Find the volume of the solid formed by rotating the region enclosed by x=8y
 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 \( x = 8y \) and \( x = y^3 \) about the \( y \)-axis, we'll use the **method of cylindrical shells**.

### Step 1: Set up the problem
We have the curves:
- \( x = 8y \)
- \( x = y^3 \)
- \( y \geq 0 \)

### Step 2: Find the points of intersection
To find the points where these curves intersect, we set the two equations equal to each other:
\[
8y = y^3
\]
Rearranging the equation:
\[
y^3 - 8y = 0
\]
Factor out \( y \):
\[
y(y^2 - 8) = 0
\]
This gives us:
\[
y = 0 \quad \text{or} \quad y^2 = 8
\]
So, \( y = 0 \) or \( y = \sqrt{8} = 2\sqrt{2} \).

### Step 3: Set up the integral using the cylindrical shells method
The formula for the volume using cylindrical shells is:
\[
V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dy
\]
Here:
- **Radius**: The distance from the \( y \)-axis, which is \( y \).
- **Height**: The difference between the functions \( x = 8y \) and \( x = y^3 \).

Thus, the integral is:
\[
V = 2\pi \int_{0}^{2\sqrt{2}} y \times (8y - y^3) \, dy
\]

### Step 4: Simplify the integrand and integrate
Simplify the integrand:
\[
V = 2\pi \int_{0}^{2\sqrt{2}} (8y^2 - y^4) \, dy
\]

Now, integrate term by term:
\[
V = 2\pi \left[ \frac{8y^3}{3} - \frac{y^5}{5} \right]_{0}^{2\sqrt{2}}
\]

Evaluate at the bounds:
- At \( y = 2\sqrt{2} \):
  \[
  \frac{8(2\sqrt{2})^3}{3} - \frac{(2\sqrt{2})^5}{5}
  \]
  First, calculate the powers of \( y \):
  \[
  (2\sqrt{2})^3 = 16\sqrt{2} \quad \text{and} \quad (2\sqrt{2})^5 = 128\sqrt{2}
  \]
  Plugging these into the integral gives:
  \[
  \frac{8 \times 16\sqrt{2}}{3} - \frac{128\sqrt{2}}{5}
  \]

- At \( y = 0 \), the expression is 0.

Thus:
\[
V = 2\pi \left[ \frac{128\sqrt{2}}{3} - \frac{128\sqrt{2}}{5} \right]
\]

Simplify:
\[
V = 2\pi \times \frac{128\sqrt{2}(5 - 3)}{15} = \frac{512\pi\sqrt{2}}{15}
\]

### Final Answer
The volume of the solid is:
\[
\boxed{\frac{512\pi\sqrt{2}}{15}}
\]

Would you like more details or have any questions?

Here are five related questions to explore further:
1. How do you set up the integral for volume using the disk method?
2. What changes in the setup if we rotate the region about the \(x\)-axis instead of the \(y\)-axis?
3. How does the method of cylindrical shells compare to the washer method?
4. What if the region was bounded by \(x = y^3\) and \(x = 16y\)? How would the volume change?
5. How does the volume of a solid change if you scale the region before rotation?

**Tip:** When using the method of cylindrical shells, always ensure that the radius and height of each shell are properly defined relative to the axis of rotation.

Learn how to find the volume of the solid formed by rotating the region enclosed by x = 8y and x = y^3 about the y-axis using the method of cylindrical shells. Detailed step-by-step solution provided.

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