Answer
$\dfrac{49 \pi}{8}$
Work Step by Step
Consider $V=\iint_{D} (4-x^2-y^2)-(1-x-y) \ d A$
or, $=\iint_{D} (3-x^2-y^2+x+y) \ d A$
Set $x=\dfrac{1}{2}+r \cos \theta; y= \dfrac{1}{2}+r \sin \theta$
Thus, $V=\int_{0}^{2 \pi} \int_{0}^{\sqrt {7/2}} (3-(\dfrac{1}{2}+r \cos \theta)^2-(\dfrac{1}{2}+r \sin \theta)^2+(\dfrac{1}{2}+r \cos \theta)+(\dfrac{1}{2}+r \sin \theta)) \ r \ dr \ d\theta $
or, $=2\pi \int_{0}^{\sqrt {7/2}} \dfrac{7r}{2}-r^3 \ dr$
or, $=2\pi[\dfrac{7r^2}{4}-\dfrac{r^4}{4}]_{0}^{\sqrt {7/2}} $
or, $=2 \pi [\dfrac{49}{8}-\dfrac{49}{16}] $
or, $=\dfrac{49 \pi}{8}$
