$\iint 2x-y$ $dA$ in the first quadrant and enclosed by $x=0$ $y=x$ and $x^2+y^2=4$
Since the function is enclosed in the first quadrant then $0 \leq \theta \leq \frac{\pi}{2}$ and since $y=x$ and $x=0 \to y=0$ and $0 \leq r \leq 2$
$\int_{0}^{\frac{\pi}{2}}$ $\int_{0}^{2}$ $2r\cos\theta-r\sin\theta$ $r$ $dr d\theta$ $\to \int_{0}^{2} 2r^2\cos\theta-r^2\sin\theta$ $drd\theta$ $\to \frac{2r^3}{3} \cos\theta - \frac{r^3}{3}\sin\theta \Big\vert_{0}^{2}= \frac{16}{3} \cos\theta-\frac{8}{3} \sin\theta$
$\int_0^{\frac{\pi}{2}}$ $\frac{16}{3} \sin\theta +\frac{8}{3}\cos\theta \Big\vert_{0}^{\frac{\pi}{2}} = 8\pi$