Answer
$0$
Work Step by Step
We can define the region inside the circle as follows: $
D=\left\{ (x, y) | -2 \leq x \leq 2, \ -\sqrt {4-x^2} \leq y \leq -\sqrt {4-x^2} \right\}
$
Therefore, $\iint_{D} (2x-y) \ dA=\int_{-2}^{2} \int_{- \sqrt {4-x^2} }^{ \sqrt {4-x^2} } 2x-y) \ dy \ dx \\ =\int_{-2}^{2} [2xy-\dfrac{y^2}{2}]_{ -\sqrt {4-x^2} }^{ \sqrt {4-x^2} } \ dx \\= \int_{-2}^{2} 4x \sqrt {4-x^2} dx$
Let us suppose that $a=4-x^2 \implies x dx=\dfrac{-da}{2}$
Now, $\iint_{D} (2x-y) \ dA=\int_0^0 \dfrac{-4 \sqrt a da}{2} \\ = -2 \int_0^0 \sqrt a da \\=0$
