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

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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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 $$\displaystyle=\int\int_{{{E}}}\int{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}$$
$$\displaystyle={\int_{{{r}={0}}}^{{{2}}}}{\int_{{\theta={0}}}^{{{2}\pi}}}{\int_{{{z}={r}}}^{{\sqrt{{{8}-{r}^{{{2}}}}}}}}{r}{d}{r}{d}\theta{\left.{d}{z}\right.}$$
$$\displaystyle={\int_{{{r}={0}}}^{{{2}}}}{\int_{{\theta={0}}}^{{{2}\pi}}}{r}{\left(\sqrt{{{8}-{r}^{{{2}}}}}-{r}\right)}{d}{r}{d}\theta$$
$$\displaystyle={\left({\int_{{{r}={0}}}^{{{2}}}}{\left({r}\sqrt{{{8}-{r}^{{{2}}}}}-{r}^{{{2}}}\right)}{d}{r}\right)}{\left({\int_{{\theta={0}}}^{{{2}\pi}}}{d}\theta\right)}$$
$$\displaystyle={\left[{\frac{{{1}}}{{{3}}}}{{\left(-{\left({8}-{r}^{{{2}}}\right)}^{{\frac{{3}}{{2}}}}-{r}^{{{3}}}\right]}_{{{0}}}^{{{2}}}}{{\left[\theta\right]}_{{{0}}}^{{{2}\pi}}}\right.}$$
$$\displaystyle={\left({\frac{{{1}}}{{{3}}}}{\left(-{\left({8}-{4}\right)}^{{\frac{{3}}{{2}}}}-{8}+{\left({8}\right)}^{{\frac{{3}}{{2}}}}\right)}\right)}{\left({2}\pi\right)}$$
$$\displaystyle={\left({\left(\sqrt{{{2}}}-{1}\right)}{\frac{{{16}}}{{{3}}}}\right)}{\left({2}\pi\right)}$$
$$\displaystyle={\frac{{{32}{\left(\sqrt{{{2}}}-{1}\right)}\pi}}{{{3}}}}$$