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 defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$
We have: $D$ is a circle having a radius of $1$ in the first quadrant.
Thus, we can express the region $D$ as follows:
$$D=\left\{ (x, y) | 0 \leq x \leq 1, \ 0 \leq y \leq \sqrt {1-x^2} \right\}
$$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\=\int_{0}^{1} \int_{0}^{\sqrt {1-x^2}} y dy \ dx \\ = \int_{0}^{1} [y^2/2]_{0}^\sqrt {1-x^2}) dx \\ =\int_{0}^{1} \dfrac{1-x^2}{2} \\=\int_{0}^{1} \dfrac{1}{2}-\dfrac{x^2}{2}\\= [\dfrac{x}{2}-\dfrac{x^3}{6}]_0^1 \\ = \dfrac{1}{2}-\dfrac{1}{6} \\=\dfrac{1}{3}$$
