Answer
$$\dfrac{5 \sqrt 2}{3} $$
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: $$\ Volume =\iint_{D} f(x,y) \ dA$$
Therefore, $\ Volume =\iint_{D} f(x,y) \ dA \\= \int_{-1/\sqrt 2}^{1/\sqrt 2} \int_{x^2}^{1-x^2} (3-y) \ dy \ dx \\ = \int_{-1/\sqrt 2}^{1/\sqrt 2} [3y-\dfrac{y^2}{2}]_{x^2}^{1-x^2} (3-y) \ dy \ dx \\=\int_{-1/\sqrt 2}^{1/\sqrt 2} [3(1-x^2)y-\dfrac{(1-x^2)^2}{2}-[3 (x^2)-\dfrac{(x^2)^2}{2}] \ dx \\= [\dfrac{5}{2} x- \dfrac{5}{3} x^3]_{-1/\sqrt 2}^{1/\sqrt 2} \\= \dfrac{5}{\sqrt 2}-\dfrac{5}{3 \sqrt 2} \\=\dfrac{5 \sqrt 2}{3} $
