Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the x-axis. y = V 81 – x², -8 < x < 8 dx = 2887 J-8

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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**Title: Calculating the Surface Area of a Revolved Curve**

**Objective:**
Learn how to write and evaluate the definite integral representing the area of the surface created by revolving a curve around the x-axis.

**Given Function:**
\[ y = \sqrt{81 - x^2} \]

**Interval:**
\[ -8 \leq x \leq 8 \]

**Objective:**
Calculate the surface area using the formula for surface area of revolution around the x-axis.

**Surface Area Integral:**
\[ 2\pi \int_{-8}^{8} \sqrt{81 - x^2} \, dx = 288\pi \]

**Explanation:**
- The function \( y = \sqrt{81 - x^2} \) can represent a semicircle with radius 9, due to the form of the equation.
- The integral is evaluated from \(-8\) to \(8\), which is a segment of the semicircle.
- Multiplying the integral by \(2\pi\) accounts for the revolution around the x-axis, calculating the surface area of the resulting shape.
- The evaluated integral yields a final result of \(288\pi\).

This approach provides the mathematical foundation for calculating the surface area formed when a curve is revolved around an axis, showcasing applications in geometry and calculus.
Transcribed Image Text:**Title: Calculating the Surface Area of a Revolved Curve** **Objective:** Learn how to write and evaluate the definite integral representing the area of the surface created by revolving a curve around the x-axis. **Given Function:** \[ y = \sqrt{81 - x^2} \] **Interval:** \[ -8 \leq x \leq 8 \] **Objective:** Calculate the surface area using the formula for surface area of revolution around the x-axis. **Surface Area Integral:** \[ 2\pi \int_{-8}^{8} \sqrt{81 - x^2} \, dx = 288\pi \] **Explanation:** - The function \( y = \sqrt{81 - x^2} \) can represent a semicircle with radius 9, due to the form of the equation. - The integral is evaluated from \(-8\) to \(8\), which is a segment of the semicircle. - Multiplying the integral by \(2\pi\) accounts for the revolution around the x-axis, calculating the surface area of the resulting shape. - The evaluated integral yields a final result of \(288\pi\). This approach provides the mathematical foundation for calculating the surface area formed when a curve is revolved around an axis, showcasing applications in geometry and calculus.
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