Step 1

\(\displaystyle\int\int\int{d}{v}\)

\(\displaystyle\Rightarrow\int\int\int{\left.{d}{z}\right.}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle{z}={0}\) to \(\displaystyle{z}={4}-{x}^{{{2}}}-{y}^{{{2}}}\)

\(\displaystyle\Rightarrow\int\int{\left({\int_{{{0}}}^{{{4}-{x}^{{{2}}}-{y}^{{{2}}}}}}{\left.{d}{z}\right.}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

\(\displaystyle\Rightarrow\int\int{\left({4}-{x}^{{{2}}}-{y}^{{{2}}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)

Change in folar co-ordinates, tre get

\(\displaystyle{x}={r}{\cos{\theta}}\)

\(\displaystyle{y}={r}{\sin{\theta}}\)

\(\displaystyle{r}={0}\) to \(\displaystyle{r}={2}\)

\(\displaystyle{\left.{d}{x}\right.}{\left.{d}{y}\right.}={r}{d}{r}{d}\theta\)

\(\displaystyle\theta={0}\) to \(\displaystyle\theta={2}\pi\)

\(\displaystyle\Rightarrow{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{\left({4}-{r}^{{{2}}}\right)}{r}{d}{r}{d}\theta\)

Let \(\displaystyle{4}-{r}^{{{2}}}={t}\Rightarrow{r}{d}{r}={\frac{{-{\left.{d}{t}\right.}}}{{{2}}}}\)

\(\displaystyle{t}={4}\) to \(\displaystyle{t}={4}-{4}={0}\)

\(\displaystyle\Rightarrow{\int_{{{0}}}^{{{2}\pi}}}{\left({\int_{{{4}}}^{{{0}}}}-{t}{\frac{{{\left.{d}{t}\right.}}}{{{2}}}}\right)}{d}\theta\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left({\int_{{{0}}}^{{{4}}}}{t}{\left.{d}{t}\right.}\right)}{d}\theta\)

\(\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{{\left({\frac{{{t}^{{{2}}}}}{{{2}}}}\right)}_{{{0}}}^{{{4}}}}{d}\theta\Rightarrow{\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}\pi}}}{8}{d}\theta\)

\(\displaystyle\Rightarrow{4}{\int_{{{0}}}^{{{2}\pi}}}{d}\theta\Rightarrow{4}{\left({2}\pi\right)}\)

\(\displaystyle\Rightarrow{2}\pi\)