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
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### 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.
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
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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