I would like to compute the integral $\int_D \sqrt{x^2+y^2} \mathrm{~d}(x, y)$ over the domain $$ D=\left\{(x, y) \in \mathbb{R}^2: 1 \leq x^2+y^2 \leq 4,|y| \geq|x|\right\} $$
I have sketched the conditions resulting in a disk shape and a mirrored v-shape form - leading me to conclude that the integral is 0.
However, I don't understand how I can approach this more "mathematically". Any help would be much appreciated!
I think your main problems are with the inequality $|y| \geq |x|$. Rewrite it as follows: $$ \left\{\begin{matrix} y\geq |x| \\ y\geq 0 \end{matrix}\right. \,\,\,\,\,\,\lor\,\,\,\,\,\, \left\{\begin{matrix} y\leq |x| \\ y< 0 \end{matrix}\right.$$
Hence, the domain of integration becomes:
I'm going to use polar coordinates, so $\rho \in [1, 2]$ and $\theta\in \left[\frac{\pi}{4}, \frac{3}{4}\pi\right]\cup\left[\frac{5}{4}\pi, \frac{7}{4}\pi\right]$. By simmetry, one can write: $$\iint_{D}\sqrt{x^2+y^2}=2\cdot\int_{1}^{2}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}\rho^2d\theta d\rho=2\cdot\left(\int_{1}^{2}\rho^2 d\rho\right)\cdot\left(\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}d\theta\right)=2\cdot\frac73\cdot\left(\frac{3}{4}\pi-\frac{1}{4}\pi\right)=\frac73\pi$$