Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.
y = x3,   y = 8,   x = 0;   about x = 3

To solve this problem using the method of cylindrical shells, we rotate the region bounded by the curves around the vertical line \( x = 3 \). Here are the steps to find the volume of the solid of revolution:

### Step 1: Identify the curves and the boundaries
The region is bounded by:
- \( y = x^3 \)
- \( y = 8 \)
- \( x = 0 \)

The rotation occurs about the vertical line \( x = 3 \). 

### Step 2: Formula for volume using cylindrical shells
The volume \( V \) of a solid of revolution generated by rotating a region about a vertical line \( x = a \) is given by:

\[
V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} (\text{radius}) (\text{height}) \, dy
\]

Where:
- **Radius**: The distance from the axis of rotation \( x = 3 \) to the curve \( x = f(y) \). In this case, the radius is \( r = 3 - x \).
- **Height**: The horizontal length of the region at a given \( y \). Since \( y = x^3 \), we solve for \( x \), giving \( x = \sqrt[3]{y} \).

### Step 3: Set up the integral
The bounds for \( y \) are from \( y = 0 \) to \( y = 8 \), since the region is between \( y = x^3 \) and \( y = 8 \).

The height of the cylindrical shell is the horizontal distance between the axis of rotation and the curve \( x = \sqrt[3]{y} \). So the radius is \( 3 - \sqrt[3]{y} \).

The integral for the volume becomes:

\[
V = 2\pi \int_0^8 (3 - \sqrt[3]{y})(1) \, dy
\]

### Step 4: Solve the integral
Now, let's solve the integral:

\[
V = 2\pi \int_0^8 (3 - y^{1/3}) \, dy
\]

Breaking it into two parts:

\[
V = 2\pi \left[ \int_0^8 3 \, dy - \int_0^8 y^{1/3} \, dy \right]
\]

#### First integral:
\[
\int_0^8 3 \, dy = 3y \bigg|_0^8 = 3(8 - 0) = 24
\]

#### Second integral:
\[
\int_0^8 y^{1/3} \, dy = \frac{3}{4} y^{4/3} \bigg|_0^8 = \frac{3}{4} \cdot 8^{4/3} = \frac{3}{4} \cdot 16 = 12
\]

### Step 5: Combine the results
Now, substitute the results into the expression for \( V \):

\[
V = 2\pi (24 - 12) = 2\pi \times 12 = 24\pi
\]

Thus, the volume of the solid generated by rotating the region around the line \( x = 3 \) is:

\[
\boxed{24\pi}
\]

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How do you apply the method of cylindrical shells to rotate about other axes?
2. How would the volume change if rotated around a horizontal axis instead?
3. Can the same volume be found using the disk/washer method?
4. How does the shape of the region affect the setup of the integral?
5. What are the differences between rotating around the y-axis versus a line parallel to the y-axis?

**Tip:** For problems involving volumes of revolution, always sketch the region and identify the axis of rotation first. This helps in choosing the appropriate method—whether cylindrical shells or disks/washers.

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. y = x^3, y = 8, x = 0; about x = 3

Learn how to find the volume of a solid generated by rotating the region bounded by y = x^3, y = 8, and x = 0 around the line x = 3 using the method of cylindrical shells. This solution involves integration and calculus concepts. Suitable for students in undergraduate Calculus I/II.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation