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) $$(-1,\ 1,\ -2)(\rho, \theta, \phi)=?$$

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 $$(\sqrt{6},\ 45^{\circ},\ 144.74^{\circ})$$