Step 1

Change from rectangular to cylindrical coordinates.

Let

$r\ge 0$ and

$0\le \theta \le 2\pi $
a)

$(x,y,z)=(-2,2,2)$
Use cylindrical coordinates.

$r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{(-2{)}^{3}+{2}^{3}}=2\sqrt{2}$
$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{2}{-2}\right)(-1)=\frac{3\pi}{4}$
$z=z=2$
Therefore, the required coordinates is,

$(r,\theta ,z)=(2\sqrt{2},\frac{2\pi}{4},2)$
Step 2

b)

$(x,y,z)=(-9.9\sqrt{3.6})$
Use cylindrical coordinates.

$r=\sqrt{{x}^{2}+{y}^{2}}={\sqrt{(-9{)}^{2}+(9\sqrt{3})}}^{2}=18$
$\theta =\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{9\sqrt{3}}{-9}\right)=\mathrm{arctan}(-\sqrt{3})=\frac{2\pi}{3}$
$z=z=6$
Therefore, the required coordinates is,

$(r,\theta ,z)=(18,\frac{2\pi}{3},6)$
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

${x}^{2}+{y}^{2}=36$
$16\le {x}^{2}+{y}^{2}\le 36$
$\Rightarrow 16\le {r}^{2}\le 36$ As

${x}^{2}+{y}^{2}={r}^{2}$
$\Rightarrow 4\le r\le 6$
and

$0\le \theta \le 2\pi $
and

$0\le z\le x+y+10$
$\Rightarrow 0\le z\le r\mathrm{cos}\theta +r\mathrm{sin}\theta +10$ As

$x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta $
and

$dV=dxdydz=rdrd\theta dz$
So,

$\int {\int}_{E}\int xdV={\int}_{0}^{2a}{\int}_{4}^{6}{\int}_{0}^{r\mathrm{cos}\theta +r\mathrm{sin}\theta +10}r\mathrm{cos}\theta (rdrd\theta dz)$
$={\int}_{0}^{2\pi}{\int}_{4}^{6}{\int}_{0}^{r\mathrm{cos}\theta +r\mathrm{sin}\theta +10}{r}^{2}\mathrm{cos}\theta drd\theta dz$
$={\int}_{0}^{2\pi}{\int}_{4}^{6}{r}^{2}(r\mathrm{cos}\theta +r\mathrm{sin}\theta +10)\mathrm{cos}\theta drd\theta $
$={\int}_{0}^{2\pi}{\int}_{4}^{6}({r}^{3}({\mathrm{cos}}^{2}\theta +\mathrm{sin}\theta \mathrm{cos}\theta )+10{r}^{2}\mathrm{cos}\theta )drd\theta $
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