The base of a three-dimensional figure is bound by the circle x2 + y2 = 4. Vertical cross sections that are perpendicular to the x-axis are rectangles with height equal to 2. Algebraically, find the area of each rectangle. 4 3 54 3 -24 12 3 4 5 -3 4
The base of a three-dimensional figure is bound by the circle x2 + y2 = 4. Vertical cross sections that are perpendicular to the x-axis are rectangles with height equal to 2. Algebraically, find the area of each rectangle. 4 3 54 3 -24 12 3 4 5 -3 4
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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![### Algebraic Solutions of Three-Dimensional Figures
#### Topic: Calculating Areas of Rectangles from Circular Base
---
**Problem Statement:**
The base of a three-dimensional figure is defined by the circle given by the equation \(x^2 + y^2 = 4\). Vertical cross-sections perpendicular to the x-axis form rectangles with a constant height of 2 units. Your task is to algebraically determine the area of each of these rectangles.
---
**Graphical Representation:**
The graph provided is a coordinate plane with the X-axis and Y-axis marked. A circle with a radius of 2 is drawn centered at the origin \((0,0)\). The equation of the circle is \(x^2 + y^2 = 4\).
---
**Explanation and Solution:**
1. **Understanding the Equation of the Circle:**
The equation \(x^2 + y^2 = 4\) represents a circle with a radius of 2 centered at the origin.
2. **Identifying the Boundaries:**
The circle intersects the x-axis at \(x = -2\) and \(x = 2\). Thus, the domain of \(x\) for which the circle exists is \(-2 \leq x \leq 2\).
3. **Using the Circle Equation to Find y:**
The vertical cross-sections are rectangles perpendicular to the x-axis. For any value of \(x\) in the interval \([-2, 2]\), the corresponding \(y\) values are derived from the equation \(x^2 + y^2 = 4\):
\[
y^2 = 4 - x^2
\]
Therefore, \(y\) can be expressed as:
\[
y = \pm\sqrt{4 - x^2}
\]
Consequently, the vertical distance (or length) between the top and bottom of the circle at a given \(x\) is:
\[
2\sqrt{4 - x^2}
\]
4. **Defining the Rectangle's Dimensions:**
The height of the rectangle is given as 2 units, and the width of the rectangle at each \(x\) value is \(2\sqrt{4 - x^2}\).
5. **Calculating the Area of the Rectangle:**
The area \(A\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cd77d14-523b-4d70-a9c2-c3967239ea67%2F1666a424-c863-4cde-8f7b-1c982e4d0313%2F3l8xi8d_processed.png&w=3840&q=75)
Transcribed Image Text:### Algebraic Solutions of Three-Dimensional Figures
#### Topic: Calculating Areas of Rectangles from Circular Base
---
**Problem Statement:**
The base of a three-dimensional figure is defined by the circle given by the equation \(x^2 + y^2 = 4\). Vertical cross-sections perpendicular to the x-axis form rectangles with a constant height of 2 units. Your task is to algebraically determine the area of each of these rectangles.
---
**Graphical Representation:**
The graph provided is a coordinate plane with the X-axis and Y-axis marked. A circle with a radius of 2 is drawn centered at the origin \((0,0)\). The equation of the circle is \(x^2 + y^2 = 4\).
---
**Explanation and Solution:**
1. **Understanding the Equation of the Circle:**
The equation \(x^2 + y^2 = 4\) represents a circle with a radius of 2 centered at the origin.
2. **Identifying the Boundaries:**
The circle intersects the x-axis at \(x = -2\) and \(x = 2\). Thus, the domain of \(x\) for which the circle exists is \(-2 \leq x \leq 2\).
3. **Using the Circle Equation to Find y:**
The vertical cross-sections are rectangles perpendicular to the x-axis. For any value of \(x\) in the interval \([-2, 2]\), the corresponding \(y\) values are derived from the equation \(x^2 + y^2 = 4\):
\[
y^2 = 4 - x^2
\]
Therefore, \(y\) can be expressed as:
\[
y = \pm\sqrt{4 - x^2}
\]
Consequently, the vertical distance (or length) between the top and bottom of the circle at a given \(x\) is:
\[
2\sqrt{4 - x^2}
\]
4. **Defining the Rectangle's Dimensions:**
The height of the rectangle is given as 2 units, and the width of the rectangle at each \(x\) value is \(2\sqrt{4 - x^2}\).
5. **Calculating the Area of the Rectangle:**
The area \(A\)
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