Consider the solid in xyz-space, which contains all points (x, y, z) whose z-coordinate satisfies...

Step 1
${\displaystyle \int \int \int dv}$
${\displaystyle \Rightarrow \int \int \int dzdydx}$
${\displaystyle z=0}$ to ${\displaystyle z=4-x^2-y^2}$
${\displaystyle \Rightarrow \int \int \left({\int }_0^{4-x^2-y^2}dz\right)dydx}$
${\displaystyle \Rightarrow \int \int (4-x^2-y^2)dydx}$
Change in folar co-ordinates, tre get
${\displaystyle x=r\cos \theta }$
${\displaystyle y=r\sin \theta }$
${\displaystyle r=0}$ to ${\displaystyle r=2}$
${\displaystyle dxdy=rdrd\theta }$
${\displaystyle \theta =0}$ to ${\displaystyle \theta =2\pi }$
${\displaystyle \Rightarrow {\int }_0^{2\pi }{\int }_0^2(4-r^2)rdrd\theta }$
Let ${\displaystyle 4-r^2=t\Rightarrow rdr=\frac{-dt}{2}}$
${\displaystyle t=4}$ to ${\displaystyle t=4-4=0}$
${\displaystyle \Rightarrow {\int }_0^{2\pi }({\int }_4^0-t\frac{dt}{2})d\theta \Rightarrow \frac{1}{2}{\int }_0^{2\pi }\left({\int }_0^{4}tdt\right)d\theta }$
${\displaystyle \Rightarrow \frac{1}{2}{\int }_0^{2\pi }{\left(\frac{t^2}{2}\right)}_0^{4}d\theta \Rightarrow \frac{1}{2}{\int }_0^{2\pi }8d\theta }$
${\displaystyle \Rightarrow 4{\int }_0^{2\pi }d\theta \Rightarrow 4\left(2\pi \right)}$
${\displaystyle \Rightarrow 2\pi }$

Here, take limit: ${\displaystyle 0\le 4-x^2-y^2}$
${\displaystyle x^2+y^2\le 4}$
And we know,
${\displaystyle {(4-x^2-y^2)}_{max}=4}$ when $x=0andy=0$
So, ${\displaystyle 0\le z\le 4}$
For each value of ${\displaystyle z\ge 0}$, it will be a disk of at each value of ${\displaystyle z\in R}$.
Now, volume in ${\displaystyle =\frac{\pi }{B}{r}^{2}h}$
${\displaystyle =\frac{\pi }{B}\times {\left(2\right)}^2\times \left(4\right)=16\frac{\pi }{B}}$

Step 1
Consider the solid  in xyz-space which contains all points (x,y,z) whose z-coordinate satisfies $0\le z\le 4-x^2-y^2$
Step 2
The solid is paraboloid.
We use cylindrical coordinates

$x=r\cos t,y=r\sin t,0\le z\le 4-r^2,0\le t\le 2\pi$
and $z=0gives4-r^2=0$
$\Rightarrow r^2=4$
$\Rightarrow r=2$
$\Rightarrow 0\le r\le 2$
The volume of the solid is given by
$\begin{array}{}V=\int \int {\int }_{S}dV\\ ={\int }_0^2{\int }_0^{2\pi }{\int }_0^{4-r^2}rdzdtdr\\ ={\int }_0^2{\int }_0^{2\pi }|z{|}_0^{4-r^2}rdtdr\\ ={\int }_0^2{\int }_0^{2\pi }(4-r^2)rdtdr\\ ={\int }_0^{2}r(4-r^2)[t{]}_0^{2\pi }dr\\ ={\int }_0^{2}r(4-r^2)[2\pi -0]dr\\ =2\pi {\int }_0^2(4r-r^3)dr\\ =2\pi [2r^2-\frac{r^4}{4}{]}_0^2\\ =2\pi [2\times 2^2-\frac{2^4}{4}-0]\\ =2\pi [8-4]\\ =8\pi \end{array}$
Thus, volume of the solid is $8\pi$.
Hence, option (c) is correct.