Question

Use polar coordinates to find the volume of the given

Solid Geometry

Use polar coordinates to find the volume of the given solid. Inside both the cylinder $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={4}{\quad\text{and the elepsoid}\quad}{4}{x}^{{{2}}}+{4}{y}^{{{2}}}+{z}^{{{2}}}={64}.$$

2021-08-08
Consider the cylinder,$$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={4}$$ and the ellipsoid, $$\displaystyle{4}{x}^{{{2}}}+{4}{y}^{{{2}}}+{z}^{{{2}}}={64}$$ In polar coordinates, we know that $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={r}^{{{2}}}$$ so the ellipsoid gives: $$\displaystyle{z}^{{{2}}}={64}-{4}{r}^{{{2}}}$$
$$\displaystyle\Rightarrow{z}=\pm\sqrt{{{64}-{4}{r}^{{2}}}}$$
So, the volume of the solid is given by V=$$\displaystyle{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{\left(\sqrt{{{64}-{4}{r}^{{2}}}}-{\left(-\sqrt{{{64}-{4}{r}^{{2}}}}\right)}\right)}{r}\ {d}{r}\ {d}\theta$$
=$$\displaystyle{2}{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{r}\sqrt{{{64}-{4}{r}^{{2}}}}{r}\ {d}{r}\ {d}\theta$$
to solve this integral, we substitute, $$\displaystyle{64}-{4}{r}^{{2}}={t}\ -{8}{r}{d}{r}={\left.{d}{t}\right.}\Rightarrow{r}{\left.{d}{t}\right.}={\frac{{-{1}}}{{{8}}}}{\left.{d}{t}\right.}$$
hence the indefinite integral is $$\displaystyle\int{r}\sqrt{{{64}-{4}{r}^{{2}}}}=\int\sqrt{{{t}}}\cdot{\frac{{-{1}}}{{{8}}}}{\left.{d}{t}\right.}=-{\frac{{{1}}}{{{8}}}}\cdot{\frac{{{2}}}{{{3}}}}{t}^{{{\frac{{{3}}}{{{2}}}}}}=-{\frac{{{1}}}{{{12}}}}{\left({64}-{4}{r}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}$$ so on applying the limit, the volume becomes $$\displaystyle{V}={2}{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{r}\sqrt{{{64}-{4}{r}^{{2}}}}{r}\ {d}{r}\ {d}\theta$$
= $$\displaystyle{2}{\int_{{{0}}}^{{{2}\pi}}}{{\left[-{\frac{{{1}}}{{{12}}}}{\left({64}-{4}{r}^{{{2}}}\right)}^{{{\frac{{{2}}}{{{3}}}}}}\right]}_{{0}}^{{2}}}{d}\theta$$
$$\displaystyle=-{\frac{{{1}}}{{{6}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left[{\left({64}-{4}{\left({2}\right)}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\left({64}-{4}{\left({0}\right)}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\right]}{d}\theta$$
$$\displaystyle=-{\frac{{{1}}}{{{6}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left[{\left({48}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\left({64}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\right]}{d}\theta$$
$$\displaystyle=-{\frac{{{1}}}{{{6}}}}{\left[{48}\sqrt{{48}}-{512}\right]}{{\left[\theta\right]}_{{0}}^{{{2}\pi}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{6}}}}{\left[{192}\sqrt{{3}}-{512}\right]}{\left[{2}\pi-{0}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{6}}}}{\left[{512}-{192}\sqrt{{3}}\right]}{\left[{2}\pi\right]}$$
$$\displaystyle={\frac{{{64}\pi}}{{{3}}}}{\left({8}-{3}\sqrt{{3}}\right)}$$