Find the volume of a pyramid with height 27 and rectangular base with dimensions 3 and 9 using integration.
Find the volume of a pyramid with height 27 and rectangular base with dimensions 3 and 9 using integration.
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:**
Find the volume of a pyramid with height 27 and a rectangular base with dimensions 3 and 9 using integration.
**Solution:**
To find the volume of the pyramid using integration, follow these steps:
1. **Set Up the Problem:**
- Consider a pyramid with its vertex at the origin, extending along the z-axis.
- The base lies on the plane z = 27, with dimensions 3 × 9.
2. **Determine the Position of a Cross-Section:**
- A horizontal cross-section of the pyramid is a rectangle.
- The dimensions of the rectangle at height z are \( \frac{3}{27}(27-z) \) and \( \frac{9}{27}(27-z) \).
3. **Calculate the Area of a Cross-Section:**
- The area A(z) of the rectangle at height z is:
\[
A(z) = \left(\frac{3}{27}(27-z)\right) \times \left(\frac{9}{27}(27-z)\right)
= \frac{27}{81}(27-z)^2
= \frac{1}{3}(27-z)^2
\]
4. **Integrate to Find the Volume:**
- The volume V of the pyramid is given by integrating the area of the cross-sections from the base to the apex (z = 0 to z = 27).
\[
V = \int_{0}^{27} A(z) \, dz
= \int_{0}^{27} \frac{1}{3}(27-z)^2 \, dz
\]
5. **Evaluate the Integral:**
- Solve the integral:
\[
V = \frac{1}{3} \int_{0}^{27} (27-z)^2 \, dz
\]
- Let \( u = 27-z \), then \( du = -dz \).
- Change the limits of integration accordingly: when \( z = 0 \), \( u = 27 \); when \( z = 27 \), \( u = 0 \).
\[
V = \frac{1}{3} \int_{27}^{0} u^2 \, (-du)
= \frac{1}{3} \int_{0}^{27} u^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76112a2-3370-45d7-a044-1f0297120ded%2F11f160bf-9a01-48c2-b14a-cf3094071dd9%2Fb26gd7f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find the volume of a pyramid with height 27 and a rectangular base with dimensions 3 and 9 using integration.
**Solution:**
To find the volume of the pyramid using integration, follow these steps:
1. **Set Up the Problem:**
- Consider a pyramid with its vertex at the origin, extending along the z-axis.
- The base lies on the plane z = 27, with dimensions 3 × 9.
2. **Determine the Position of a Cross-Section:**
- A horizontal cross-section of the pyramid is a rectangle.
- The dimensions of the rectangle at height z are \( \frac{3}{27}(27-z) \) and \( \frac{9}{27}(27-z) \).
3. **Calculate the Area of a Cross-Section:**
- The area A(z) of the rectangle at height z is:
\[
A(z) = \left(\frac{3}{27}(27-z)\right) \times \left(\frac{9}{27}(27-z)\right)
= \frac{27}{81}(27-z)^2
= \frac{1}{3}(27-z)^2
\]
4. **Integrate to Find the Volume:**
- The volume V of the pyramid is given by integrating the area of the cross-sections from the base to the apex (z = 0 to z = 27).
\[
V = \int_{0}^{27} A(z) \, dz
= \int_{0}^{27} \frac{1}{3}(27-z)^2 \, dz
\]
5. **Evaluate the Integral:**
- Solve the integral:
\[
V = \frac{1}{3} \int_{0}^{27} (27-z)^2 \, dz
\]
- Let \( u = 27-z \), then \( du = -dz \).
- Change the limits of integration accordingly: when \( z = 0 \), \( u = 27 \); when \( z = 27 \), \( u = 0 \).
\[
V = \frac{1}{3} \int_{27}^{0} u^2 \, (-du)
= \frac{1}{3} \int_{0}^{27} u^
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