Answer
$\iint_R (2x-y)dA=\frac{16-12\sqrt{2}}{3}$
Work Step by Step
In polar coordinates, the circle $x^2+y^2=4$ is given by $r=2$, the line $x=0$ is given by $\theta=\frac{\pi}{2}$, and the line $y=x$ is given by $\theta=\frac{\pi}{4}$.
Then, $R=\{(r,\theta)|0\leq r\leq 2,\frac{\pi}{4}\leq \theta\leq \frac{\pi}{2}\}$
Evaluate the given integral:
$\iint_R(2x-y)dA=\int_0^2\int_{\pi/4}^{\pi/2}(2r\cos\theta -r\sin\theta)rd\theta dr$
$=\int_0^2(2r\sin\theta+r\cos\theta)r]_{\pi/4}^{\pi/2} dr$
$=\int_0^2(2r\cdot 1+r\cdot 0)r-(2r\cdot \frac{\sqrt{2}}{2}+r\frac{\sqrt{2}}{2})rdr$
$=\int_0^2(\frac{4-3\sqrt{2}}{2})r^2dr$
$=\frac{4-3\sqrt{2}}{6}r^3]_0^2$
$=\frac{4-3\sqrt{2}}{6}\cdot 2^3-0$
$=\frac{16-12\sqrt{2}}{3}$
