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
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Chapter1: Functions And Models
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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.
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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