# Change from rectangular to spherical coordinates. (Let \rho\geq0,\ 0\leq\theta\leq\2\pi, and 0\leq\phi\leq\pi.)a) (0,\ -8,\ 0)(\rho,\theta, \phi)=?b) (-1,\ 1,\ -2)(\rho, \theta, \phi)=?

Change from rectangular to spherical coordinates. (Let $$\displaystyle\rho\geq{0},\ {0}\leq\theta\leq{2}\pi$$, and $$\displaystyle{0}\leq\phi\leq\pi.{)}$$
a) $$\displaystyle{\left({0},\ -{8},\ {0}\right)}{\left(\rho,\theta,\phi\right)}=?$$
b) $$\displaystyle{\left(-{1},\ {1},\ -{2}\right)}{\left(\rho,\theta,\phi\right)}=?$$

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Step 1
a) Consider the rectangular coordinate $$\displaystyle{\left({0},\ -{8},\ {0}\right)}$$
The objective is to convert the rectangular coordinate to spherical coordinate $$\displaystyle{\left(\rho,\theta,\phi\right)}$$
Here, $$\displaystyle\rho\geq{0},\ {0}\leq\theta\leq{2}\pi,\ {0}\leq\phi\leq\pi$$
Consider the following statement:
The point P in $$\displaystyle{\mathbb{{{R}}}}^{{{3}}}$$ with rectangular coordinates $$\displaystyle{\left({x},{y},{z}\right)}$$ and spheerical coordinates $$\displaystyle{\left(\rho,\theta\phi\right)}$$
The relation between rectangular coordinates $$\displaystyle{\left({x},{y},{z}\right)}$$ and spherical coordinates $$\displaystyle{\left(\rho,\theta\phi\right)}$$ is,
$$\displaystyle{x}=\rho{\sin{\phi}}{\cos{\theta}}$$
$$\displaystyle{y}=\rho{\sin{\phi}}{\sin{\theta}}$$
$$\displaystyle{z}=\rho{\cos{\phi}}$$
Step 2
The rectangular coordinate is $$\displaystyle{\left({x},{y},{z}\right)}={\left({0},\ -{8},\ {0}\right)}$$
That is, $$\displaystyle{x}={0},\ {y}=-{8},\ {z}={0}.$$
Thus, from the statement,
$$\displaystyle\rho=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}+{z}^{{{2}}}}}$$
$$\displaystyle=\sqrt{{{0}^{{{2}}}+{\left(-{8}\right)}^{{{2}}}+{0}^{{{2}}}}}$$
$$\displaystyle{\left\lbrace{x}={0},\ {y}=-{8},{z}={0}\right\rbrace}$$
$$\displaystyle=\sqrt{{{64}}}$$
$$\displaystyle={8}$$
From the statement,
$$\displaystyle{z}=\rho{\cos{\phi}}$$
$$\displaystyle{0}={8}{\cos{\phi}}$$
$$\displaystyle{\left\lbrace{z}={0},\rho={8}\right\rbrace}$$
$$\displaystyle{\cos{\phi}}={0}$$
$$\displaystyle\phi={\frac{{\pi}}{{{2}}}}$$
$$\displaystyle{\left\lbrace{0}\leq\phi\leq\pi\right\rbrace}$$
Step 3
From the statement,
$$\displaystyle{y}=\rho{\sin{\phi}}{\sin{\theta}}$$
$$\displaystyle{\sin{\theta}}={\frac{{{y}}}{{\rho{\sin{\phi}}}}}$$
Substitute $$\displaystyle{y}=-{8},\rho={8},\phi={\frac{{\pi}}{{{2}}}}$$
$$\displaystyle{\sin{\theta}}={\frac{{-{8}}}{{{8}{\sin{{\frac{{\pi}}{{{2}}}}}}}}}$$
$$\displaystyle=-{1}$$
$$\left\{\sin\frac{\pi}{2}=1\right\}$$
$$\displaystyle\theta={\frac{{{3}\pi}}{{{2}}}}$$
$$\left\{0\leq\theta\leq\frac{\pi}{2}\right\}$$
Hence, the spherical coordinate of the point $$\displaystyle{\left({0},\ -{8},\ {0}\right)}$$ is $$\displaystyle{\left({8},\ {\frac{{{3}\pi}}{{{2}}}},\ {\frac{{\pi}}{{{2}}}}\right)}$$
Step 4
Consider the rectangular coordinate $$\displaystyle{\left(-{1},\ {1},\ -{2}\right)}$$
The objective is to convert the rectangular coordinate to spherical coordinate $$\displaystyle{\left(\rho,\theta,\phi\right)}$$.
