Answer
$\dfrac{16}{3}$
Work Step by Step
The volume under the surface is given by :$z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $\ Volume ; V=\iint_{D} f(x,y) \space dA$
Our aim is to calculate the volume of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | 0 \leq x \leq 2, \ 0 \leq x \leq 2y \right\}
$
Therefore, $ 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 $
Set $4-y^2 = u $ and $2y dy = - du$
Now, $Volume= - \int_{4}^{0} u^{1/2} du \\=-\dfrac{2}{3}[ u^{3/2}]_0^4 \\= \dfrac{2}{3} \times [ (4)^{3/2}] \\=\dfrac{16}{3}$
