Answer
$\dfrac{\pi}{2}$
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: $$\ Volume =\iint_{D} f(x,y) \ dA$$
$f(x,y) =1-x^2-y^2$
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.
Thus, we can express the region $D$ using the point of intersection as follows:
$$D=\left\{ (x, y) | -1 \leq x \leq 1 , -\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}\right\}$$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA\\=\iint_{D} (1-x^2-y^2) \ dA \\= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx $$
Now, we will use a calculator to compute the integral:
$\ Volume = \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx \approx \dfrac{\pi}{2}$