The region formed by y=3x-x3,y=2, and y-axis is revolved around the y-axis.

To solve this problem, we will find the volume of the solid generated by revolving the region bounded by the curve \( y = 3x - x^3 \), the line \( y = 2 \), and the y-axis around the y-axis using the **method of cylindrical shells**.

### Step 1: Express \( x \) as a function of \( y \)
We are revolving around the y-axis, so we need to express \( x \) in terms of \( y \).

The given equation is:
\[
y = 3x - x^3
\]
We need to solve this for \( x \). However, since this is a cubic equation, solving analytically is challenging. Let's focus on the bounds and the setup for integration.

### Step 2: Determine the bounds of integration
From the given region, we know the region is bounded by \( y = 2 \) and the y-axis (where \( x = 0 \)). The curve \( y = 3x - x^3 \) will intersect \( y = 2 \), and we need to determine the value of \( x \) where this happens. To do this, we solve the equation:
\[
3x - x^3 = 2
\]
\[
x^3 - 3x + 2 = 0
\]
Factoring gives:
\[
(x - 1)(x^2 + x - 2) = 0
\]
This gives \( x = 1 \) as the solution within the region of interest (since \( x^2 + x - 2 = 0 \) gives no relevant real roots for our region).

Thus, the region goes from \( x = 0 \) to \( x = 1 \).

### Step 3: Set up the cylindrical shell method
The formula for the volume of a solid of revolution using the method of cylindrical shells is:
\[
V = 2\pi \int_{x=a}^{x=b} x \cdot f(x) \, dx
\]
In this case, we revolve around the y-axis, so the radius of a cylindrical shell is \( x \), and the height is given by the difference between the curve \( y = 3x - x^3 \) and the line \( y = 2 \). Thus, the height of the shell is:
\[
\text{Height} = (3x - x^3) - 2
\]
Now, set up the integral for the volume:
\[
V = 2\pi \int_{0}^{1} x \left( (3x - x^3) - 2 \right) \, dx
\]
Simplifying the expression inside the integral:
\[
V = 2\pi \int_{0}^{1} x \left( 3x - x^3 - 2 \right) \, dx
\]
\[
V = 2\pi \int_{0}^{1} x \left( 3x - 2 - x^3 \right) \, dx
\]
\[
V = 2\pi \int_{0}^{1} \left( 3x^2 - 2x - x^4 \right) \, dx
\]

### Step 4: Compute the integral
Now, calculate the integral term by term:
\[
V = 2\pi \left[ \int_{0}^{1} 3x^2 \, dx - \int_{0}^{1} 2x \, dx - \int_{0}^{1} x^4 \, dx \right]
\]
Each integral is straightforward:
\[
\int 3x^2 \, dx = x^3 \quad \text{so} \quad \left[ x^3 \right]_{0}^{1} = 1
\]
\[
\int 2x \, dx = x^2 \quad \text{so} \quad \left[ x^2 \right]_{0}^{1} = 1
\]
\[
\int x^4 \, dx = \frac{x^5}{5} \quad \text{so} \quad \left[ \frac{x^5}{5} \right]_{0}^{1} = \frac{1}{5}
\]

Thus, the volume is:
\[
V = 2\pi \left( 1 - 1 - \frac{1}{5} \right)
\]
\[
V = 2\pi \left( -\frac{1}{5} \right)
\]
\[
V = -\frac{2\pi}{5}
\]

However, since volume cannot be negative, we take the absolute value:
\[
V = \frac{2\pi}{5}
\]

### Final Answer:
The volume of the solid formed by revolving the region around the y-axis is \( \frac{2\pi}{5} \) cubic units.

---

Would you like further clarification or more details? 

Here are some additional related questions:
1. How does the method of cylindrical shells differ from the disk/washer method?
2. What is the geometric interpretation of the function \( y = 3x - x^3 \)?
3. How can we generalize the volume calculation to other bounds?
4. What happens to the volume if the region is revolved around the x-axis instead?
5. Can this problem be solved using parametric equations?

**Tip**: When setting up volume integrals, always sketch the region first to correctly identify the bounds and the method of revolution.

The region formed by y=3x-x^3, y=2, and y-axis is revolved around the y-axis.

Solve the volume of a solid formed by revolving the region bounded by y=3x - x^3, y=2, and the y-axis around the y-axis using the cylindrical shell method. This step-by-step guide covers calculus techniques for volume calculations and algebraic factoring of cubic equations.

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