Change from rectangular to spherical coordinates. (Let 𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋, and 0 ≤ 𝜑 ≤ 𝜋.)

(a)

(0, 6, −6)

(𝜌, 𝜃, 𝜑) =

(b)

(−6, 6, 6

6

)

(𝜌, 𝜃, 𝜑) =

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

---

### (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**:  
\[
(\rho, \theta, \phi) = \left(6\sqrt{3}, \frac{3\pi}{4}, \frac{\pi}{3}\right).
\]

---

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!

Change from rectangular to spherical coordinates. (Let 𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋, and 0 ≤ 𝜑 ≤ 𝜋.) (a) (0, 6, −6) (𝜌, 𝜃, 𝜑) = ? (b) (−6, 6, 6) (𝜌, 𝜃, 𝜑) = ?

Learn how to convert rectangular coordinates to spherical coordinates with step-by-step solutions for (0, 6, -6) and (-6, 6, 6). This guide covers the key formulas, including 𝜌 = √(x² + y² + z²), 𝜃 = arctan(y / x), and 𝜑 = arccos(z / 𝜌). Suitable for undergraduate students and enthusiasts of 3D geometry.

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