Could you help me solve this?
$$\iint_{M}\!\cos\left(\sqrt{x^2+y^2}\right)\,dxdy;$$
$M: \frac{\pi^2}{4}\leq x^2+y^2\leq 4\pi^2$
I know that the region would look like this and I need to solve it as difference of two regions. Lets say $C=A\cup B$, then $A = C – B$
I was able to get the integrals, but don't know how to solve them.
$$\int_{0}^{2\pi}\!\int_{0}^{\sqrt{4\pi^2-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx – \int_{0}^{\pi/2}\!\int_{0}^{\sqrt{\frac{\pi^2}{4}-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx$$
Best Answer
Change to polar coordinates. Thus $\sqrt{x^2+y^2}$ gets replaced by $r$. and $dx\,dy$ gets replaced by $r\,dr\,d\theta$. So we end up needing to integrate $r\cos r\, dr\,d\theta$ over our region.
Note that the limits of integration for $r$ are $\frac{\pi}{2}$ and $2\pi$. If the region you are interested in is just the first quadrant region, as in the picture, then $\theta$ will run from $0$ to $\frac{\pi}{2}$.
The integration with respect to $r$ is done using integration by parts. Let $u=r$ and $dv=\cos r\,dr$. Then $du=dr$ and we can take $v=\sin r$. So $$\int r\cos r\,dr=r\sin r-\int \sin r\, dr=r\sin r +\cos r+C.$$
Added: So for the first quadrant region, our integral is $$\int_{\theta=0}^{\pi/2} \int_{r=\pi/2}^{2\pi} r\cos r\,dr\,d\theta.$$ The integration with respect to $\theta$ will be easy, since there is no dependence on $\theta$.