4) Region R is bound by the function r(x) = 9-x² and the x-axis. a) Sketch Region R & the Volume of Solid formed by revolving region R about the x-axis. b) Determine the volume of this solid.
4) Region R is bound by the function r(x) = 9-x² and the x-axis. a) Sketch Region R & the Volume of Solid formed by revolving region R about the x-axis. b) Determine the volume of this solid.
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 4: Volume of a Solid Formed by Revolving a Region
#### Given:
Region \( R \) is bound by the function \( r(x) = 9 - x^2 \) and the x-axis.
#### Tasks:
a) **Sketch Region \( R \)** and the **Volume of Solid** formed by revolving region \( R \) about the x-axis.
b) **Determine the volume of this solid.**
#### Steps to Approach:
1. **Identify the Region \( R \):**
- The function \( r(x) = 9 - x^2 \) is a downward-facing parabola with its vertex at \( (0, 9) \).
- The x-axis forms the boundary by \( y = 0 \).
2. **Sketch the Region \( R \):**
- Plot the parabola by identifying key points such as the vertex and the x-intercepts where \( 9 - x^2 = 0 \) which are \( x = -3 \) and \( x = 3 \).
3. **Form the Solid by Revolving Region \( R \) about the x-axis:**
- The revolution forms a three-dimensional solid, specifically a solid of revolution. The shape formed is similar to a paraboloid.
4. **Determine the Volume:**
- Using the disk method, the volume \( V \) of the solid of revolution can be found using the integral formula:
\[
V = \pi \int_{-3}^{3} (9 - x^2)^2 dx
\]
- Solve the integral and multiply by \( \pi \).
### Detailed Explanation of the Graph:
- The graph should include the function \( r(x) = 9 - x^2 \), marked clearly with the vertex and intercepts.
- Indicate the region under the curve and above the x-axis, bounded between \( x = -3 \) and \( x = 3 \).
### Calculating the Volume:
To be filled in by the student:
---
This transcription is intended for an educational website, providing clear instructions and steps to help students understand and solve the problem of determining the volume of a solid formed by revolving a given region about the x-axis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec4afb63-02f6-48e1-bd91-d1418f4294a6%2F0bdeb8a9-0d4b-4785-8e5a-60fc8b203a8c%2Fqz0qikg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Problem 4: Volume of a Solid Formed by Revolving a Region
#### Given:
Region \( R \) is bound by the function \( r(x) = 9 - x^2 \) and the x-axis.
#### Tasks:
a) **Sketch Region \( R \)** and the **Volume of Solid** formed by revolving region \( R \) about the x-axis.
b) **Determine the volume of this solid.**
#### Steps to Approach:
1. **Identify the Region \( R \):**
- The function \( r(x) = 9 - x^2 \) is a downward-facing parabola with its vertex at \( (0, 9) \).
- The x-axis forms the boundary by \( y = 0 \).
2. **Sketch the Region \( R \):**
- Plot the parabola by identifying key points such as the vertex and the x-intercepts where \( 9 - x^2 = 0 \) which are \( x = -3 \) and \( x = 3 \).
3. **Form the Solid by Revolving Region \( R \) about the x-axis:**
- The revolution forms a three-dimensional solid, specifically a solid of revolution. The shape formed is similar to a paraboloid.
4. **Determine the Volume:**
- Using the disk method, the volume \( V \) of the solid of revolution can be found using the integral formula:
\[
V = \pi \int_{-3}^{3} (9 - x^2)^2 dx
\]
- Solve the integral and multiply by \( \pi \).
### Detailed Explanation of the Graph:
- The graph should include the function \( r(x) = 9 - x^2 \), marked clearly with the vertex and intercepts.
- Indicate the region under the curve and above the x-axis, bounded between \( x = -3 \) and \( x = 3 \).
