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A region bounded by f(x) = √√4 — x², y = 0, and x = 0 is shown below. Find the volume of the solid formed by revolving the region about the x-axis. 4. 3 2 1 543-2-11' 1 2 3 4 5 -2 V = 3- 4+ -5 OV = 32 T O V = 16 T 32 T 16 O V= π 3 X

A region bounded by f(x) = √√4 — x², y = 0, and x = 0 is shown below. Find the volume of the solid formed by revolving the region about the x-axis. 4. 3 2 1 543-2-11' 1 2 3 4 5 -2 V = 3- 4+ -5 OV = 32 T O V = 16 T 32 T 16 O V= π 3 X

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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### Volume of a Solid of Revolution #### Problem Statement: A region bounded by \( f(x) = \sqrt{4 - x^2} \), \( y = 0 \), and \( x = 0 \) is shown below. Find the volume of the solid formed by revolving the region about the x-axis. #### Diagram Explanation: The diagram shows a semicircular region in the first quadrant bounded by the function \( f(x) = \sqrt{4 - x^2} \), the x-axis (line \( y = 0 \)), and the y-axis (line \( x = 0 \)). The semicircle is centered at the origin and has a radius of 2 units along the x-axis. It extends from \( x = 0 \) to \( x = 2 \). ![Diagram](image.png) ##### Options for the Volume (V): - \( \circ \quad V = 32\pi \) - \( \circ \quad V = 16\pi \) - \( \circ \quad V = \frac{32}{3}\pi \) - \( \circ \quad V = \frac{16}{3}\pi \) ### Solution: To find the volume of the solid formed by revolving the given region around the x-axis, we use the disk method: The volume \( V \) is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] For our given function \( f(x) = \sqrt{4 - x^2} \), and the bounds for x are from 0 to 2: \[ V = \pi \int_{0}^{2} (\sqrt{4 - x^2})^2 \, dx \] Simplify the integrand: \[ V = \pi \int_{0}^{2} (4 - x^2) \, dx \] Evaluate the integral: \[ V = \pi \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} \] \[ V = \pi \left[ \left( 4(2) - \frac{(2)^3}{3} \right) - (4(0) - \frac{(0)^3}{3}) \right] \] \[ V = \pi \left[ 8 - \frac{8}{3} \
Transcribed Image Text:### Volume of a Solid of Revolution #### Problem Statement: A region bounded by \( f(x) = \sqrt{4 - x^2} \), \( y = 0 \), and \( x = 0 \) is shown below. Find the volume of the solid formed by revolving the region about the x-axis. #### Diagram Explanation: The diagram shows a semicircular region in the first quadrant bounded by the function \( f(x) = \sqrt{4 - x^2} \), the x-axis (line \( y = 0 \)), and the y-axis (line \( x = 0 \)). The semicircle is centered at the origin and has a radius of 2 units along the x-axis. It extends from \( x = 0 \) to \( x = 2 \). ![Diagram](image.png) ##### Options for the Volume (V): - \( \circ \quad V = 32\pi \) - \( \circ \quad V = 16\pi \) - \( \circ \quad V = \frac{32}{3}\pi \) - \( \circ \quad V = \frac{16}{3}\pi \) ### Solution: To find the volume of the solid formed by revolving the given region around the x-axis, we use the disk method: The volume \( V \) is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] For our given function \( f(x) = \sqrt{4 - x^2} \), and the bounds for x are from 0 to 2: \[ V = \pi \int_{0}^{2} (\sqrt{4 - x^2})^2 \, dx \] Simplify the integrand: \[ V = \pi \int_{0}^{2} (4 - x^2) \, dx \] Evaluate the integral: \[ V = \pi \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} \] \[ V = \pi \left[ \left( 4(2) - \frac{(2)^3}{3} \right) - (4(0) - \frac{(0)^3}{3}) \right] \] \[ V = \pi \left[ 8 - \frac{8}{3} \
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