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
![**Finding the Area of the Shaded Region Inside a Circle**
*Problem Statement:*
Find the area of the region shaded below that is inside of a circle of radius 4.
*Diagram Interpretation:*
In the provided graph, we have a circle centered at the origin (0, 0) with a radius of 4 units. Along with the circle, there is a shaded region that starts from the positive x-axis at x = 2 and extends to the right up to the edge of the circle. This shaded region is confined within the first quadrant of the circle.
*Detailed Explanation:*
1. **Circle Description:**
- Radius (\(r\)): 4 units
- Center: Origin (0, 0)
2. **Shaded Region:**
- The area of concern starts at \(x = 2\) and extends to \(x = 4\) (the radius of the circle).
- The shaded region is between the vertical lines \(x = 2\) and \(x = 4\), and extends upward to meet the circle's edge.
*Geometric Approach:*
To find the area of the shaded region, it's essential to compute the integral of the area beneath the circle’s curve from \(x = 2\) to \(x = 4\). The integral setup is as follows:
1. **Circle Equation:**
\[ x^2 + y^2 = r^2 \]
In this case, \( r = 4 \), so the equation becomes:
\[ x^2 + y^2 = 16 \]
2. **Solving for \( y \):**
\[ y = \sqrt{16 - x^2} \]
3. **Integral Setup:**
To find the area under the curve \( y = \sqrt{16 - x^2} \) from \( x = 2 \) to \( x = 4 \):
\[ \text{Area} = \int_{2}^{4} \sqrt{16 - x^2} \, dx \]
*Calculating the Integral:*
\[ \int_{2}^{4} \sqrt{16 - x^2} \, dx \]
This integral represents the area of the shaded region.
By solving this integral using appropriate techniques or numerical methods, one can pinpoint the exact area of the shaded region within the given circle](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F776a07ce-21ba-443e-aa37-00f610ca7adb%2Febaab8c5-4bb7-4012-9533-b7ffec463dbb%2Fa2qk4wp_processed.png&w=3840&q=75)
Transcribed Image Text:**Finding the Area of the Shaded Region Inside a Circle**
*Problem Statement:*
Find the area of the region shaded below that is inside of a circle of radius 4.
*Diagram Interpretation:*
In the provided graph, we have a circle centered at the origin (0, 0) with a radius of 4 units. Along with the circle, there is a shaded region that starts from the positive x-axis at x = 2 and extends to the right up to the edge of the circle. This shaded region is confined within the first quadrant of the circle.
*Detailed Explanation:*
1. **Circle Description:**
- Radius (\(r\)): 4 units
- Center: Origin (0, 0)
2. **Shaded Region:**
- The area of concern starts at \(x = 2\) and extends to \(x = 4\) (the radius of the circle).
- The shaded region is between the vertical lines \(x = 2\) and \(x = 4\), and extends upward to meet the circle's edge.
*Geometric Approach:*
To find the area of the shaded region, it's essential to compute the integral of the area beneath the circle’s curve from \(x = 2\) to \(x = 4\). The integral setup is as follows:
1. **Circle Equation:**
\[ x^2 + y^2 = r^2 \]
In this case, \( r = 4 \), so the equation becomes:
\[ x^2 + y^2 = 16 \]
2. **Solving for \( y \):**
\[ y = \sqrt{16 - x^2} \]
3. **Integral Setup:**
To find the area under the curve \( y = \sqrt{16 - x^2} \) from \( x = 2 \) to \( x = 4 \):
\[ \text{Area} = \int_{2}^{4} \sqrt{16 - x^2} \, dx \]
*Calculating the Integral:*
\[ \int_{2}^{4} \sqrt{16 - x^2} \, dx \]
This integral represents the area of the shaded region.
By solving this integral using appropriate techniques or numerical methods, one can pinpoint the exact area of the shaded region within the given circle
![**Arc Length of a Parametric Curve**
**Problem Statement:**
Setup an integral that will compute the arc length of the following parametric curve. Then, evaluate the integral using the Midpoint Approximation with \( n = 5 \).
\[ x = 4 + 5t, \quad y = 7 - 2t^{\frac{5}{3}} \quad \text{for} \quad 0 \le t \le 10 \]
**Steps to Solve:**
1. **Find the derivatives of \( x \) and \( y \) with respect to \( t \):**
\[
\frac{dx}{dt} = \frac{d}{dt}(4 + 5t) = 5
\]
\[
\frac{dy}{dt} = \frac{d}{dt}\left(7 - 2t^{\frac{5}{3}}\right) = -2 \cdot \frac{5}{3} \cdot t^{\frac{2}{3}} = -\frac{10}{3} t^{\frac{2}{3}}
\]
2. **Setup the arc length integral formula:**
\[
\text{Arc Length} = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
\]
Substitute the derivatives found:
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{(5)^2 + \left( -\frac{10}{3} t^{\frac{2}{3}} \right)^2} \, dt
\]
Simplify inside the integral:
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{25 + \left( \frac{100}{9} t^{\frac{4}{3}} \right)} \, dt
\]
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{25 + \frac{100}{9} t^{\frac{4}{3}}} \, dt
\]
3. **Evaluate the integral using the Midpoint Approximation with \( n = 5 \):**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F776a07ce-21ba-443e-aa37-00f610ca7adb%2Febaab8c5-4bb7-4012-9533-b7ffec463dbb%2F9oeqgsa_processed.png&w=3840&q=75)
Transcribed Image Text:**Arc Length of a Parametric Curve**
**Problem Statement:**
Setup an integral that will compute the arc length of the following parametric curve. Then, evaluate the integral using the Midpoint Approximation with \( n = 5 \).
\[ x = 4 + 5t, \quad y = 7 - 2t^{\frac{5}{3}} \quad \text{for} \quad 0 \le t \le 10 \]
**Steps to Solve:**
1. **Find the derivatives of \( x \) and \( y \) with respect to \( t \):**
\[
\frac{dx}{dt} = \frac{d}{dt}(4 + 5t) = 5
\]
\[
\frac{dy}{dt} = \frac{d}{dt}\left(7 - 2t^{\frac{5}{3}}\right) = -2 \cdot \frac{5}{3} \cdot t^{\frac{2}{3}} = -\frac{10}{3} t^{\frac{2}{3}}
\]
2. **Setup the arc length integral formula:**
\[
\text{Arc Length} = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
\]
Substitute the derivatives found:
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{(5)^2 + \left( -\frac{10}{3} t^{\frac{2}{3}} \right)^2} \, dt
\]
Simplify inside the integral:
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{25 + \left( \frac{100}{9} t^{\frac{4}{3}} \right)} \, dt
\]
\[
\text{Arc Length} = \int_{0}^{10} \sqrt{25 + \frac{100}{9} t^{\frac{4}{3}}} \, dt
\]
3. **Evaluate the integral using the Midpoint Approximation with \( n = 5 \):**
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