Change from rectangular to cylindrical coordinates. (Let r\geq0 and 0\leq\theta\l

nagasenaz 2021-08-18 Answered
Change from rectangular to cylindrical coordinates. (Let \(\displaystyle{r}\geq{0}\) and \(\displaystyle{0}\leq\theta\leq{2}\pi\).)
a) \(\displaystyle{\left(-{2},{2},{2}\right)}\)
b) \(\displaystyle{\left(-{9},{9}\sqrt{{{3},{6}}}\right)}\)
c) Use cylindrical coordinates.
Evaluate
\(\displaystyle\int\int\int_{{{E}}}{x}{d}{V}\)
where E is enclosed by the planes \(\displaystyle{z}={0}\) and
\(\displaystyle{z}={x}+{y}+{10}\)
and by the cylinders
\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={16}\) and \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={36}\)
d) Use cylindrical coordinates.
Find the volume of the solid that is enclosed by the cone
\(\displaystyle{z}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\)
and the sphere
\(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}+{z}^{{{2}}}={8}\).

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Expert Answer

Luvottoq
Answered 2021-08-19 Author has 13710 answers
Step 1
Change from rectangular to cylindrical coordinates.
Let \(\displaystyle{r}\geq{0}\) and \(\displaystyle{0}\leq\theta\leq{2}\pi\)
a) \(\displaystyle{\left({x},{y},{z}\right)}={\left(-{2},{2},{2}\right)}\)
Use cylindrical coordinates.
\(\displaystyle{r}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}=\sqrt{{{\left(-{2}\right)}^{{{3}}}+{2}^{{{3}}}}}={2}\sqrt{{{2}}}\)
\(\displaystyle\theta={\arctan{{\left({\frac{{{y}}}{{{x}}}}\right)}}}={\arctan{{\left({\frac{{{2}}}{{-{2}}}}\right)}}}{\left(-{1}\right)}={\frac{{{3}\pi}}{{{4}}}}\)
\(\displaystyle{z}={z}={2}\)
Therefore, the required coordinates is,
\(\displaystyle{\left({r},\theta,{z}\right)}={\left({2}\sqrt{{{2}}},{\frac{{{2}\pi}}{{{4}}}},{2}\right)}\)
Step 2
b) \(\displaystyle{\left({x},{y},{z}\right)}={\left(-{9.9}\sqrt{{{3.6}}}\right)}\)
Use cylindrical coordinates.
\(\displaystyle{r}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}=\sqrt{{{\left(-{9}\right)}^{{{2}}}+{\left({9}\sqrt{{{3}}}\right)}}}^{{{2}}}={18}\)
\(\displaystyle\theta={\arctan{{\left({\frac{{{y}}}{{{x}}}}\right)}}}={\arctan{{\left({\frac{{{9}\sqrt{{{3}}}}}{{-{9}}}}\right)}}}={\arctan{{\left(-\sqrt{{{3}}}\right)}}}={\frac{{{2}\pi}}{{{3}}}}\)
\(\displaystyle{z}={z}={6}\)
Therefore, the required coordinates is,
\(\displaystyle{\left({r},\theta,{z}\right)}={\left({18},{\frac{{{2}\pi}}{{{3}}}},{6}\right)}\)
Step 3
c) Use cylindrical coordinates, to evaluate \(\displaystyle\int\int_{{{E}}}\int{x}{d}{V}\)
Where E is enclosed by the planes \(\displaystyle{z}={0}\) and \(\displaystyle{z}={x}+{y}+{10}\) and by the cylinders \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={16}\) and \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={36}\)
\(\displaystyle{16}\leq{x}^{{{2}}}+{y}^{{{2}}}\leq{36}\)
\(\displaystyle\Rightarrow{16}\leq{r}^{{{2}}}\leq{36}\) As \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={r}^{{{2}}}\)
\(\displaystyle\Rightarrow{4}\leq{r}\leq{6}\)
and \(\displaystyle{0}\leq\theta\leq{2}\pi\)
and \(\displaystyle{0}\leq{z}\leq{x}+{y}+{10}\)
\(\displaystyle\Rightarrow{0}\leq{z}\leq{r}{\cos{\theta}}+{r}{\sin{\theta}}+{10}\) As \(\displaystyle{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}}\)
and \(\displaystyle{d}{V}={\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}={r}{d}{r}{d}\theta{\left.{d}{z}\right.}\)
