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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Set up the appropirate intergal
![**Problem Statement:**
Find the surface area for the object obtained by rotating \( y = 7x + 2 \) about the x-axis over the interval \([2, 5]\).
**Solution Explanation:**
To find the surface area of a solid of revolution obtained by rotating a function \( y = f(x) \) around the x-axis, we use the formula:
\[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2 } \, dx \]
where:
- \( f(x) \) is the function being rotated,
- \( f'(x) \) is the derivative of \( f(x) \),
- \( [a, b] \) is the interval over which the rotation occurs.
Given \( y = 7x + 2 \):
1. Find the derivative \( f'(x) \):
\[ f'(x) = 7 \]
2. Substitute \( f(x) \) and \( f'(x) \) into the formula:
\[ A = 2\pi \int_{2}^{5} (7x + 2) \sqrt{1 + (7)^2} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \sqrt{1 + 49} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \sqrt{50} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \cdot \sqrt{50} \, dx \]
\[ = 2\pi \sqrt{50} \int_{2}^{5} (7x + 2) \, dx \]
3. Compute the integral:
\[
\int_{2}^{5} (7x + 2) \, dx = \left[ \frac{7x^2}{2} + 2x \right]_{2}^{5}
\]
Evaluate the definite integral:
\[
\left[ \frac{7(5)^2}{2} + 2(5) \right] - \left[ \frac{7(2)^2}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3d9715-5083-4150-a4a6-02c5c7b14656%2F4b1bb87b-e224-4e38-8aa7-b3bddd81917b%2Flce6wkd_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the surface area for the object obtained by rotating \( y = 7x + 2 \) about the x-axis over the interval \([2, 5]\).
**Solution Explanation:**
To find the surface area of a solid of revolution obtained by rotating a function \( y = f(x) \) around the x-axis, we use the formula:
\[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2 } \, dx \]
where:
- \( f(x) \) is the function being rotated,
- \( f'(x) \) is the derivative of \( f(x) \),
- \( [a, b] \) is the interval over which the rotation occurs.
Given \( y = 7x + 2 \):
1. Find the derivative \( f'(x) \):
\[ f'(x) = 7 \]
2. Substitute \( f(x) \) and \( f'(x) \) into the formula:
\[ A = 2\pi \int_{2}^{5} (7x + 2) \sqrt{1 + (7)^2} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \sqrt{1 + 49} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \sqrt{50} \, dx \]
\[ = 2\pi \int_{2}^{5} (7x + 2) \cdot \sqrt{50} \, dx \]
\[ = 2\pi \sqrt{50} \int_{2}^{5} (7x + 2) \, dx \]
3. Compute the integral:
\[
\int_{2}^{5} (7x + 2) \, dx = \left[ \frac{7x^2}{2} + 2x \right]_{2}^{5}
\]
Evaluate the definite integral:
\[
\left[ \frac{7(5)^2}{2} + 2(5) \right] - \left[ \frac{7(2)^2}{
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