# Change from rectangular to cylindrical coordinates. (Let r\geq0 and 0\leq\theta\leq2\pi.) a) (-2, 2, 2) b) (-9,9\sqrt{3,6}) c) Use cylindrical coordin

Change from rectangular to cylindrical coordinates. (Let $r\ge 0$ and $0\le \theta \le 2\pi$.)
a) $\left(-2,2,2\right)$
b) $\left(-9,9\sqrt{3,6}\right)$
c) Use cylindrical coordinates.
Evaluate
$\int \int {\int }_{E}xdV$
where E is enclosed by the planes $z=0$ and
$z=x+y+10$
and by the cylinders
${x}^{2}+{y}^{2}=16$ and ${x}^{2}+{y}^{2}=36$
d) Use cylindrical coordinates.
Find the volume of the solid that is enclosed by the cone
$z=\sqrt{{x}^{2}+{y}^{2}}$
and the sphere
${x}^{2}+{y}^{2}+{z}^{2}=8$.
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Step 1
Change from rectangular to cylindrical coordinates.
Let $r\ge 0$ and $0\le \theta \le 2\pi$
a) $\left(x,y,z\right)=\left(-2,2,2\right)$
Use cylindrical coordinates.
$r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{\left(-2{\right)}^{3}+{2}^{3}}=2\sqrt{2}$
$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{2}{-2}\right)\left(-1\right)=\frac{3\pi }{4}$
$z=z=2$
Therefore, the required coordinates is,
$\left(r,\theta ,z\right)=\left(2\sqrt{2},\frac{2\pi }{4},2\right)$
Step 2
b) $\left(x,y,z\right)=\left(-9.9\sqrt{3.6}\right)$
Use cylindrical coordinates.
$r=\sqrt{{x}^{2}+{y}^{2}}={\sqrt{\left(-9{\right)}^{2}+\left(9\sqrt{3}\right)}}^{2}=18$
$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{9\sqrt{3}}{-9}\right)=\mathrm{arctan}\left(-\sqrt{3}\right)=\frac{2\pi }{3}$
$z=z=6$
Therefore, the required coordinates is,
$\left(r,\theta ,z\right)=\left(18,\frac{2\pi }{3},6\right)$
Step 3
c) Use cylindrical coordinates, to evaluate $\int {\int }_{E}\int xdV$
Where E is enclosed by the planes $z=0$ and $z=x+y+10$ and by the cylinders ${x}^{2}+{y}^{2}=16$ and ${x}^{2}+{y}^{2}=36$
$16\le {x}^{2}+{y}^{2}\le 36$
$⇒16\le {r}^{2}\le 36$ As ${x}^{2}+{y}^{2}={r}^{2}$
$⇒4\le r\le 6$
and $0\le \theta \le 2\pi$
and $0\le z\le x+y+10$
$⇒0\le z\le r\mathrm{cos}\theta +r\mathrm{sin}\theta +10$ As $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$
and $dV=dxdydz=rdrd\theta dz$
So, $\int {\int }_{E}\int xdV={\int }_{0}^{2a}{\int }_{4}^{6}{\int }_{0}^{r\mathrm{cos}\theta +r\mathrm{sin}\theta +10}r\mathrm{cos}\theta \left(rdrd\theta dz\right)$
$={\int }_{0}^{2\pi }{\int }_{4}^{6}{\int }_{0}^{r\mathrm{cos}\theta +r\mathrm{sin}\theta +10}{r}^{2}\mathrm{cos}\theta drd\theta dz$
$={\int }_{0}^{2\pi }{\int }_{4}^{6}{r}^{2}\left(r\mathrm{cos}\theta +r\mathrm{sin}\theta +10\right)\mathrm{cos}\theta drd\theta$
$={\int }_{0}^{2\pi }{\int }_{4}^{6}\left({r}^{3}\left({\mathrm{cos}}^{2}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \right)+10{r}^{2}\mathrm{cos}\theta \right)drd\theta$