Answer
$\dfrac{1}{3}$
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.
We have $D$ is a circle having radius $1$ in the first quadrant.
The domain $D$ can be expressed as follows:$D=\left\{ (x, y) | 0 \leq x \leq 1, \ 0 \leq y \leq \sqrt {1-x^2} \right\}$
Now, $\ Volume =\iint_{D} f(x,y) \ dA \\=\int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} y dy \ dx \\ = \int_{0}^{1} (1/2)[y^2]_{0}^\sqrt {1-x^2}) dx \\ =\int_{0}^{1} \dfrac{1-x^2}{2} \\=\int_{0}^{1} \dfrac{1}{2}- \int_{0}^{1}
\dfrac{x^2}{2}\\= \dfrac{1}{2} \times [x]_0^1 -\dfrac{1}{6} \times [x^3]_0^1 \\ = \dfrac{1}{2}-\dfrac{1}{6} \\=\dfrac{1}{3}$
