+r \cos \theta \sqrt{\frac{4}{\cos^2\theta}-r^2} \;dr\; d\theta$$ Integrating w.r.t $r$ $$\int_{\pi}^{2\pi}\left[\frac{r^2}{2}\sin \theta-\frac{1}{3}\cos \theta \left(\frac{4}{\cos^2\theta}-r^2\right)^{\frac{3}{2}}\; \right]_0^2\; d\theta$$
Substituting for $r$ and simplifying then gives the second integral w.r.t. $\theta$
$$\int_{\pi}^{2\pi} 2\sin \theta +\frac{8}{3} \sec^2\theta \;-\frac{8}{3}\tan^3\theta\cos\theta \; d\theta $$