### Calculating the Volume:
To be filled in by the student:
---
This transcription is intended for an educational website, providing clear instructions and steps to help students understand and solve the problem of determining the volume of a solid formed by revolving a given region about the x-axis.
![**Title: Calculus: Volume by Revolution**
---
**Problem 5: Volume by Revolution of Region R**
Region \( R \) is bounded by the functions \( f(x) = 8 - x^2 \) and \( g(x) = -x^2 \).
### Part (a): Sketch Region \( R \) and the Volume of the Solid
1. **Sketch the Functions**:
- Plot the function \( f(x) = 8 - x^2 \), which is a downward-facing parabola with vertex at \( (0, 8) \).
- Plot the function \( g(x) = -x^2 \), which is an upward-facing parabola with vertex at \( (0, 0) \).
2. **Find the Intersection Points**:
- Set \( f(x) = g(x) \):
\[
8 - x^2 = -x^2
\]
- Solving for \( x \):
\[
8 = 0
\]
This results in solving \( x = \pm \sqrt{8} \).
3. **Shade the Bounded Region**:
- Shade the region between the two curves from \( x = -\sqrt{8} \) to \( x = \sqrt{8} \).
4. **Visualization of Volume by Revolution**:
- Revolve the shaded region about the x-axis to form the solid.
### Part (b): Determine the Volume of the Solid
To find the volume of the solid formed by revolving region \( R \) about the x-axis:
1. **Set Up the Integral**:
- Use the washer method.
- The volume \( V \) is given by:
\[
V = \pi \int_{- \sqrt{8}}^{ \sqrt{8}} [(8 - x^2)^2 - (-x^2)^2] \, dx
\]
- Simplify the integrand:
\[
V = \pi \int_{- \sqrt{8}}^{ \sqrt{8}} [(8 - x^2)^2 - x^4] \, dx
\]
2. **Evaluate the Integral**:
- Compute the integral step-by-step, ensuring proper algebraic manipulation and integration techniques.
By following these steps, one can derive the exact volume of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec4afb63-02f6-48e1-bd91-d1418f4294a6%2F0bdeb8a9-0d4b-4785-8e5a-60fc8b203a8c%2Fszx91v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculus: Volume by Revolution**
---
**Problem 5: Volume by Revolution of Region R**
Region \( R \) is bounded by the functions \( f(x) = 8 - x^2 \) and \( g(x) = -x^2 \).
### Part (a): Sketch Region \( R \) and the Volume of the Solid
1. **Sketch the Functions**:
- Plot the function \( f(x) = 8 - x^2 \), which is a downward-facing parabola with vertex at \( (0, 8) \).
- Plot the function \( g(x) = -x^2 \), which is an upward-facing parabola with vertex at \( (0, 0) \).
2. **Find the Intersection Points**:
- Set \( f(x) = g(x) \):
\[
8 - x^2 = -x^2
\]
- Solving for \( x \):
\[
8 = 0
\]
This results in solving \( x = \pm \sqrt{8} \).
3. **Shade the Bounded Region**:
- Shade the region between the two curves from \( x = -\sqrt{8} \) to \( x = \sqrt{8} \).
4. **Visualization of Volume by Revolution**:
- Revolve the shaded region about the x-axis to form the solid.
### Part (b): Determine the Volume of the Solid
To find the volume of the solid formed by revolving region \( R \) about the x-axis:
1. **Set Up the Integral**:
- Use the washer method.
- The volume \( V \) is given by:
\[
V = \pi \int_{- \sqrt{8}}^{ \sqrt{8}} [(8 - x^2)^2 - (-x^2)^2] \, dx
\]
- Simplify the integrand:
\[
V = \pi \int_{- \sqrt{8}}^{ \sqrt{8}} [(8 - x^2)^2 - x^4] \, dx
\]
2. **Evaluate the Integral**:
- Compute the integral step-by-step, ensuring proper algebraic manipulation and integration techniques.
By following these steps, one can derive the exact volume of
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