Use a triple integral to find the volume of the solid bounded by the parabolic cylinder y = 8x² and the planes z = 0, z = 8 and y = 4.
Use a triple integral to find the volume of the solid bounded by the parabolic cylinder y = 8x² and the planes z = 0, z = 8 and y = 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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![**Problem Statement:**
Use a triple integral to find the volume of the solid bounded by the parabolic cylinder \( y = 8x^2 \) and the planes \( z = 0, z = 8, \) and \( y = 4 \).
**Solution Explanation:**
This problem involves calculating the volume of a solid using a triple integral. The solid is defined by the following boundaries:
- **Parabolic Cylinder**: \( y = 8x^2 \)
- **Planes**:
- \( z = 0 \) (the base of the solid)
- \( z = 8 \) (the top of the solid)
- \( y = 4 \) (one of the vertical boundaries)
**Approach:**
1. **Identify the Region in the xy-plane**:
- The parabolic cylinder and vertical boundary in the xy-plane are defined by \( y = 8x^2 \) and \( y = 4 \).
- This indicates that \( x \) ranges such that \( 0 \leq 8x^2 \leq 4 \).
2. **Determine Limits of Integration**:
- In the xy-plane, the parabola intersects \( y = 4 \) at \((-\frac{1}{2}, 4)\) and \((\frac{1}{2}, 4)\).
- Thus, \( x \) ranges from \(-\frac{1}{2}\) to \(\frac{1}{2}\).
- For each \( x \), \( y \) varies from \( 8x^2 \) up to \( 4 \).
- The \( z \) coordinate varies from \( 0 \) to \( 8 \).
3. **Set Up the Triple Integral**:
\[
\int_{-\frac{1}{2}}^{\frac{1}{2}} \int_{8x^2}^{4} \int_{0}^{8} dz \, dy \, dx
\]
4. **Compute the Integral**:
- Integrate with respect to \( z \): \(\int_{0}^{8} dz \) gives \( \left[ z \right]_{0}^{8} = 8 \).
- Integrate the resulting expression with respect to \( y \).
- Finally, integrate with respect to \( x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3444a3f-7d5c-4474-ab08-7930ed9e066d%2Fa18fad1f-76c4-485c-9ae5-844739be7824%2Fvmz7kf9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use a triple integral to find the volume of the solid bounded by the parabolic cylinder \( y = 8x^2 \) and the planes \( z = 0, z = 8, \) and \( y = 4 \).
**Solution Explanation:**
This problem involves calculating the volume of a solid using a triple integral. The solid is defined by the following boundaries:
- **Parabolic Cylinder**: \( y = 8x^2 \)
- **Planes**:
- \( z = 0 \) (the base of the solid)
- \( z = 8 \) (the top of the solid)
- \( y = 4 \) (one of the vertical boundaries)
**Approach:**
1. **Identify the Region in the xy-plane**:
- The parabolic cylinder and vertical boundary in the xy-plane are defined by \( y = 8x^2 \) and \( y = 4 \).
- This indicates that \( x \) ranges such that \( 0 \leq 8x^2 \leq 4 \).
2. **Determine Limits of Integration**:
- In the xy-plane, the parabola intersects \( y = 4 \) at \((-\frac{1}{2}, 4)\) and \((\frac{1}{2}, 4)\).
- Thus, \( x \) ranges from \(-\frac{1}{2}\) to \(\frac{1}{2}\).
- For each \( x \), \( y \) varies from \( 8x^2 \) up to \( 4 \).
- The \( z \) coordinate varies from \( 0 \) to \( 8 \).
3. **Set Up the Triple Integral**:
\[
\int_{-\frac{1}{2}}^{\frac{1}{2}} \int_{8x^2}^{4} \int_{0}^{8} dz \, dy \, dx
\]
4. **Compute the Integral**:
- Integrate with respect to \( z \): \(\int_{0}^{8} dz \) gives \( \left[ z \right]_{0}^{8} = 8 \).
- Integrate the resulting expression with respect to \( y \).
- Finally, integrate with respect to \( x
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