Answer
$$\dfrac{64}{3}$$
Work Step by Step
Our first step is to find the points of intersection of the parabolas $y=x^2-1$ and $y=1-x^2$ in the $xy$-plane.
Now, $x^2-1=1-x^2 \implies x=\pm 1$
For any x, y in the above region, the plane $z=2-x-y$ is below the plane $z=10+2x+2y$. Therefore, the volume of the solid will be the difference of the solid below the plane $z= 10+2x+2y$ and the solid below the plane $z=2-x-y$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\= \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (10+2x+2y) \ dy \ dx - \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (2-x-y) \ dy \ dx\\ = \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (8+3x+3y) \ dy \ dx \\= \int_{-1}^{1} (8y+3xy+\dfrac{3y^2}{2})|_{x^2-1}^{1-x^2} dx \\=\int_{-1}^{1} (16-16x^2+6x -6x^3) dx\\=[16x-\dfrac{16x^3}{3}+3x^2-\dfrac{3x^4}{2}]_{-1}^{1}\\=\dfrac{64}{3}$$
