ind the volume of the solid formed by rotating the region enclosed by x=6y and x=y^3 (with y>=0) about the y axis

To find the volume of the solid formed by rotating the region enclosed by the curves \( x = 6y \) and \( x = y^3 \) (with \( y \geq 0 \)) about the \( y \)-axis, we can use the **shell method**. This method is particularly useful when the region is rotated about the \( y \)-axis, and we express the curves in terms of \( y \).

### Step-by-step solution:

1. **Identify the region:**
   The region is enclosed by the curves:
   - \( x = 6y \)
   - \( x = y^3 \)

   We need to determine the points of intersection between these two curves. To find where the curves intersect, we set \( 6y = y^3 \) and solve for \( y \).

   \[
   6y = y^3
   \]
   Dividing both sides by \( y \) (assuming \( y \neq 0 \)):

   \[
   6 = y^2
   \]

   So,
   \[
   y = \sqrt{6}
   \]

   The points of intersection are at \( y = 0 \) and \( y = \sqrt{6} \).

2. **Volume by the Shell Method:**
   Using the shell method, the volume \( V \) is given by the integral:
   
   \[
   V = 2\pi \int_{a}^{b} \left( \text{radius} \times \text{height} \right) \, dy
   \]
   
   Here, the **radius** of the shell is \( y \), and the **height** is the difference between the two functions, \( x = 6y \) and \( x = y^3 \), which describes the horizontal distance between the curves.

   The limits of integration are from \( y = 0 \) to \( y = \sqrt{6} \).

   Therefore, the volume integral is:

   \[
   V = 2\pi \int_0^{\sqrt{6}} y \left(6y - y^3\right) \, dy
   \]

3. **Simplify and integrate:**

   First, simplify the integrand:

   \[
   V = 2\pi \int_0^{\sqrt{6}} y \left(6y - y^3\right) \, dy = 2\pi \int_0^{\sqrt{6}} \left(6y^2 - y^4\right) \, dy
   \]

   Now, integrate term by term:

   \[
   V = 2\pi \left[ \int_0^{\sqrt{6}} 6y^2 \, dy - \int_0^{\sqrt{6}} y^4 \, dy \right]
   \]

   The integrals of \( y^2 \) and \( y^4 \) are:

   \[
   \int y^2 \, dy = \frac{y^3}{3}, \quad \int y^4 \, dy = \frac{y^5}{5}
   \]

   Substituting these into the equation:

   \[
   V = 2\pi \left[ 6 \times \frac{y^3}{3} \Bigg|_0^{\sqrt{6}} - \frac{y^5}{5} \Bigg|_0^{\sqrt{6}} \right]
   \]

4. **Evaluate the integrals:**

   Now, evaluate each term at the upper limit \( y = \sqrt{6} \) and the lower limit \( y = 0 \):

   \[
   6 \times \frac{(\sqrt{6})^3}{3} = 6 \times \frac{6\sqrt{6}}{3} = 12\sqrt{6}
   \]
   and
   \[
   \frac{(\sqrt{6})^5}{5} = \frac{6\sqrt{6}}{5}
   \]

   So the volume becomes:

   \[
   V = 2\pi \left( 12\sqrt{6} - \frac{6\sqrt{6}}{5} \right)
   \]

   Factor out \( 6\sqrt{6} \):

   \[
   V = 2\pi \times 6\sqrt{6} \left( 2 - \frac{1}{5} \right)
   \]

   Simplify the expression inside the parentheses:

   \[
   2 - \frac{1}{5} = \frac{10}{5} - \frac{1}{5} = \frac{9}{5}
   \]

   So the volume is:

   \[
   V = 2\pi \times 6\sqrt{6} \times \frac{9}{5} = \frac{108\pi\sqrt{6}}{5}
   \]

Thus, the volume of the solid is:

\[
V = \frac{108\pi\sqrt{6}}{5}
\]

### Would you like more details or have any questions about this process?

Here are 5 related questions:
1. How do you determine the method to use for finding the volume of a solid of revolution?
2. What is the difference between the disk/washer method and the shell method?
3. Can you apply the disk method to the same problem?
4. How do you find the points of intersection between two curves?
5. What other types of rotations (besides about the \( y \)-axis) could be considered?

**Tip:** Always sketch the region and the axis of rotation to visualize the solid when solving volume problems.

Learn how to find the volume of a solid formed by rotating the region enclosed by x=6y and x=y^3 (y>=0) about the y-axis using the shell method. Includes step-by-step solution and integration process.

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