Answer
$\dfrac{64}{3}$
Work Step by Step
We will compute the points of intersection of the parabolas $y=x^2-1$ and $y=1-x^2$ in the $xy$-plane.
Now, $x^2-1=1-x^2$ or, $ x=\pm 1$
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.
The domain $D$ can be expressed as follows:
For any x, y in the above region the plane $z=2-x-y$ is below the plane $z=10+2x+2y$.
The volume of the solid is the difference of the solid below the plane $z= 10+2x+2y$ and the solid below the plane $z=2-x-y$
$\ V =\iint_{D} f(x,y) \ dA \\= \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (10+2x+2y) \ dy \ dx - \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (2-x-y) \ dy \ dx\\ = \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (8+3x+3y) \ dy \ dx \\= \int_{-1}^{1} (8y+3xy+\dfrac{3y^2}{2})|_{x^2-1}^{1-x^2} dx \\=\int_{-1}^{1} (16-16x^2+6x -6x^3) dx\\=[16x-\dfrac{16}{3} x^3 +3x^2-\dfrac{3}{2} x^4]_{-1}^{1}\\=\dfrac{64}{3}$
