Sketch the region whose area is given by the definite integral. 6 6 L√36. -4 -4 36x² dx Need Help? -2 -2 y 12 10 Read It 8 4 2 y 5 4 3 2 2 2 Then use a geometric formula to evaluate the integral (a > 0, r> 0). 4 Watch It 4 Master It -4 -4 -2 -2 y 6 5 4 3 2 y 35 30 25 20 15 10 5 2 2 4 4
Sketch the region whose area is given by the definite integral. 6 6 L√36. -4 -4 36x² dx Need Help? -2 -2 y 12 10 Read It 8 4 2 y 5 4 3 2 2 2 Then use a geometric formula to evaluate the integral (a > 0, r> 0). 4 Watch It 4 Master It -4 -4 -2 -2 y 6 5 4 3 2 y 35 30 25 20 15 10 5 2 2 4 4
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
![The problem involves sketching the region whose area is given by the definite integral:
\[
\int_{-6}^{6} \sqrt{36 - x^2} \, dx
\]
This integral represents the area of a semicircle. The expression \(\sqrt{36 - x^2}\) corresponds to the top half of a circle with a radius of 6, centered at the origin.
### Diagrams Description
1. **Top Left Graph**:
- The graph shows a semicircle with a radius of 6 units, centered on the origin along the x-axis. The semicircle spans from -6 to 6 on the x-axis, and from 0 to 6 on the y-axis. This accurately represents the area described by the integral.
2. **Top Right Graph**:
- This graph shows a smaller semicircle centered at the origin, spanning from -2 to 2 on the x-axis, and from 0 to 2 on the y-axis. This does not represent the integral correctly.
3. **Bottom Left Graph**:
- This graph shows a narrow ellipse-like shape centered at the origin, spanning from -6 to 6 on the x-axis, and from 0 to 6 on the y-axis. This is not the correct representation.
4. **Bottom Right Graph**:
- This graph shows a tall parabolic shape, spanning from -6 to 6 on the x-axis, and from 0 to approximately 36 on the y-axis. This is not a correct representation.
The correct diagram is the **Top Left Graph**, which depicts a semicircle of radius 6, matching the integral.
### Evaluation
To evaluate the integral geometrically, the formula for the area of a semicircle can be used:
\[
\text{Area} = \frac{1}{2} \pi r^2
\]
Given \( r = 6 \), the area is:
\[
\text{Area} = \frac{1}{2} \pi (6)^2 = 18\pi
\]
### Assistance
For additional help, options are available:
- **Read It**: Provides textual explanation.
- **Watch It**: Offers a video demonstration.
- **Master It**: Engages with practice problems.
If you have any questions or need further clarification, click on "Need Help?"](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a2391e0-5e95-4f9c-a142-12a2224b3252%2Fd4d75083-7df6-41e1-83f0-abe877c377f2%2F37g2qw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The problem involves sketching the region whose area is given by the definite integral:
\[
\int_{-6}^{6} \sqrt{36 - x^2} \, dx
\]
This integral represents the area of a semicircle. The expression \(\sqrt{36 - x^2}\) corresponds to the top half of a circle with a radius of 6, centered at the origin.
### Diagrams Description
1. **Top Left Graph**:
- The graph shows a semicircle with a radius of 6 units, centered on the origin along the x-axis. The semicircle spans from -6 to 6 on the x-axis, and from 0 to 6 on the y-axis. This accurately represents the area described by the integral.
2. **Top Right Graph**:
- This graph shows a smaller semicircle centered at the origin, spanning from -2 to 2 on the x-axis, and from 0 to 2 on the y-axis. This does not represent the integral correctly.
3. **Bottom Left Graph**:
- This graph shows a narrow ellipse-like shape centered at the origin, spanning from -6 to 6 on the x-axis, and from 0 to 6 on the y-axis. This is not the correct representation.
4. **Bottom Right Graph**:
- This graph shows a tall parabolic shape, spanning from -6 to 6 on the x-axis, and from 0 to approximately 36 on the y-axis. This is not a correct representation.
The correct diagram is the **Top Left Graph**, which depicts a semicircle of radius 6, matching the integral.
### Evaluation
To evaluate the integral geometrically, the formula for the area of a semicircle can be used:
\[
\text{Area} = \frac{1}{2} \pi r^2
\]
Given \( r = 6 \), the area is:
\[
\text{Area} = \frac{1}{2} \pi (6)^2 = 18\pi
\]
### Assistance
For additional help, options are available:
- **Read It**: Provides textual explanation.
- **Watch It**: Offers a video demonstration.
- **Master It**: Engages with practice problems.
If you have any questions or need further clarification, click on "Need Help?"
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