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Let's walk through how to find the volume of the solid formed by rotating the region enclosed by the given curves around the **y-axis**.

### Problem Setup
The two equations provided are:  
1. \(x = 5y\)  
2. \(y^3 = x\), with \(y \geq 0\).  

We are rotating the region **about the y-axis**, so we will use the **shell method** to find the volume.

---

### Shell Method Formula:
The shell method for rotation around the y-axis is:
\[
V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy
\]
- **Radius**: Since we're rotating around the y-axis, the radius for each shell is simply \(x\), the horizontal distance from the y-axis.
- **Height**: The height of the shell is given by the difference between the two x-values for the given y-value. 

---

### Step 1: Express \(x\) in terms of \(y\)
- From the equation \(x = 5y\), the first curve is already expressed as \(x = 5y\).
- From \(y^3 = x\), rearrange to express \(x\) as:
  \[
  x = y^3
  \]

---

### Step 2: Identify Limits of Integration
We need the region where these two curves intersect. Set \(x = 5y\) equal to \(x = y^3\):
\[
5y = y^3 \implies y^3 - 5y = 0 \implies y(y^2 - 5) = 0
\]
So, the solutions are:
\[
y = 0 \quad \text{and} \quad y = \sqrt{5}
\]
Thus, our limits of integration will be from \(y = 0\) to \(y = \sqrt{5}\).

---

### Step 3: Write the Height of the Shell
For a fixed value of \(y\), the height of the shell is the difference in the x-values:
\[
\text{Height} = 5y - y^3
\]

---

### Step 4: Set Up the Integral
Using the shell method formula:
\[
V = 2\pi \int_0^{\sqrt{5}} y \left( 5y - y^3 \right) \, dy
\]

---

### Step 5: Solve the Integral
First, expand the integrand:
\[
y(5y - y^3) = 5y^2 - y^4
\]
Now integrate term by term:
\[
\int_0^{\sqrt{5}} \left( 5y^2 - y^4 \right) \, dy = \left[ \frac{5y^3}{3} - \frac{y^5}{5} \right]_0^{\sqrt{5}}
\]
Evaluate at the limits:
1. When \(y = \sqrt{5}\):
   \[
   \frac{5(\sqrt{5})^3}{3} - \frac{(\sqrt{5})^5}{5} = \frac{5(5\sqrt{5})}{3} - \frac{25\sqrt{5}}{5}
   \]
   Simplify:
   \[
   \frac{25\sqrt{5}}{3} - 5\sqrt{5} = \frac{25\sqrt{5} - 15\sqrt{5}}{3} = \frac{10\sqrt{5}}{3}
   \]
2. When \(y = 0\), the result is 0.

Thus, the value of the integral is:
\[
\frac{10\sqrt{5}}{3}
\]

---

### Step 6: Multiply by \(2\pi\)
\[
V = 2\pi \cdot \frac{10\sqrt{5}}{3} = \frac{20\pi \sqrt{5}}{3}
\]

---

### Final Answer
The volume of the solid formed by rotating the region enclosed by the given curves about the y-axis is:
\[
V = \frac{20\pi \sqrt{5}}{3}
\]

---

Do you have any questions or need further clarification?

---

### 5 Related Questions:
1. How does the disk method differ from the shell method?
2. What would change if we rotated around the x-axis instead of the y-axis?
3. How can we visualize the solid formed by this rotation?
4. Can the volume be computed using polar coordinates instead?
5. What happens to the integral if the limits are incorrectly determined?

### Tip:
When using the shell method, always double-check which variable (x or y) should serve as the radius to avoid setup errors!

Find the volume formed by rotating about the y-axis the region enclosed by: x = 5y and y^3 = x with y ≥ 0.

Learn how to calculate the volume of a solid formed by rotating the region enclosed by x = 5y and y^3 = x about the y-axis. This step-by-step solution uses the shell method and covers key calculus concepts, suitable for college-level calculus.

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