Answer
$\dfrac{16}{3}$
Work Step by Step
Since, $z \gt 0$ because it is in the first quadrant, so we have: $z=f(x,y)=\sqrt {4-y^2}$
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$
$D=\left\{ (x, y) | 0 \leq x \leq 2, \ 0 \leq x \leq 2y \right\}
$
Now, $V =\iint_{D} f(x,y) \ dA \\=\int_{0}^{2} \int_{0}^{2y} \sqrt{4-y^2} \ dx \ dy \\ = \int_{0}^{2} 2y \sqrt{4-y^2} \ dy $
Consider $4-y^2=a$ and $ -da= 2y dy$
Now, $V=\int_{4}^{0} -\sqrt a da \\=-[\dfrac{2a^{3/2}}{3}]_0^4 \\= \dfrac{2}{3} \times [ (4)^{3/2}] \\=\dfrac{16}{3}$