I'm trying to evaluate the following double integral:
$\int_F \arctan(\frac{x}{y})\ dxdy, F:=\{(x,y)\in\mathbb{R^2}:1\leq x^2+y^2\leq4, |y|\leq|x|\}$.
By switching to polar coordinates and splitting I get:
$\int_F \arctan(\frac{x}{y})\ dxdy= \int_0^{\pi/4}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta + \int_{3\pi/4}^{5\pi/4}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta + \int_{7\pi/4}^{2\pi}(\int_{r=1}^{r=2}\arctan(\tan\theta)r dr)d\theta = \int_0^{\pi/4}\theta d\theta\int_{r=1}^{r=2}rdr+ \int_{3\pi/4}^{5\pi/4}\theta d\theta\int_{r=1}^{r=2}rdr + \int_{7\pi/4}^{2\pi}\theta d\theta\int_{r=1}^{r=2}rdr = \frac{1}{4}(\frac{\pi^2}{16}\cdot 3)+\frac{1}{4}(\frac{25\pi^2}{16}-\frac{9\pi^2}{16})\cdot 3+ \frac{1}{4}(4\pi^2-\frac{49\pi^2}{16})\cdot 3= \frac{3\pi^2}{64}+\frac{48\pi^2}{64}+\frac{45\pi^2}{64}=\frac{96\pi^2}{64}=\frac{3\pi^2}{2}$.
Can someone double-check this?
Thanks a lot.