Find the area of the surface generated when the curve f(x) = 9 tanx on [] Area = (Type an integer or decimal rounded to three decimal places as needed.) is revolved about the x-axis.
Find the area of the surface generated when the curve f(x) = 9 tanx on [] Area = (Type an integer or decimal rounded to three decimal places as needed.) is revolved about the x-axis.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Problem Statement for Surface Area Calculation**
Find the area of the surface generated when the curve \( f(x) = 9 \tan x \) on the interval \(\left[ 0, \frac{\pi}{6} \right] \) is revolved about the x-axis.
**Input Section**
Area = [Input Box]
(Type an integer or decimal rounded to three decimal places as needed.)
---
This task involves calculating the surface area of a solid of revolution. Specifically, it requires us to find the surface area generated by revolving the function \( f(x) = 9 \tan x \) around the x-axis from \( x = 0 \) to \( x = \frac{\pi}{6} \).
To achieve this, we use the surface area formula for a solid of revolution around the x-axis:
\[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx \]
where \( f'(x) \) is the derivative of the function \( f(x) \). In this case:
1. Compute the derivative \( f'(x) \) of \( f(x) = 9 \tan x \).
2. Substitute \( f(x) \) and \( f'(x) \) into the surface area formula.
3. Compute the integral over the interval \(\left[ 0, \frac{\pi}{6} \right] \).
4. Finally, input the computed surface area rounded to three decimal places.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F511d779c-bf10-4315-98b1-4f0038c55113%2Fb3c2ae67-cf54-47b5-ae03-8aa66429b8d4%2Fygxqplo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement for Surface Area Calculation**
Find the area of the surface generated when the curve \( f(x) = 9 \tan x \) on the interval \(\left[ 0, \frac{\pi}{6} \right] \) is revolved about the x-axis.
**Input Section**
Area = [Input Box]
(Type an integer or decimal rounded to three decimal places as needed.)
---
This task involves calculating the surface area of a solid of revolution. Specifically, it requires us to find the surface area generated by revolving the function \( f(x) = 9 \tan x \) around the x-axis from \( x = 0 \) to \( x = \frac{\pi}{6} \).
To achieve this, we use the surface area formula for a solid of revolution around the x-axis:
\[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx \]
where \( f'(x) \) is the derivative of the function \( f(x) \). In this case:
1. Compute the derivative \( f'(x) \) of \( f(x) = 9 \tan x \).
2. Substitute \( f(x) \) and \( f'(x) \) into the surface area formula.
3. Compute the integral over the interval \(\left[ 0, \frac{\pi}{6} \right] \).
4. Finally, input the computed surface area rounded to three decimal places.
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