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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Question
![### Problem Statement
**Objective:**
Find the area of the surface generated when the given curve is revolved about the y-axis.
**Given Curve:**
\[ y = (8x)^{\frac{1}{3}} \]
**Range:**
For \( 0 \leq y \leq 4 \)
### Explanation
To solve this problem, we need to calculate the surface area of the curve when it is revolved around the y-axis. This calculation typically involves the use of integral calculus and specific surface area formulas designed for curves rotated about an axis.
### Steps to Solve:
1. **Express \( x \) in terms of \( y \):**
Given equation: \( y = (8x)^{\frac{1}{3}} \)
To express \( x \) in terms of \( y \):
\[ y^3 = 8x \]
\[ x = \frac{y^3}{8} \]
2. **Find the derivative \( \frac{dx}{dy} \):**
\[ x = \frac{y^3}{8} \]
\[ \frac{dx}{dy} = \frac{3y^2}{8} \]
3. **Surface Area Formula for rotation about the y-axis:**
The surface area \( S \) of a curve \( x = f(y) \) rotated around the y-axis from \( y = a \) to \( y = b \) is given by:
\[ S = 2\pi \int_{a}^{b} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]
4. **Substitute \( x \) and \( \frac{dx}{dy} \) into the formula:**
\[ x = \frac{y^3}{8} \]
\[ \frac{dx}{dy} = \frac{3y^2}{8} \]
Thus,
\[ 1 + \left( \frac{dx}{dy} \right)^2 = 1 + \left( \frac{3y^2}{8} \right)^2 = 1 + \frac{9y^4}{64} \]
5. **Simplify and integrate:**
Substitute these into the surface area formula](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff74ed3d3-57eb-4d2e-87f6-d577352010e1%2F29691abe-972b-40ad-a572-cbee0eeba6f6%2Fprgxndh_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Objective:**
Find the area of the surface generated when the given curve is revolved about the y-axis.
**Given Curve:**
\[ y = (8x)^{\frac{1}{3}} \]
**Range:**
For \( 0 \leq y \leq 4 \)
### Explanation
To solve this problem, we need to calculate the surface area of the curve when it is revolved around the y-axis. This calculation typically involves the use of integral calculus and specific surface area formulas designed for curves rotated about an axis.
### Steps to Solve:
1. **Express \( x \) in terms of \( y \):**
Given equation: \( y = (8x)^{\frac{1}{3}} \)
To express \( x \) in terms of \( y \):
\[ y^3 = 8x \]
\[ x = \frac{y^3}{8} \]
2. **Find the derivative \( \frac{dx}{dy} \):**
\[ x = \frac{y^3}{8} \]
\[ \frac{dx}{dy} = \frac{3y^2}{8} \]
3. **Surface Area Formula for rotation about the y-axis:**
The surface area \( S \) of a curve \( x = f(y) \) rotated around the y-axis from \( y = a \) to \( y = b \) is given by:
\[ S = 2\pi \int_{a}^{b} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]
4. **Substitute \( x \) and \( \frac{dx}{dy} \) into the formula:**
\[ x = \frac{y^3}{8} \]
\[ \frac{dx}{dy} = \frac{3y^2}{8} \]
Thus,
\[ 1 + \left( \frac{dx}{dy} \right)^2 = 1 + \left( \frac{3y^2}{8} \right)^2 = 1 + \frac{9y^4}{64} \]
5. **Simplify and integrate:**
Substitute these into the surface area formula
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