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.
Next, we have: $x^2-1=1-x^2 $ and $x=\pm 1$
Since, for any x, y-plane in the above region, the plane $z=2-x-y$ is below the plane $z=10+2x+2y$, so, the volume of the solid becomes: the difference of the solid below the plane $z= 10+2x+2y$ and the solid below the plane $z=2-x-y$ .
Now, $\ Volume ; V=\iint_{D} f(x,y) \ dA $
or, $= \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 $
or, $ = \int_{-1}^{1} \int_{x^2-1}^{1-x^2} (8+3x+3y) \ dy \ dx $
or, $= \int_{-1}^{1} (8y+3xy+\dfrac{3y^2}{2})|_{x^2-1}^{1-x^2} dx $
So, $Volume=\int_{-1}^{1} (16-16x^2+6x -6x^3) \ dx\\=[16x-\dfrac{16x^3}{3}+3x^2-\dfrac{3x^4}{2}]_{-1}^{1}\\=\dfrac{64}{3}$
