Determine the volume of the solid generated by rotating the region bounded by y = root(x, 3) and y=pie/4 that lies in first quadrant along the y-axis.

To find the volume of the solid generated by rotating the region bounded by \( y = \sqrt[3]{x} \) and \( y = \frac{\pi}{4} \) in the first quadrant around the y-axis, we will use the **method of cylindrical shells**.

### Step-by-Step Solution

1. **Understand the Region Bounded by the Curves:**

   - The region is in the first quadrant and is bounded by:
     - The curve \( y = \sqrt[3]{x} \)
     - The horizontal line \( y = \frac{\pi}{4} \)

   - We are rotating this region around the **y-axis**.

2. **Express \( x \) in terms of \( y \):**

   The equation \( y = \sqrt[3]{x} \) can be rewritten as:

   \[
   x = y^3
   \]

3. **Find the Interval for \( y \):**

   - The region is bounded below by \( y = 0 \) (since it's in the first quadrant) and above by \( y = \frac{\pi}{4} \).

   So, \( y \) ranges from \( 0 \) to \( \frac{\pi}{4} \).

4. **Volume of Solid by Cylindrical Shells:**

   The volume \( V \) of the solid generated by rotating the region around the y-axis can be found using the method of cylindrical shells. The formula for the volume of a solid of revolution about the y-axis using cylindrical shells is:

   \[
   V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
   \]

   However, since we are revolving around the y-axis and our function is \( x = y^3 \), we rewrite the formula in terms of \( y \):

   \[
   V = 2\pi \int_{0}^{\frac{\pi}{4}} x \cdot (dy)
   \]

   Since \( x = y^3 \), we substitute this into the integral:

   \[
   V = 2\pi \int_{0}^{\frac{\pi}{4}} y^3 \cdot y \, dy = 2\pi \int_{0}^{\frac{\pi}{4}} y^4 \, dy
   \]

5. **Evaluate the Integral:**

   \[
   V = 2\pi \int_{0}^{\frac{\pi}{4}} y^4 \, dy
   \]

   Find the antiderivative of \( y^4 \):

   \[
   \int y^4 \, dy = \frac{y^5}{5}
   \]

   Thus, the volume \( V \) is:

   \[
   V = 2\pi \left[ \frac{y^5}{5} \right]_{0}^{\frac{\pi}{4}}
   \]

   Evaluate the definite integral:

   \[
   V = 2\pi \left( \frac{\left( \frac{\pi}{4} \right)^5}{5} - \frac{0^5}{5} \right)
   \]

   Simplify:

   \[
   V = 2\pi \left( \frac{\left( \frac{\pi}{4} \right)^5}{5} \right)
   \]

   \[
   V = 2\pi \cdot \frac{\pi^5}{4^5 \cdot 5}
   \]

   \[
   V = \frac{2\pi^6}{4^5 \cdot 5}
   \]

6. **Simplify the Expression:**

   \[
   4^5 = 1024
   \]

   Therefore, the volume \( V \) is:

   \[
   V = \frac{2\pi^6}{1024 \cdot 5} = \frac{2\pi^6}{5120}
   \]

   Simplify the fraction:

   \[
   V = \frac{\pi^6}{2560}
   \]

### Conclusion

The volume of the solid generated by rotating the region bounded by \( y = \sqrt[3]{x} \) and \( y = \frac{\pi}{4} \) about the y-axis is:

\[
\boxed{\frac{\pi^6}{2560}}
\]

Would you like more details or have any questions? Here are some related questions to consider:

1. How would the volume change if the rotation were about the x-axis instead?
2. What if the region were rotated around the line \( y = 1 \) instead of the y-axis?
3. How does the volume calculation change if the function is \( y = \sqrt[4]{x} \) instead of \( y = \sqrt[3]{x} \)?
4. What is the surface area of the solid formed by this rotation?
5. How would you set up the integral if you wanted to use the disk method instead of the shell method?

**Tip:** When rotating regions around an axis, visualize the shape and consider whether the disk, washer, or shell method is most appropriate for the given scenario.

Calculate the volume of the solid generated by rotating the region bounded by y = ∛(x) and y = π/4 along the y-axis using the method of cylindrical shells. Includes a detailed step-by-step solution and integral calculus application.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation