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