Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Conversion of Rectangular (Cartesian) Coordinates to Spherical Coordinates
**Problem Statement:**
Convert the following from rectangular (Cartesian) coordinates to spherical coordinates:
\[
(-\sqrt{3}, -3, 2)
\]
**Choices:**
1. \(\left( 4, \frac{5\pi}{6}, \frac{\pi}{3} \right)\)
2. \(\left( 4, \frac{\pi}{3}, \frac{4\pi}{3} \right)\)
3. \(\left( 4, \frac{4\pi}{3}, \frac{\pi}{3} \right)\)
4. \(\left( 4, \frac{\pi}{3}, \frac{7\pi}{6} \right)\)
5. \(\left( 4, \frac{2\pi}{3}, \frac{\pi}{3} \right)\)
### Explanation:
To convert from Cartesian coordinates \((x, y, z)\) to spherical coordinates \((r, \theta, \phi)\), we use the following formulas:
1. \( r = \sqrt{x^2 + y^2 + z^2} \)
2. \( \theta = \arccos\left(\frac{z}{r}\right) \)
3. \( \phi = \arctan2(y, x) \)
Where:
- \( r \) is the radial distance,
- \( \theta \) is the polar angle (measured from the positive z-axis),
- \( \phi \) is the azimuthal angle (measured from the positive x-axis in the xy-plane).
We apply these formulas to the given Cartesian coordinates \((- \sqrt{3}, -3, 2)\):
1. Calculate \( r \):
\[
r = \sqrt{(-\sqrt{3})^2 + (-3)^2 + 2^2} = \sqrt{3 + 9 + 4} = \sqrt{16} = 4
\]
2. Calculate \( \theta \):
\[
\theta = \arccos\left(\frac{2}{4}\right) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}
\]
3. Calculate \( \phi \):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bda9c1d-08ea-450e-a130-838d16131279%2F6e003ebb-0a31-45f9-92cb-e7bb330c27dd%2Fsjndxe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Conversion of Rectangular (Cartesian) Coordinates to Spherical Coordinates
**Problem Statement:**
Convert the following from rectangular (Cartesian) coordinates to spherical coordinates:
\[
(-\sqrt{3}, -3, 2)
\]
**Choices:**
1. \(\left( 4, \frac{5\pi}{6}, \frac{\pi}{3} \right)\)
2. \(\left( 4, \frac{\pi}{3}, \frac{4\pi}{3} \right)\)
3. \(\left( 4, \frac{4\pi}{3}, \frac{\pi}{3} \right)\)
4. \(\left( 4, \frac{\pi}{3}, \frac{7\pi}{6} \right)\)
5. \(\left( 4, \frac{2\pi}{3}, \frac{\pi}{3} \right)\)
### Explanation:
To convert from Cartesian coordinates \((x, y, z)\) to spherical coordinates \((r, \theta, \phi)\), we use the following formulas:
1. \( r = \sqrt{x^2 + y^2 + z^2} \)
2. \( \theta = \arccos\left(\frac{z}{r}\right) \)
3. \( \phi = \arctan2(y, x) \)
Where:
- \( r \) is the radial distance,
- \( \theta \) is the polar angle (measured from the positive z-axis),
- \( \phi \) is the azimuthal angle (measured from the positive x-axis in the xy-plane).
We apply these formulas to the given Cartesian coordinates \((- \sqrt{3}, -3, 2)\):
1. Calculate \( r \):
\[
r = \sqrt{(-\sqrt{3})^2 + (-3)^2 + 2^2} = \sqrt{3 + 9 + 4} = \sqrt{16} = 4
\]
2. Calculate \( \theta \):
\[
\theta = \arccos\left(\frac{2}{4}\right) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}
\]
3. Calculate \( \phi \):
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