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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 = √36 - x², -2 ≤x≤2 A 2π dx =

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 = √36 - x², -2 ≤x≤2 A 2π dx =

Calculus: Early Transcendentals
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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### Problem 6: Surface Area of Revolved Curve **Problem Statement:** 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. Given Curve Function: \[ y = \sqrt{36 - x^2}, \quad -2 \leq x \leq 2 \] Integral Representation: \[ 2\pi \int_{-2}^{2} \left( \sqrt{36 - x^2} \right) dx = \] **Explanation:** To find the surface area generated by revolving the given curve \( y = \sqrt{36 - x^2} \) about the x-axis on the interval \([-2, 2]\), we set up the definite integral as shown. The given function is a semi-circle of radius 6, and we will revolve it around the x-axis to obtain the surface area. The integral evaluates the area with respect to \(x\) within the specified limits \([-2, 2]\).
Transcribed Image Text:### Problem 6: Surface Area of Revolved Curve **Problem Statement:** 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. Given Curve Function: \[ y = \sqrt{36 - x^2}, \quad -2 \leq x \leq 2 \] Integral Representation: \[ 2\pi \int_{-2}^{2} \left( \sqrt{36 - x^2} \right) dx = \] **Explanation:** To find the surface area generated by revolving the given curve \( y = \sqrt{36 - x^2} \) about the x-axis on the interval \([-2, 2]\), we set up the definite integral as shown. The given function is a semi-circle of radius 6, and we will revolve it around the x-axis to obtain the surface area. The integral evaluates the area with respect to \(x\) within the specified limits \([-2, 2]\).
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