Consider the following statement:
The point P in $$\displaystyle{\mathbb{{{R}}}}^{{{3}}}$$ with rectangular coordinates $$\displaystyle{\left({x},{y},{z}\right)}$$ and spherical coordinates $$\displaystyle{\left(\rho,\theta,\phi\right)}$$.
The relation between rectangular coordinates $$\displaystyle{\left({x},{y},{z}\right)}$$ and spherical coordinates $$\displaystyle{\left(\rho,\theta,\phi\right)}$$ is,
$$\displaystyle{x}=\rho{\sin{\phi}}{\cos{\theta}}$$
$$\displaystyle{y}=\rho{\sin{\phi}}{\sin{\theta}}$$
$$\displaystyle{z}=\rho{\cos{\phi}}$$
The rectangular coordinate is $$\displaystyle{\left({x},{y},{z}\right)}={\left(-{1},\ {1},\ -{2}\right)}$$
That is, $$\displaystyle{x}=-{1},\ {y}={1},\ {z}=-{2}.$$
Thus, from the statement,
$$\displaystyle\rho=\sqrt{{{x}^{{{2}}}+{y}^{{{2}}}+{z}^{{{2}}}}}$$
$$\displaystyle=\sqrt{{{\left(-{1}\right)}^{{{2}}}+{\left({1}\right)}^{{{2}}}+{\left(-{2}\right)}^{{{2}}}}}$$
$$\displaystyle{\left\lbrace{x}=-{1},\ {y}={1},\ {z}=-{2}\right\rbrace}$$
$$\displaystyle=\sqrt{{{1}+{1}+{4}}}$$
$$\displaystyle=\sqrt{{{6}}}$$
From the statement,
$$\displaystyle{z}=\rho{\cos{\phi}}$$
$$\displaystyle-{2}=\sqrt{{{6}}}{\cos{\phi}}$$
$$\displaystyle{\cos{\phi}}={\frac{{-{2}}}{{\sqrt{{{6}}}}}}$$
$$\displaystyle\phi={{\cos}^{{-{1}}}{\left({\frac{{-{2}}}{{\sqrt{{{6}}}}}}\right)}}$$
$$\displaystyle\phi={144.74}^{{\circ}}$$
$$\displaystyle{\left\lbrace{0}\leq\phi\leq\pi\right\rbrace}$$
From the statement,
$$\displaystyle{y}=\rho{\sin{\phi}}{\sin{\theta}}$$
$$\displaystyle{\sin{\theta}}={\frac{{{y}}}{{\rho{\sin{\phi}}}}}$$
Substitute $$\displaystyle{y}={1},\rho=\sqrt{{{6}}},\phi={144.74}^{{\circ}}$$
$$\displaystyle{\sin{\theta}}={\frac{{{1}}}{{\sqrt{{{6}}}{\left({144.74}^{{\circ}}\right)}}}}$$
$$\displaystyle={\frac{{{1}}}{{\sqrt{{{6}}}{\left({0.577287}\right)}}}}$$
$$\displaystyle{\sin{\theta}}={0.7071}$$
$$\displaystyle\theta={{\sin}^{{-{1}}}{\left({0.7071}\right)}}$$
$$\displaystyle={45}^{{\circ}}$$
$$\left\{0\leq\theta\leq\frac{\pi}{2}\right\}$$
Hence, the spherical coordinate of the point $$\displaystyle{\left(-{1},\ {1},\ -{2}\right)}$$ is $$\displaystyle{\left(\sqrt{{{6}}},\ {45}^{{\circ}},\ {144.74}^{{\circ}}\right)}$$