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
  
Incorrect: Your answer is incorrect.

To find the volume of the solid generated by rotating the region bounded by the curves \( y = x^3 \), \( y = 8 \), and \( x = 0 \) about the vertical axis \( x = 3 \), we'll use the method of cylindrical shells.

### Step-by-Step Process:

1. **Set Up the Shell Radius and Height:**
   - The height of a cylindrical shell is the difference between the upper and lower functions in terms of \( x \). In this case, it's the difference between \( y = 8 \) and \( y = x^3 \).
     \[
     \text{Height of the shell} = 8 - x^3
     \]
   - The radius of a shell is the horizontal distance from the axis of rotation \( x = 3 \) to the point \( x \) in the region. Since the axis of rotation is vertical, the distance from the axis to a point is \( 3 - x \).
     \[
     \text{Radius of the shell} = 3 - x
     \]

2. **Determine the Limits of Integration:**
   - The region of integration is bounded by \( x = 0 \) (the y-axis) and the value of \( x \) where the curve \( y = x^3 \) intersects \( y = 8 \). To find the upper bound, set \( x^3 = 8 \):
     \[
     x = 2
     \]
   - So the bounds for \( x \) are from 0 to 2.

3. **Set Up the Volume Integral:**
   - The formula for the volume of a solid of revolution using cylindrical shells is:
     \[
     V = 2\pi \int_{a}^{b} (\text{radius}) \times (\text{height}) \, dx
     \]
     Substituting the radius \( 3 - x \), the height \( 8 - x^3 \), and the bounds from 0 to 2:
     \[
     V = 2\pi \int_0^2 (3 - x)(8 - x^3) \, dx
     \]

4. **Expand and Simplify the Integral:**
   First, expand the product:
   \[
   (3 - x)(8 - x^3) = 24 - 3x^3 - 8x + x^4
   \]
   So the integral becomes:
   \[
   V = 2\pi \int_0^2 (24 - 3x^3 - 8x + x^4) \, dx
   \]

5. **Integrate Term by Term:**
   Now, integrate each term:
   \[
   \int_0^2 24 \, dx = 24x \Big|_0^2 = 48
   \]
   \[
   \int_0^2 -3x^3 \, dx = -3 \left( \frac{x^4}{4} \right) \Big|_0^2 = -3 \left( \frac{16}{4} \right) = -12
   \]
   \[
   \int_0^2 -8x \, dx = -8 \left( \frac{x^2}{2} \right) \Big|_0^2 = -8 \left( \frac{4}{2} \right) = -16
   \]
   \[
   \int_0^2 x^4 \, dx = \frac{x^5}{5} \Big|_0^2 = \frac{32}{5}
   \]

6. **Combine the Results:**
   Adding all the integrals together:
   \[
   48 - 12 - 16 + \frac{32}{5} = 20 + \frac{32}{5} = \frac{100}{5} + \frac{32}{5} = \frac{132}{5}
   \]

7. **Final Volume:**
   Now, multiply by \( 2\pi \) to get the volume:
   \[
   V = 2\pi \times \frac{132}{5} = \frac{264\pi}{5}
   \]

### Final Answer:
The volume of the solid is:
\[
V = \frac{264\pi}{5}
\]

Would you like further details on any of the steps? Here are some related questions:

1. How do you apply the method of cylindrical shells for other types of curves?
2. What is the difference between the disk/washer method and the shell method?
3. How would the integral change if the region was rotated around a different axis?
4. Can you provide a geometric interpretation of the cylindrical shell method?
5. How would you calculate the volume if the curve was rotated about the y-axis?

**Tip:** Always double-check the limits of integration and the setup of your radius and height when using cylindrical shells!

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

Explore how to calculate the volume of a solid generated by rotating the region bounded by y = x^3, y = 8, and x = 0 about x = 3 using the method of cylindrical shells. This detailed solution walks through the setup of the integral, expansion, and integration of terms. Perfect for students learning Calculus and volumes of revolution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation