The area in the first quadrant bounded by x² = 8y, the line y = 2 and the y-axis is revolved about the line y = 2. What is the volume generated?

We are asked to find the volume of the solid generated when the region bounded by the parabola \(x^2 = 8y\), the line \(y = 2\), and the y-axis is revolved about the line \(y = 2\).

### Step 1: Analyze the region
The parabola equation \(x^2 = 8y\) can be rewritten as:
\[
y = \frac{x^2}{8}
\]
This describes a parabola that opens upwards.

The region is bounded by:
1. The parabola \(x^2 = 8y\) or \(y = \frac{x^2}{8}\),
2. The line \(y = 2\),
3. The y-axis (which is \(x = 0\)).

We are revolving the region around the line \(y = 2\), so we will apply the **method of cylindrical shells**.

### Step 2: Method of cylindrical shells
The cylindrical shells method is appropriate here because we are revolving the region around a horizontal line, and the integral will be in terms of \(y\).

The general formula for the volume using cylindrical shells is:
\[
V = \int_{a}^{b} 2\pi (r_{\text{shell}})(h_{\text{shell}}) \, dy
\]
Where:
- \(r_{\text{shell}}\) is the radius of a typical shell, and
- \(h_{\text{shell}}\) is the height of the shell.

### Step 3: Set up the integral
1. **Height of the shell**:
   The height of the shell is the distance between the y-axis (which is \(x = 0\)) and the parabola \(x = \sqrt{8y}\). So:
   \[
   h_{\text{shell}} = \sqrt{8y}
   \]

2. **Radius of the shell**:
   The radius of the shell is the distance from the line \(y = 2\) to a typical point at height \(y\), which is:
   \[
   r_{\text{shell}} = 2 - y
   \]

### Step 4: Limits of integration
The region of interest is from \(y = 0\) to \(y = 2\), so the limits of integration are \(y = 0\) and \(y = 2\).

### Step 5: Compute the volume
Now, we set up the integral for the volume:
\[
V = \int_{0}^{2} 2\pi (2 - y)(\sqrt{8y}) \, dy
\]

Simplifying the expression:
\[
V = 2\pi \int_{0}^{2} (2 - y) \sqrt{8y} \, dy
\]
We will now evaluate this integral step by step.

### Step 6: Solve the integral
We first break the integral into two parts:
\[
V = 2\pi \left[ \int_{0}^{2} 2 \sqrt{8y} \, dy - \int_{0}^{2} y \sqrt{8y} \, dy \right]
\]

For the first part:
\[
\int 2 \sqrt{8y} \, dy = 2 \times \int \sqrt{8y} \, dy
\]
We can factor out the constant \( \sqrt{8} \):
\[
= 4\sqrt{2} \int_0^2 \sqrt{y} \, dy = 4\sqrt{2} \cdot \frac{2}{3} y^{3/2} \Big|_0^2
\]
Substituting \(y = 2\) into the result:
\[
4\sqrt{2} \cdot \frac{2}{3} \cdot (2^{3/2}) = \frac{32}{3}
\]

For the second part:
\[
\int y \sqrt{8y} \, dy = \sqrt{8} \int_0^2 y^{3/2} \, dy = 2\sqrt{2} \cdot \frac{2}{5} y^{5/2} \Big|_0^2
\]
Substituting \(y = 2\):
\[
2\sqrt{2} \cdot \frac{2}{5} \cdot (2^{5/2}) = \frac{64}{5}
\]

### Step 7: Final volume calculation
Now, putting it all together:
\[
V = 2\pi \left(\frac{32}{3} - \frac{64}{5}\right)
\]
To subtract these, find a common denominator:
\[
\frac{32}{3} - \frac{64}{5} = \frac{160}{15} - \frac{192}{15} = \frac{-32}{15}
\]

Thus, the final volume is:
\[
V = 2\pi \times \frac{-32}{15} = \frac{-64\pi}{15}
\]
This is the volume of the solid generated.

Would you like more details or have any further questions?

Here are 5 related questions for deeper understanding:
1. How would the volume change if the region were revolved around the x-axis instead of \(y = 2\)?
2. Can you derive the formula for the volume of revolution using the disk method instead?
3. What if the parabola equation was \(x^2 = 4y\) instead of \(x^2 = 8y\)?
4. How would the limits of integration change if the region extended beyond \(y = 2\)?
5. How does the cylindrical shells method compare to the disk method for different problems?

**Tip:** When using cylindrical shells, always carefully identify the radius and height of each shell to avoid mistakes in setting up the integral.

Explore how to find the volume of the solid generated by revolving the region bounded by the parabola x² = 8y, the line y = 2, and the y-axis about the line y = 2. This problem uses the method of cylindrical shells and involves solving integrals. Suitable for students studying undergraduate calculus or AP Calculus BC.

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