Compute the surface area of revolution about the x-axis over the interval [0, n] for y = 8 sin(x). (Use symbolic notation and fractions where needed.) S =
Compute the surface area of revolution about the x-axis over the interval [0, n] for y = 8 sin(x). (Use symbolic notation and fractions where needed.) S =
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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![To compute the surface area of revolution about the x-axis for the function \( y = 8 \sin(x) \) over the interval \([0, \pi]\), we use the formula for the surface area \( S \) of a solid of revolution:
\[
S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]
For the function \( y = 8 \sin(x) \):
- The derivative \( \frac{dy}{dx} = 8 \cos(x) \).
Substitute \( y \) and \( \frac{dy}{dx} \) into the formula:
\[
S = \int_{0}^{\pi} 2\pi (8 \sin(x)) \sqrt{1 + (8 \cos(x))^2} \, dx
\]
This integral represents the surface area of the solid formed by revolving the curve \( y = 8 \sin(x) \) around the x-axis from \( x = 0 \) to \( x = \pi \).
The solution involves evaluating this integral, which may require numerical methods or further analytical techniques depending on the complexity of the expression inside the integral.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa297a8fb-6644-4f66-925a-05650f7247e5%2F1678394b-a768-4673-a767-89b1df2b9be0%2Fx7hopg6_processed.png&w=3840&q=75)
Transcribed Image Text:To compute the surface area of revolution about the x-axis for the function \( y = 8 \sin(x) \) over the interval \([0, \pi]\), we use the formula for the surface area \( S \) of a solid of revolution:
\[
S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]
For the function \( y = 8 \sin(x) \):
- The derivative \( \frac{dy}{dx} = 8 \cos(x) \).
Substitute \( y \) and \( \frac{dy}{dx} \) into the formula:
\[
S = \int_{0}^{\pi} 2\pi (8 \sin(x)) \sqrt{1 + (8 \cos(x))^2} \, dx
\]
This integral represents the surface area of the solid formed by revolving the curve \( y = 8 \sin(x) \) around the x-axis from \( x = 0 \) to \( x = \pi \).
The solution involves evaluating this integral, which may require numerical methods or further analytical techniques depending on the complexity of the expression inside the integral.
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