Use the disk method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. y = ex, y = 0, x = 0, X=6

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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### Finding the Volume of a Solid Using the Disk Method

In this exercise, we aim to determine the volume of the solid generated by revolving the region bounded by the given graphs around the x-axis. The equations defining the region are:

\[ y = e^{-x} \]
\[ y = 0 \]
\[ x = 0 \]
\[ x = 6 \]

#### Steps for Using the Disk Method

1. **Identify the Bounded Region:**
   - The region is bounded by the curve \( y = e^{-x} \) from \( x = 0 \) to \( x = 6 \) and by the lines \( y = 0 \) (the x-axis), \( x = 0 \) (the y-axis), and \( x = 6 \) (a vertical line).

2. **Set Up the Integral:**
   - To use the disk method, we revolve the region around the x-axis, generating disks with radii equal to \( y = e^{-x} \). The volume of each infinitesimally small disk is \(\pi y^2 \, dx\).

3. **Integral Formulation:**
   - The volume \( V \) is given by:
     \[
     V = \pi \int_{0}^{6} \left( e^{-x} \right)^2 \, dx
     \]
   
4. **Simplify and Compute the Integral:**
   - Simplify the integrand:
     \[
     \left( e^{-x} \right)^2 = e^{-2x}
     \]
   - The volume integral becomes:
     \[
     V = \pi \int_{0}^{6} e^{-2x} \, dx
     \]

#### Diagram Explanation
- **Graph:**
  - The provided graph shows an exponential decay function \( y = e^{-x} \), a horizontal line at \( y = 0 \), and vertical lines at \( x = 0 \) and \( x = 6 \).
  - The shaded area between \( x = 0 \) and \( x = 6 \) under the curve \( y = e^{-x} \) is revolved around the x-axis to create the solid.

Using these steps, you can compute the volume of the solid generated.
Transcribed Image Text:--- ### Finding the Volume of a Solid Using the Disk Method In this exercise, we aim to determine the volume of the solid generated by revolving the region bounded by the given graphs around the x-axis. The equations defining the region are: \[ y = e^{-x} \] \[ y = 0 \] \[ x = 0 \] \[ x = 6 \] #### Steps for Using the Disk Method 1. **Identify the Bounded Region:** - The region is bounded by the curve \( y = e^{-x} \) from \( x = 0 \) to \( x = 6 \) and by the lines \( y = 0 \) (the x-axis), \( x = 0 \) (the y-axis), and \( x = 6 \) (a vertical line). 2. **Set Up the Integral:** - To use the disk method, we revolve the region around the x-axis, generating disks with radii equal to \( y = e^{-x} \). The volume of each infinitesimally small disk is \(\pi y^2 \, dx\). 3. **Integral Formulation:** - The volume \( V \) is given by: \[ V = \pi \int_{0}^{6} \left( e^{-x} \right)^2 \, dx \] 4. **Simplify and Compute the Integral:** - Simplify the integrand: \[ \left( e^{-x} \right)^2 = e^{-2x} \] - The volume integral becomes: \[ V = \pi \int_{0}^{6} e^{-2x} \, dx \] #### Diagram Explanation - **Graph:** - The provided graph shows an exponential decay function \( y = e^{-x} \), a horizontal line at \( y = 0 \), and vertical lines at \( x = 0 \) and \( x = 6 \). - The shaded area between \( x = 0 \) and \( x = 6 \) under the curve \( y = e^{-x} \) is revolved around the x-axis to create the solid. Using these steps, you can compute the volume of the solid generated.
### Calculus: Volume of Solids of Revolution Using the Shell Method

**Problem Statement:**
Use the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis:

\[ y = \frac{1}{x^2}, \quad y = 0, \quad x = 5, \quad \textcolor{red}{x = 8} \]

### Explanation:

To solve for the volume of the solid generated by revolving the given region about the y-axis using the shell method, follow these steps:

1. **Identify the region of integration:**
   - The region is bounded by the function \( y = \frac{1}{x^2} \), the x-axis \( y = 0 \), and the vertical lines \( x = 5 \) and \( x = 8 \).

2. **Set up the volume integral using the shell method:**
   - The shell method involves integrating along the axis perpendicular to the axis of rotation (in this case, the x-axis).
   - The radius of the shell at a point \( x \) is the distance from the y-axis, which is \( x \).
   - The height of the shell is given by the function, \( \frac{1}{x^2} \).
   - The thickness of each shell is \( dx \).

3. **Formulate the integral:**
   - The volume \( V \) is given by the integral:
   \[
   V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx
   \]
   - In this case, \( a = 5 \), \( b = 8 \), radius \( = x \), height \( = \frac{1}{x^2} \):

   \[
   V = \int_{5}^{8} 2\pi x \cdot \frac{1}{x^2} \, dx
   \]

4. **Simplify and evaluate the integral:**
   - The integrand simplifies to:
   \[
   V = \int_{5}^{8} 2\pi \cdot \frac{1}{x} \, dx
   \]

   - This can be evaluated as:
   \[
   V = 2\pi \int_{5
Transcribed Image Text:### Calculus: Volume of Solids of Revolution Using the Shell Method **Problem Statement:** Use the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis: \[ y = \frac{1}{x^2}, \quad y = 0, \quad x = 5, \quad \textcolor{red}{x = 8} \] ### Explanation: To solve for the volume of the solid generated by revolving the given region about the y-axis using the shell method, follow these steps: 1. **Identify the region of integration:** - The region is bounded by the function \( y = \frac{1}{x^2} \), the x-axis \( y = 0 \), and the vertical lines \( x = 5 \) and \( x = 8 \). 2. **Set up the volume integral using the shell method:** - The shell method involves integrating along the axis perpendicular to the axis of rotation (in this case, the x-axis). - The radius of the shell at a point \( x \) is the distance from the y-axis, which is \( x \). - The height of the shell is given by the function, \( \frac{1}{x^2} \). - The thickness of each shell is \( dx \). 3. **Formulate the integral:** - The volume \( V \) is given by the integral: \[ V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx \] - In this case, \( a = 5 \), \( b = 8 \), radius \( = x \), height \( = \frac{1}{x^2} \): \[ V = \int_{5}^{8} 2\pi x \cdot \frac{1}{x^2} \, dx \] 4. **Simplify and evaluate the integral:** - The integrand simplifies to: \[ V = \int_{5}^{8} 2\pi \cdot \frac{1}{x} \, dx \] - This can be evaluated as: \[ V = 2\pi \int_{5
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