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)=?

CoormaBak9 2021-08-21 Answered
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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Expert Answer

Clara Reese
Answered 2021-08-22 Author has 16881 answers

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 \(\displaystyle{\left(\sqrt{{{6}}},\ {45}^{{\circ}},\ {144.74}^{{\circ}}\right)}\)

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