find the solids by revolving the region y =x^3 bounded by y=8 and x=0 around the line y=8

We are tasked with finding the solid of revolution formed by rotating the region bounded by the curve \( y = x^3 \), the vertical line \( x = 0 \), and the horizontal line \( y = 8 \), around the line \( y = 8 \). This can be done using the **method of cylindrical shells**.

### Step-by-Step Solution:

1. **Equation of the curve:**
   The curve is given as \( y = x^3 \), which we can rewrite in terms of \( x \) as:
   \[
   x = \sqrt[3]{y}
   \]
   since \( y = x^3 \), and we will revolve this region around \( y = 8 \).

2. **Bounds of integration:**
   The region is bounded by \( y = 8 \) at the top, and the curve touches the \( y \)-axis at \( x = 0 \). Therefore, the bounds of integration for \( y \) range from \( y = 0 \) to \( y = 8 \).

3. **Cylindrical shell method:**
   For a small element at \( y \), the radius of the cylindrical shell is the distance from the line \( y = 8 \) to the current value of \( y \). The radius is thus \( 8 - y \), and the height of the shell is the corresponding \( x \)-value, which is \( \sqrt[3]{y} \).

   The volume of a thin cylindrical shell at \( y \) is given by:
   \[
   dV = 2\pi \cdot \text{(radius)} \cdot \text{(height)} \cdot \text{(thickness)}
   \]
   Therefore, for each \( y \), we have:
   \[
   dV = 2\pi \cdot (8 - y) \cdot \sqrt[3]{y} \cdot dy
   \]

4. **Set up the integral:**
   To find the total volume, integrate this expression from \( y = 0 \) to \( y = 8 \):
   \[
   V = \int_0^8 2\pi (8 - y) \cdot \sqrt[3]{y} \, dy
   \]

5. **Solve the integral:**
   To solve this, break it into two parts:
   \[
   V = 2\pi \int_0^8 (8 \cdot \sqrt[3]{y} - y \cdot \sqrt[3]{y}) \, dy
   \]
   Simplifying:
   \[
   V = 2\pi \left( 8 \int_0^8 \sqrt[3]{y} \, dy - \int_0^8 y^{4/3} \, dy \right)
   \]

   Now, evaluate each integral:
   1. \( \int_0^8 y^{1/3} \, dy \)
   2. \( \int_0^8 y^{4/3} \, dy \)

### Solve the first integral:
\[
\int_0^8 y^{1/3} \, dy = \left[ \frac{3}{4} y^{4/3} \right]_0^8 = \frac{3}{4} \cdot 8^{4/3} = \frac{3}{4} \cdot 16 = 12
\]

### Solve the second integral:
\[
\int_0^8 y^{4/3} \, dy = \left[ \frac{3}{7} y^{7/3} \right]_0^8 = \frac{3}{7} \cdot 8^{7/3} = \frac{3}{7} \cdot 128 = \frac{384}{7}
\]

6. **Final volume expression:**
   Substituting back into the original expression:
   \[
   V = 2\pi \left( 8 \cdot 12 - \frac{384}{7} \right) = 2\pi \left( 96 - \frac{384}{7} \right) = 2\pi \cdot \frac{288}{7} = \frac{576\pi}{7}
   \]

Thus, the volume of the solid of revolution is:
\[
V = \frac{576\pi}{7} \, \text{cubic units}.
\]

---

Would you like more details on any step or further explanations?

Here are some additional questions to consider:

1. What is the difference between the cylindrical shell and disk methods for solving solids of revolution?
2. How do we adjust the bounds when revolving around different lines, like \( y = -8 \)?
3. How would the problem change if the curve were \( y = x^2 \) instead of \( y = x^3 \)?
4. What would happen if we revolved the region around the \( x \)-axis instead of \( y = 8 \)?
5. How does the integral setup change if we use the disk method in this problem?

**Tip:** When using cylindrical shells, always ensure the radius is measured as the perpendicular distance from the axis of rotation.

find the solids by revolving the region y = x^3 bounded by y=8 and x=0 around the line y=8

Learn how to calculate the volume of a solid of revolution by revolving the region bounded by y = x^3, y = 8, and x = 0 around the line y = 8. This problem demonstrates the use of the cylindrical shell method and is suitable for advanced calculus students in Grades 11-12 or college level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation