	 𝑥=𝑦^3−2𝑦^2,𝑥=0,𝑥=9. use shells to find the volume generated by rotating the regions between the given curve and  𝑦=0
  around the  𝑥-axis.

To solve this problem using the method of cylindrical shells, we will:

1. Express the function \( x = y^3 - 2y^2 \) in terms of \( y \).
2. Set up the integral using the formula for the volume of a solid of revolution generated by rotating the region around the \( x \)-axis.

### Step 1: Defining the function

The equation given is \( x = y^3 - 2y^2 \), and we are rotating the region between this curve and \( y = 0 \) around the \( x \)-axis.

The volume using cylindrical shells for a curve rotated around the \( x \)-axis is given by:
\[
V = \int_{y_1}^{y_2} 2\pi y \cdot (f(y)) \, dy
\]
where \( f(y) = x \) is the function in terms of \( y \), and \( y_1 \) and \( y_2 \) are the bounds of \( y \) over which we will integrate.

### Step 2: Determining the bounds

From the condition \( x = 0 \), we set:
\[
y^3 - 2y^2 = 0
\]
Factoring the equation:
\[
y^2(y - 2) = 0
\]
This gives \( y = 0 \) and \( y = 2 \). Therefore, the region of interest is between \( y = 0 \) and \( y = 2 \).

We are also given \( x = 9 \), so we'll determine the \( y \)-value that corresponds to \( x = 9 \). Setting \( y^3 - 2y^2 = 9 \), we can solve:
\[
y^3 - 2y^2 - 9 = 0
\]
Testing possible solutions, we find \( y = 3 \) satisfies this equation.

Thus, the region of interest is from \( y = 0 \) to \( y = 3 \).

### Step 3: Setting up the integral

Now, we set up the integral for the volume of the solid. The volume is:
\[
V = 2\pi \int_{0}^{3} y \cdot (y^3 - 2y^2) \, dy
\]

### Step 4: Solving the integral

We will now compute the integral:
\[
V = 2\pi \int_{0}^{3} y(y^3 - 2y^2) \, dy
= 2\pi \int_{0}^{3} (y^4 - 2y^3) \, dy
\]

Now integrate each term:
\[
V = 2\pi \left[ \frac{y^5}{5} - \frac{2y^4}{4} \right]_{0}^{3}
= 2\pi \left[ \frac{3^5}{5} - \frac{2 \cdot 3^4}{4} \right]
\]

Simplifying:
\[
V = 2\pi \left[ \frac{243}{5} - \frac{2 \cdot 81}{4} \right]
= 2\pi \left[ \frac{243}{5} - 40.5 \right]
\]

Converting to a common denominator:
\[
V = 2\pi \left[ \frac{243}{5} - \frac{202.5}{5} \right]
= 2\pi \left[ \frac{40.5}{5} \right]
= 2\pi \cdot 8.1
= 16.2\pi
\]

Thus, the volume is:
\[
V = 16.2\pi \, \text{cubic units}
\]

### Final Answer:
The volume of the solid generated by rotating the region between the curve \( x = y^3 - 2y^2 \) and \( y = 0 \) around the \( x \)-axis is \( 16.2\pi \, \text{cubic units} \).

---

Would you like any more details, or do you have any questions?

Here are some related questions:
1. How would the solution change if the region was rotated around the \( y \)-axis?
2. How can you find the surface area of the solid instead of the volume?
3. What happens to the volume if the curve is translated vertically?
4. Can the method of disks/washers be used here instead of shells?
5. How would the volume change if the curve \( x = y^3 - 2y^2 \) was altered?

**Tip:** When using the method of cylindrical shells, always ensure that the variable of integration matches the axis of rotation for accurate results.

𝑥=𝑦^3−2𝑦^2,𝑥=0,𝑥=9. Use shells to find the volume generated by rotating the regions between the given curve and 𝑦=0 around the 𝑥-axis.

Find the volume of a solid generated by rotating the curve x = y^3 - 2y^2 between y=0 and y=3 around the x-axis using the method of cylindrical shells. A detailed solution involves setting up and solving the integral. Suitable for students in Grades 11-12 studying Calculus I.

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