Answer
$$0$$
Work Step by Step
Our aim is to calculate the area of the given surface.
The domain $D$ can be expressed as follows:
$
D=\left\{ (x, y) | -2 \leq x \leq 2 , \ -\sqrt {4-x^2} \leq y \leq -\sqrt {4-x^2} \right\}
$
Now, $A=\iint_{D} (2x-y) \ dA=\int_{-2}^{2} \int_{- \sqrt {4-x^2} }^{ \sqrt {4-x^2} } 2x-y) \ dy \ dx \\ =\int_{-2}^{2} [2xy-\dfrac{y^2}{2}]_{ -\sqrt {4-x^2} }^{ \sqrt {4-x^2} } \ dx \\= \int_{-2}^{2} 4x \sqrt {4-x^2} dx$
Set $4-x^2 =u$ and $\dfrac{-1}{2} du = x dx$
So $A=\iint_{D} (2x-y) \ dA=\int_0^0 \dfrac{-4u^{1/2} du}{2} \\ = -2 \int_0^0 u^{1/2} du \\=0$
