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# 000000 JUUUO Find the area of the region enclosed by the graphs of y = √9-x² √9-x² and -0.1x + 1. y =

# 000000 JUUUO Find the area of the region enclosed by the graphs of y = √9-x² √9-x² and -0.1x + 1. y =

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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Question
**Problem:** Find the area of the region enclosed by the graphs of \( y = \sqrt{9 - x^2} \) and \( y = -0.1x + 1 \). **Provided Answer:** The area is 0.352. **Explanation:** To find the area of the region enclosed by the graphs, you need to: 1. **Graph the Equations:** - \( y = \sqrt{9 - x^2} \): This represents the upper semicircle centered at the origin with a radius of 3. - \( y = -0.1x + 1 \): This represents a straight line with a negative slope of -0.1 and y-intercept at 1. 2. **Determine the Points of Intersection:** - Solve the equations simultaneously \( y = \sqrt{9 - x^2} \) and \( y = -0.1x + 1 \) to find the points where the semicircle intersects the line. 3. **Set Up the Integral:** - Once the points of intersection (let's say at \( x = a \) and \( x = b \)) are found, set up the definite integral with appropriate limits. - Compute the integral \( \int_a^b (\sqrt{9 - x^2} - (-0.1x + 1)) \, dx \). 4. **Calculate the Definite Integral:** - Find the area under the semicircle and the area under the line, then take the difference to get the enclosed area. **Graphical Representation:** There is a small icon of a graphing calculator in the image, likely indicating that the user might need to use a graphing calculator to visualize the graphs and assist in the calculations. In this particular case, the computed area is 0.352 square units.
Transcribed Image Text:**Problem:** Find the area of the region enclosed by the graphs of \( y = \sqrt{9 - x^2} \) and \( y = -0.1x + 1 \). **Provided Answer:** The area is 0.352. **Explanation:** To find the area of the region enclosed by the graphs, you need to: 1. **Graph the Equations:** - \( y = \sqrt{9 - x^2} \): This represents the upper semicircle centered at the origin with a radius of 3. - \( y = -0.1x + 1 \): This represents a straight line with a negative slope of -0.1 and y-intercept at 1. 2. **Determine the Points of Intersection:** - Solve the equations simultaneously \( y = \sqrt{9 - x^2} \) and \( y = -0.1x + 1 \) to find the points where the semicircle intersects the line. 3. **Set Up the Integral:** - Once the points of intersection (let's say at \( x = a \) and \( x = b \)) are found, set up the definite integral with appropriate limits. - Compute the integral \( \int_a^b (\sqrt{9 - x^2} - (-0.1x + 1)) \, dx \). 4. **Calculate the Definite Integral:** - Find the area under the semicircle and the area under the line, then take the difference to get the enclosed area. **Graphical Representation:** There is a small icon of a graphing calculator in the image, likely indicating that the user might need to use a graphing calculator to visualize the graphs and assist in the calculations. In this particular case, the computed area is 0.352 square units.
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