Question

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 \(\rho\geq0,\ 0\leq\theta\leq 2\pi\), and \(0\leq\phi\leq\pi.)\)
a) \((0,\ -8,\ 0)(\rho,\theta, \phi)=?\)
b) \((-1,\ 1,\ -2)(\rho, \theta, \phi)=?\)

Answers (1)

2021-05-15

Step 1
a) Consider the rectangular coordinate \((0,\ -8,\ 0)\)
The objective is to convert the rectangular coordinate to spherical coordinate \((\rho, \theta,\phi)\)
Here, \(\rho\geq0,\ 0\leq\theta\leq2\pi,\ 0\leq\phi\leq\pi\)
Consider the following statement:
The point P in \(\mathbb{R}^{3}\) with rectangular coordinates \((x,y,z)\) and spheerical coordinates \((\rho,\theta\phi)\)
The relation between rectangular coordinates \((x,y,z)\) and spherical coordinates \((\rho,\theta\phi)\) is,
\(x=\rho\sin\phi\cos\theta\)
\(y=\rho\sin\phi\sin\theta\)
\(z=\rho\cos\phi\)
Step 2
The rectangular coordinate is \((x,y,z)=(0,\ -8,\ 0)\)
That is, \(x=0,\ y=-8,\ z=0.\)
Thus, from the statement,
\(\rho=\sqrt{x^{2}+y^{2}+z^{2}}\)
\(=\sqrt{0^{2}+(-8)^{2}+0^{2}}\)
\(\{x=0,\ y=-8,z=0\}\)
\(=\sqrt{64}\)
\(=8\)
From the statement,
\(z=\rho\cos\phi\)
\(0=8\cos\phi\)
\(\{z=0,\rho=8\}\)
\(\cos\phi=0\)
\(\phi=\frac{\pi}{2}\)
\(\{0\leq\phi\leq\pi\}\)
Step 3
From the statement,
\(y=\rho\sin\phi\sin\theta\)
\(\sin\theta=\frac{y}{\rho\sin\phi}\)
Substitute \(y=-8,\rho=8,\phi=\frac{\pi}{2}\)
\(\sin\theta=\frac{-8}{8\sin\frac{\pi}{2}}\)
\(=-1\)
\(\left\{\sin\frac{\pi}{2}=1\right\}\)
\(\theta=\frac{3\pi}{2}\)
\(\left\{0\leq\theta\leq\frac{\pi}{2}\right\}\)
Hence, the spherical coordinate of the point \((0,\ -8,\ 0)\) is \(\left(8,\ \frac{3\pi}{2},\ \frac{\pi}{2}\right)\)
Step 4
Consider the rectangular coordinate \((-1,\ 1,\ -2)\)
The objective is to convert the rectangular coordinate to spherical coordinate \((\rho,\theta,\phi)\).
Consider the following statement:
The point P in \(\mathbb{R}^{3}\) with rectangular coordinates \((x,y,z)\) and spherical coordinates \((\rho,\theta,\phi)\).
The relation between rectangular coordinates \((x,y,z)\) and spherical coordinates \((\rho,\theta,\phi)\) is,
\(x=\rho\sin\phi\cos\theta\)
\(y=\rho\sin\phi\sin\theta\)
\(z=\rho\cos\phi\)
The rectangular coordinate is \((x,y,z)=(-1,\ 1,\ -2)\)
That is, \(x=-1,\ y=1,\ z=-2.\)
Thus, from the statement,
\(\rho=\sqrt{x^{2}+y^{2}+z^{2}}\)
\(=\sqrt{(-1)^{2}+(1)^{2}+(-2)^{2}}\)
\(\{x=-1,\ y=1,\ z=-2\}\)
\(=\sqrt{1+1+4}\)
\(=\sqrt{6}\)
From the statement,
\(z=\rho\cos\phi\)
\(-2=\sqrt{6}\cos\phi\)
\(\cos\phi=\frac{-2}{\sqrt{6}}\)
\(\phi=\cos^{-1}\left(\frac{-2}{\sqrt{6}}\right)\)
\(\phi=144.74^{\circ}\)
\(\{0\leq\phi\leq\pi\}\)
From the statement,
\(y=\rho\sin\phi\sin\theta\)
\(\sin\theta=\frac{y}{\rho\sin\phi}\)
Substitute \(y=1,\rho=\sqrt{6},\phi=144.74^{\circ}\)
\(\sin\theta=\frac{1}{\sqrt{6}(144.74^{\circ})}\)
\(=\frac{1}{\sqrt{6}(0.577287)}\)
\(\sin\theta=0.7071\)
\(\theta=\sin^{-1}(0.7071)\)
\(=45^{\circ}\)
\(\left\{0\leq\theta\leq\frac{\pi}{2}\right\}\)
Hence, the spherical coordinate of the point \((-1,\ 1,\ -2)\) is \((\sqrt{6},\ 45^{\circ},\ 144.74^{\circ})\)

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