So, \(\displaystyle\int\int_{{{E}}}\int{x}{d}{V}={\int_{{{0}}}^{{{2}{a}}}}{\int_{{{4}}}^{{{6}}}}{\int_{{{0}}}^{{{r}{\cos{\theta}}+{r}{\sin{\theta}}+{10}}}}{r}{\cos{\theta}}{\left({r}{d}{r}{d}\theta{\left.{d}{z}\right.}\right)}\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{\int_{{{4}}}^{{{6}}}}{\int_{{{0}}}^{{{r}{\cos{\theta}}+{r}{\sin{\theta}}+{10}}}}{r}^{{{2}}}{\cos{\theta}}{d}{r}{d}\theta{\left.{d}{z}\right.}\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{\int_{{{4}}}^{{{6}}}}{r}^{{{2}}}{\left({r}{\cos{\theta}}+{r}{\sin{\theta}}+{10}\right)}{\cos{\theta}}{d}{r}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{\int_{{{4}}}^{{{6}}}}{\left({r}^{{{3}}}{\left({{\cos}^{{{2}}}\theta}+{\sin{\theta}}{\cos{\theta}}\right)}+{10}{r}^{{{2}}}{\cos{\theta}}\right)}{d}{r}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{{\left({\frac{{{1}}}{{{4}}}}{r}^{{{4}}}{\left({{\cos}^{{{2}}}\theta}+{\sin{\theta}}{\cos{\theta}}\right)}+{\frac{{{10}}}{{{3}}}}{r}^{{{3}}}{\cos{\theta}}\right)}_{{{4}}}^{{{6}}}}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{\left({\frac{{{1}}}{{{4}}}}{\left({6}^{{{4}}}-{4}^{{{4}}}\right)}{\left({{\cos}^{{{2}}}\theta}+{\sin{\theta}}{\cos{\theta}}\right)}+{\frac{{{10}}}{{{3}}}}{\left({6}^{{{3}}}-{4}^{{{3}}}\right)}{\cos{\theta}}\right)}{d}\theta\)
\(\displaystyle={\int_{{{0}}}^{{{2}\pi}}}{\left({260}{\left({{\cos}^{{{2}}}\theta}+{\sin{\theta}}{\cos{\theta}}\right)}+{\frac{{{1520}}}{{{3}}}}{\cos{\theta}}\right)}{d}\theta\)
\(\displaystyle={260}\pi\)
Step 4
d) Use cylindrical coordinates.
Find the volume of the solid that is enclosed by the cone
\(\displaystyle{z}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\) and the sphere \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}+{z}^{{{2}}}={8}\)
Use cylindrical polar coordinates,
\(\displaystyle{x}={r}{\cos{\theta}},{y}={r}{\sin{\theta}},{z}={z}\)
\(\displaystyle\Rightarrow{x}^{{{2}}}+{y}^{{{2}}}={r}^{{{2}}},{d}{V}={\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}={r}{d}{r}{d}\theta{\left.{d}{z}\right.}\)
Limits:
As \(\displaystyle{z}=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}}}\) and \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}+{z}^{{{2}}}={8}\)
\(\displaystyle\Rightarrow{2}{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}={8}\)
\(\displaystyle\Rightarrow{x}^{{{2}}}+{y}^{{{2}}}={4}\)
\(\displaystyle{r}^{{{2}}}={4}\)
\(\displaystyle\Rightarrow{r}={2}\)
So, \(\displaystyle{E}=\le{f}{t}{\left\lbrace{\left({r},\theta,{z}\right)}:{0}\leq{r}\leq{2},\ {0}\leq\theta\leq{2}\pi,\ {r}\leq{z}\leq\sqrt{{{8}-{r}^{{{2}}}}}{r}{i}{g}{h}{t}\right\rbrace}\)
Volume \(\displaystyle=\int\int_{{{E}}}\int{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
\(\displaystyle={\int_{{{r}={0}}}^{{{2}}}}{\int_{{\theta={0}}}^{{{2}\pi}}}{\int_{{{z}={r}}}^{{\sqrt{{{8}-{r}^{{{2}}}}}}}}{r}{d}{r}{d}\theta{\left.{d}{z}\right.}\)
\(\displaystyle={\int_{{{r}={0}}}^{{{2}}}}{\int_{{\theta={0}}}^{{{2}\pi}}}{r}{\left(\sqrt{{{8}-{r}^{{{2}}}}}-{r}\right)}{d}{r}{d}\theta\)
\(\displaystyle={\left({\int_{{{r}={0}}}^{{{2}}}}{\left({r}\sqrt{{{8}-{r}^{{{2}}}}}-{r}^{{{2}}}\right)}{d}{r}\right)}{\left({\int_{{\theta={0}}}^{{{2}\pi}}}{d}\theta\right)}\)
\(\displaystyle={\left[{\frac{{{1}}}{{{3}}}}{{\left(-{\left({8}-{r}^{{{2}}}\right)}^{{\frac{{3}}{{2}}}}-{r}^{{{3}}}\right]}_{{{0}}}^{{{2}}}}{{\left[\theta\right]}_{{{0}}}^{{{2}\pi}}}\right.}\)
\(\displaystyle={\left({\frac{{{1}}}{{{3}}}}{\left(-{\left({8}-{4}\right)}^{{\frac{{3}}{{2}}}}-{8}+{\left({8}\right)}^{{\frac{{3}}{{2}}}}\right)}\right)}{\left({2}\pi\right)}\)
\(\displaystyle={\left({\left(\sqrt{{{2}}}-{1}\right)}{\frac{{{16}}}{{{3}}}}\right)}{\left({2}\pi\right)}\)
\(\displaystyle={\frac{{{32}{\left(\sqrt{{{2}}}-{1}\right)}\pi}}{{{3}}}}\)
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