The base of a three-dimensional figure is bound by the circle x² + y2 = 1. Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 6. Algebraically, find the area of each rectangle.. -2 -2 2
The base of a three-dimensional figure is bound by the circle x² + y2 = 1. Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 6. Algebraically, find the area of each rectangle.. -2 -2 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Description:**
The base of a three-dimensional figure is bound by the circle \( x^2 + y^2 = 1 \). Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 6. Algebraically, find the area of each rectangle.
**Graph Explanation:**
The diagram shows a circle centered at the origin on the coordinate plane, with radius 1. The circle is represented by the equation \( x^2 + y^2 = 1 \).
- The x-axis and y-axis are shown.
- The circle extends from -1 to 1 on both the x and y axes.
- A shaded region within the circle represents the area under consideration for the problem.
**Solution:**
1. **Circle Equation:**
The equation of the circle is \( x^2 + y^2 = 1 \).
2. **Cross-sectional Rectangles:**
Each vertical cross section perpendicular to the y-axis forms a rectangle.
3. **Width of the Rectangle:**
For a given y-value, the bounds on x within the circle are determined by solving \( x^2 = 1 - y^2 \).
Hence, \( x = \pm \sqrt{1 - y^2} \).
The width of the rectangle is \( 2\sqrt{1 - y^2} \).
4. **Height of the Rectangle:**
The height of each rectangle is given as 6.
5. **Area of Rectangle:**
The area \( A \) of each rectangle is given by:
\[
A = \text{width} \times \text{height} = 2\sqrt{1 - y^2} \times 6 = 12\sqrt{1 - y^2}
\]
This solution provides a step-by-step approach to finding the area of the rectangles described in the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa18a6723-a2ab-43ae-9c9a-35233fc8cb89%2F0591f749-6be5-48e2-a0dd-6721dcb8b16b%2Fzn6ici_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
The base of a three-dimensional figure is bound by the circle \( x^2 + y^2 = 1 \). Vertical cross sections that are perpendicular to the y-axis are rectangles with height equal to 6. Algebraically, find the area of each rectangle.
**Graph Explanation:**
The diagram shows a circle centered at the origin on the coordinate plane, with radius 1. The circle is represented by the equation \( x^2 + y^2 = 1 \).
- The x-axis and y-axis are shown.
- The circle extends from -1 to 1 on both the x and y axes.
- A shaded region within the circle represents the area under consideration for the problem.
**Solution:**
1. **Circle Equation:**
The equation of the circle is \( x^2 + y^2 = 1 \).
2. **Cross-sectional Rectangles:**
Each vertical cross section perpendicular to the y-axis forms a rectangle.
3. **Width of the Rectangle:**
For a given y-value, the bounds on x within the circle are determined by solving \( x^2 = 1 - y^2 \).
Hence, \( x = \pm \sqrt{1 - y^2} \).
The width of the rectangle is \( 2\sqrt{1 - y^2} \).
4. **Height of the Rectangle:**
The height of each rectangle is given as 6.
5. **Area of Rectangle:**
The area \( A \) of each rectangle is given by:
\[
A = \text{width} \times \text{height} = 2\sqrt{1 - y^2} \times 6 = 12\sqrt{1 - y^2}
\]
This solution provides a step-by-step approach to finding the area of the rectangles described in the problem.
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