Answer
$0$
Work Step by Step
The region $D$ can be defined as: $
D=\left\{ (x, y) | -2 \leq x \leq 2, \ -\sqrt {4-x^2} \leq y \leq -\sqrt {4-x^2} \right\}
$
Now, $\iint_{D} (2x-y) \ dA=\int_{-2}^{2} \int_{- \sqrt {4-x^2} }^{ \sqrt {4-x^2} } (2x-y) \ dy \ dx \\ =\int_{-2}^{2} [2xy-\dfrac{y^2}{2}]_{ -\sqrt {4-x^2} }^{ \sqrt {4-x^2} } \ dx \\= \int_{-2}^{2} 4x \sqrt {4-x^2} dx$
Substitute $a=4-x^2$ and $\dfrac{-da}{2}= x \ dx$
Thus, $$\iint_{D} (2x-y) \ dA=\int_0^0 \dfrac{-4}{2} \times \sqrt a da\\ = -2 \int_0^0 \sqrt a da \\=0$$
