Answer
$\dfrac{1}{3}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $$\ Volume =\iint_{D} f(x,y) \space dA$$
We have $D$ is a circle having a radius of $1$ in the first quadrant, so, the region $D$ can be defined as: $D=\left\{ (x, y) | 0 \leq x \leq 1, \ 0 \leq y \leq \sqrt {1-x^2} \right\}$
Now, $V=\int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} y dy \ dx \\ = \dfrac{1}{2} \times \int_{0}^{1} [y^2]_{0}^\sqrt {1-x^2}) dx \\ =\dfrac{1}{2} \times \int_{0}^{1} (1-x^2) \\=\int_{0}^{1} \dfrac{1}{2}-\dfrac{x^2}{2}\\= [\dfrac{x}{2}-\dfrac{x^3}{6}]_0^1\\=[\dfrac{1}{2}-0]-[\dfrac{(1)^3}{6}-0] \\ = \dfrac{1}{2}-\dfrac{1}{6} \\=\dfrac{1}{3}$
