Question

Use polar coordinates to find the volume of the given

Solid Geometry
ANSWERED
asked 2021-08-07

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

Expert Answers (1)

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 \(\displaystyle{V}={2}{\int_{{{0}}}^{{{2}\pi}}}{\int_{{{0}}}^{{{2}}}}{r}\sqrt{{{64}-{4}{r}^{{2}}}}{r}\ {d}{r}\ {d}\theta\)
= \(\displaystyle{2}{\int_{{{0}}}^{{{2}\pi}}}{{\left[-{\frac{{{1}}}{{{12}}}}{\left({64}-{4}{r}^{{{2}}}\right)}^{{{\frac{{{2}}}{{{3}}}}}}\right]}_{{0}}^{{2}}}{d}\theta\)
\(\displaystyle=-{\frac{{{1}}}{{{6}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left[{\left({64}-{4}{\left({2}\right)}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\left({64}-{4}{\left({0}\right)}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\right]}{d}\theta\)
\(\displaystyle=-{\frac{{{1}}}{{{6}}}}{\int_{{{0}}}^{{{2}\pi}}}{\left[{\left({48}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\left({64}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\right]}{d}\theta\)
\(\displaystyle=-{\frac{{{1}}}{{{6}}}}{\left[{48}\sqrt{{48}}-{512}\right]}{{\left[\theta\right]}_{{0}}^{{{2}\pi}}}\)
\(\displaystyle=-{\frac{{{1}}}{{{6}}}}{\left[{192}\sqrt{{3}}-{512}\right]}{\left[{2}\pi-{0}\right]}\)
\(\displaystyle={\frac{{{1}}}{{{6}}}}{\left[{512}-{192}\sqrt{{3}}\right]}{\left[{2}\pi\right]}\)
\(\displaystyle={\frac{{{64}\pi}}{{{3}}}}{\left({8}-{3}\sqrt{{3}}\right)}\)
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