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