$\int\int_D \frac{|y|}{(x^2+y^2)^2} dxdy \space\space\space D:=\{(x, y) \in R^2: 1\le x^2+y^2 \le 4, |y| <= \sqrt{3}x \}$
This looks like an integral that should be easy to solve by switching to polar coordinates, sketching the domain I get that $D:= \{(\rho,\theta) \in R^2 :1\le \rho\le 2, \theta \in[-\frac{\pi}{3}, \frac{\pi}{3}]$
$\int_{-\frac{\pi}{3}}^\frac{\pi}{3} \int_1^2 \rho \frac{|\rho sin(\theta)|}{\rho^4} dpd\theta $
This is where I get confused. I can simplify $\rho$, but one of them is inside the absolute value. Since $\rho$ is the length, I guess that it can be moved outside of the absolute value considering that it will always be positive. If so, I get that the first integral is $[-\frac{1}{p}]_1^2 = \frac{1}{2} $ leaving me with $\int_{\frac{-\pi}{3}}^\frac{\pi}{3} |sin(\theta)|d\theta $.
I was wondering if I can split the integral in $\int_{-\frac{\pi}{3}}^0 -sin(\theta) d\theta + \int_0^\frac{\pi}{3} sin(\theta)d\theta $
If so, the end result of the entire integral is $\frac{1}{2}$, right?
Yes, $\rho$ is positive.
\begin{align} \int_{-\frac{\pi}3}^\frac{\pi}3 \int_1^2 \frac{|\sin \theta|}{\rho^2}\,\,d\rho d\theta &=\int_1^2 \rho^{-2} \,d\rho\int_{-\frac{\pi}3}^\frac{\pi}3 |\sin \theta|\, d\theta \\ &=\left[-\frac12+1 \right]\left[ 2 \int_0^\frac{\pi}3 \sin \theta \, d\theta\right]\\ &= \int_0^\frac{\pi}3 \sin \theta \, d\theta \\ &= -\cos \theta|_0^\frac{\pi}3 \\ &= -\frac12 + 1\\ &= \frac12 \end{align}
You are right, the final value is $\frac12$.