Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Volume of a Solid of Revolution**

**Problem Statement:**

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.

The equations and boundaries are:
\[ y = \frac{1}{\sqrt{5x + 6}} \]
\[ y = 0 \]
\[ x = 0 \]
\[ x = 9 \]

**Step-by-Step Explanation:**

1. **Identify the Bounded Region:**
   - The curve given is \( y = \frac{1}{\sqrt{5x + 6}} \).
   - The line \( y = 0 \) represents the x-axis.
   - The vertical lines \( x = 0 \) and \( x = 9 \) act as the left and right boundaries, respectively.

2. **Revolving Around the x-axis:**
   - To find the volume of the solid generated by revolving this bounded region about the x-axis, we use the disk method.

3. **Volume Using Disk Method:**
   - The formula for the volume \( V \) of the solid of revolution using the disk method is:
   \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
   - Here, \( f(x) = \frac{1}{\sqrt{5x + 6}} \), \( a = 0 \), and \( b = 9 \).

4. **Set Up the Integral:**
   \[ V = \pi \int_{0}^{9} \left( \frac{1}{\sqrt{5x + 6}} \right)^2 \, dx \]

5. **Simplify the Integrand:**
   \[ \left( \frac{1}{\sqrt{5x + 6}} \right)^2 = \frac{1}{5x + 6} \]
   - Now our integral becomes:
   \[ V = \pi \int_{0}^{9} \frac{1}{5x + 6} \, dx \]

6. **Integral Calculation:**
   - To integrate \( \frac{1}{5x + 6} \), we can use the natural logarithm (ln) rule. The integral of \( \frac{1}{u} \) is \( \ln{|u|} \), where \( u
Transcribed Image Text:**Volume of a Solid of Revolution** **Problem Statement:** Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. The equations and boundaries are: \[ y = \frac{1}{\sqrt{5x + 6}} \] \[ y = 0 \] \[ x = 0 \] \[ x = 9 \] **Step-by-Step Explanation:** 1. **Identify the Bounded Region:** - The curve given is \( y = \frac{1}{\sqrt{5x + 6}} \). - The line \( y = 0 \) represents the x-axis. - The vertical lines \( x = 0 \) and \( x = 9 \) act as the left and right boundaries, respectively. 2. **Revolving Around the x-axis:** - To find the volume of the solid generated by revolving this bounded region about the x-axis, we use the disk method. 3. **Volume Using Disk Method:** - The formula for the volume \( V \) of the solid of revolution using the disk method is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] - Here, \( f(x) = \frac{1}{\sqrt{5x + 6}} \), \( a = 0 \), and \( b = 9 \). 4. **Set Up the Integral:** \[ V = \pi \int_{0}^{9} \left( \frac{1}{\sqrt{5x + 6}} \right)^2 \, dx \] 5. **Simplify the Integrand:** \[ \left( \frac{1}{\sqrt{5x + 6}} \right)^2 = \frac{1}{5x + 6} \] - Now our integral becomes: \[ V = \pi \int_{0}^{9} \frac{1}{5x + 6} \, dx \] 6. **Integral Calculation:** - To integrate \( \frac{1}{5x + 6} \), we can use the natural logarithm (ln) rule. The integral of \( \frac{1}{u} \) is \( \ln{|u|} \), where \( u
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