To convert from rectangular coordinates $(x, y, z)$ to spherical coordinates $(\rho, \theta, \phi)$, we use the following formulas:

1. $\rho = \sqrt{x^2 + y^2 + z^2}$
2. $\theta = \arctan\left(\frac{y}{x}\right)$ (adjusted for the quadrant of $(x, y)$)
3. $\phi = \arccos\left(\frac{z}{\rho}\right)$

(a) Convert $(x, y, z) = (0, 6, -6)$:

1. Compute $\rho$: $\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{0^2 + 6^2 + (-6)^2} = \sqrt{0 + 36 + 36} = \sqrt{72} = 6\sqrt{2}.$

2. Compute $\theta$: Since $x = 0$ and $y = 6$, $\theta = \arctan\left(\frac{y}{x}\right)$ is undefined. However, in this case, $\theta = \frac{\pi}{2}$ (positive y-axis).

3. Compute $\phi$: $\phi = \arccos\left(\frac{z}{\rho}\right) = \arccos\left(\frac{-6}{6\sqrt{2}}\right) = \arccos\left(-\frac{1}{\sqrt{2}}\right).$ $-\frac{1}{\sqrt{2}}$ corresponds to an angle of $\phi = \frac{3\pi}{4}$.

Result:
$(\rho, \theta, \phi) = \left(6\sqrt{2}, \frac{\pi}{2}, \frac{3\pi}{4}\right).$

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(b) Convert $(x, y, z) = (-6, 6, 6)$:

1. Compute $\rho$: $\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{(-6)^2 + 6^2 + 6^2} = \sqrt{36 + 36 + 36} = \sqrt{108} = 6\sqrt{3}.$

2. Compute $\theta$: $\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{6}{-6}\right) = \arctan(-1).$ Since the point $(-6, 6)$ is in the second quadrant, $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

3. Compute $\phi$: $\phi = \arccos\left(\frac{z}{\rho}\right) = \arccos\left(\frac{6}{6\sqrt{3}}\right) = \arccos\left(\frac{1}{\sqrt{3}}\right).$ $\frac{1}{\sqrt{3}}$ corresponds to an angle of $\phi = \frac{\pi}{3}$.

Result:
$(\rho, \theta, \phi) = \left(6\sqrt{3}, \frac{3\pi}{4}, \frac{\pi}{3}\right).$

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Let me know if you want further explanations or verification!

Related Questions:

1. How do you interpret spherical coordinates geometrically?
2. What are the advantages of spherical coordinates in certain applications?
3. How do you determine the correct quadrant for $\theta$?
4. How are spherical coordinates used in physics, such as for electric fields?
5. What are the differences between cylindrical and spherical coordinates?

Tip: Always double-check which quadrant your point lies in when calculating $\theta$ to avoid errors!