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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![**Problem 5: Calculation of Surface Area of a Sphere**
Objective: To determine the surface area of a sphere with a radius of 4 using integration techniques.
### Explanation
To find the surface area \( S \) of a sphere using calculus, we use the formula:
\[
S = \int_0^{2\pi} \int_0^{\pi} r^2 \sin \theta \, d\theta \, d\phi
\]
### Where:
- \( r \) is the radius of the sphere.
- \( \theta \) is the polar angle (0 to \( \pi \)).
- \( \phi \) is the azimuthal angle (0 to \( 2\pi \)).
- For a sphere of radius 4, substitute \( r = 4 \).
Substituting \( r = 4 \), the integral becomes:
\[
S = \int_0^{2\pi} \int_0^{\pi} 4^2 \sin \theta \, d\theta \, d\phi
\]
\[
S = \int_0^{2\pi} \int_0^{\pi} 16 \sin \theta \, d\theta \, d\phi
\]
### Solving the Integral
1. **Inner Integral**: Integrate \( 16 \sin \theta \) with respect to \( \theta \):
\[
\int_0^{\pi} 16 \sin \theta \, d\theta = 16 [-\cos \theta]_0^{\pi}
\]
\[
= 16 [(-\cos \pi) - (-\cos 0)]
\]
\[
= 16 [1 - (-1)] = 16 \times 2 = 32
\]
2. **Outer Integral**: Integrate the result with respect to \( \phi \):
\[
\int_0^{2\pi} 32 \, d\phi = 32 \phi \Big|_0^{2\pi}
\]
\[
= 32 \times 2\pi = 64\pi
\]
Thus, the surface area of the sphere with radius 4 is \( 64\pi \) square units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65e9754d-e7bf-4951-a913-18b478a6d2d4%2F6f3c2392-27ba-4a8f-88a1-5a5ba8baa7e4%2Fckrl6ieh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 5: Calculation of Surface Area of a Sphere**
Objective: To determine the surface area of a sphere with a radius of 4 using integration techniques.
### Explanation
To find the surface area \( S \) of a sphere using calculus, we use the formula:
\[
S = \int_0^{2\pi} \int_0^{\pi} r^2 \sin \theta \, d\theta \, d\phi
\]
### Where:
- \( r \) is the radius of the sphere.
- \( \theta \) is the polar angle (0 to \( \pi \)).
- \( \phi \) is the azimuthal angle (0 to \( 2\pi \)).
- For a sphere of radius 4, substitute \( r = 4 \).
Substituting \( r = 4 \), the integral becomes:
\[
S = \int_0^{2\pi} \int_0^{\pi} 4^2 \sin \theta \, d\theta \, d\phi
\]
\[
S = \int_0^{2\pi} \int_0^{\pi} 16 \sin \theta \, d\theta \, d\phi
\]
### Solving the Integral
1. **Inner Integral**: Integrate \( 16 \sin \theta \) with respect to \( \theta \):
\[
\int_0^{\pi} 16 \sin \theta \, d\theta = 16 [-\cos \theta]_0^{\pi}
\]
\[
= 16 [(-\cos \pi) - (-\cos 0)]
\]
\[
= 16 [1 - (-1)] = 16 \times 2 = 32
\]
2. **Outer Integral**: Integrate the result with respect to \( \phi \):
\[
\int_0^{2\pi} 32 \, d\phi = 32 \phi \Big|_0^{2\pi}
\]
\[
= 32 \times 2\pi = 64\pi
\]
Thus, the surface area of the sphere with radius 4 is \( 64\pi \) square units.
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