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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![**Calculating the Surface Area of a Rotated Curve**
To find the surface area of the surface obtained by rotating the graph of \( y = 4 \sin(x) \) for \( 0 \leq x \leq \frac{\pi}{6} \) around the x-axis, we need to express it as an integral.
**Integral Options:**
1. \[
2\pi \int_{0}^{\frac{\pi}{6}} 4 \sin(x) \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
2. \[
2\pi \int_{0}^{\frac{\pi}{6}} 4 \sin(x) \left( 1 + 4 \cos(x) \right) \, dx
\]
3. \[
4 \int_{0}^{\frac{\pi}{6}} \sin(x) \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
4. \[
2\pi \int_{0}^{\frac{\pi}{6}} \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
5. \[
\int_{0}^{\frac{\pi}{6}} \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
**Explanation of the Correct Integral:**
The surface area A of the surface obtained by rotating a curve \( y = f(x) \) around the x-axis is given by:
\[
A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
Given \( y = 4 \sin(x) \), we need to compute the term \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = 4 \cos(x)
\]
Substituting \( \frac{dy}{dx} = 4 \cos(x) \) into \( \left( \frac{dy}{dx} \right)^2 \):
\[
\left( \frac{dy}{dx} \right)^2 = (4 \cos(x))^2 = 16 \cos^2 (x)
\]
Now,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8d403f6-f90e-48f4-b6e8-074bb9cf2c2e%2Fde0a2682-f7ae-44b9-8db4-f1b2cc23d212%2Flby0lgk_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the Surface Area of a Rotated Curve**
To find the surface area of the surface obtained by rotating the graph of \( y = 4 \sin(x) \) for \( 0 \leq x \leq \frac{\pi}{6} \) around the x-axis, we need to express it as an integral.
**Integral Options:**
1. \[
2\pi \int_{0}^{\frac{\pi}{6}} 4 \sin(x) \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
2. \[
2\pi \int_{0}^{\frac{\pi}{6}} 4 \sin(x) \left( 1 + 4 \cos(x) \right) \, dx
\]
3. \[
4 \int_{0}^{\frac{\pi}{6}} \sin(x) \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
4. \[
2\pi \int_{0}^{\frac{\pi}{6}} \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
5. \[
\int_{0}^{\frac{\pi}{6}} \sqrt{1 + 16 \cos^2 (x)} \, dx
\]
**Explanation of the Correct Integral:**
The surface area A of the surface obtained by rotating a curve \( y = f(x) \) around the x-axis is given by:
\[
A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
Given \( y = 4 \sin(x) \), we need to compute the term \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = 4 \cos(x)
\]
Substituting \( \frac{dy}{dx} = 4 \cos(x) \) into \( \left( \frac{dy}{dx} \right)^2 \):
\[
\left( \frac{dy}{dx} \right)^2 = (4 \cos(x))^2 = 16 \cos^2 (x)
\]
Now,
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