Answer
$\approx \dfrac{\pi}{2}$
Work Step by Step
$f(x,y) =1-x^2-y^2$ and the plane and paraboloid intersect along the circle $x^2+y^2=1$ and $D$ is the projection of the surface on the xy-plane.
The volume under the surface $z=f(x,y)$ and above the region $D$ in the $xy$-plane can be expressed as follows: $$\ Volume =\iint_{D} f(x,y) \ dA$$
and the region $D$ using the point of intersection can be expressed as follows: $D=\left\{ (x, y) | -1 \leq x \leq 1 , -\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}\right\}$
Now, $\ Volume; V =\iint_{D} f(x,y) \ dA=\iint_{D} (1-x^2-y^2) \ dA $
or, $= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx $
The integral can be computed by using a calculator as follows:
$\ Volume; V = \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx \\ \approx \dfrac{\pi}{2}$
