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We are tasked with finding the area of the surface generated by revolving the curve \( x = \frac{y^3}{2} \) for \( 0 \leq y \leq 3 \) about the y-axis.

### Formula for Surface Area of Revolution:
To compute the surface area when a curve is revolved around the y-axis, we use the formula:
\[
A = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]
Where:
- \( x = f(y) \) is the function in terms of \( y \),
- \( \frac{dx}{dy} \) is the derivative of \( x \) with respect to \( y \),
- \( [a, b] \) is the range of \( y \)-values (here, 0 to 3).

### Step-by-step Solution:

1. **Function and Derivative**:
   The curve is given as:
   \[
   x = \frac{y^3}{2}
   \]
   Differentiate \( x \) with respect to \( y \):
   \[
   \frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^3}{2} \right) = \frac{3y^2}{2}
   \]

2. **Set up the Integral**:
   Now, substitute \( x = \frac{y^3}{2} \) and \( \frac{dx}{dy} = \frac{3y^2}{2} \) into the surface area formula:
   \[
   A = 2\pi \int_0^3 \frac{y^3}{2} \sqrt{1 + \left(\frac{3y^2}{2}\right)^2} \, dy
   \]
   Simplifying the square term:
   \[
   1 + \left( \frac{3y^2}{2} \right)^2 = 1 + \frac{9y^4}{4} = \frac{4 + 9y^4}{4}
   \]
   Thus, the integral becomes:
   \[
   A = 2\pi \int_0^3 \frac{y^3}{2} \sqrt{\frac{4 + 9y^4}{4}} \, dy = \pi \int_0^3 y^3 \sqrt{\frac{4 + 9y^4}{4}} \, dy
   \]
   Factor the square root:
   \[
   A = \pi \int_0^3 y^3 \cdot \frac{\sqrt{4 + 9y^4}}{2} \, dy
   \]
   Which simplifies to:
   \[
   A = \frac{\pi}{2} \int_0^3 y^3 \sqrt{4 + 9y^4} \, dy
   \]

Now, we can solve this integral either through numerical methods or by further simplifying, but we’ll use a numerical approach to compute the exact value.

Let me calculate the value of the integral.The surface area of the surface generated by revolving the curve \( x = \frac{y^3}{2} \), for \( 0 \leq y \leq 3 \), about the y-axis is approximately **577.04 square units**.

Would you like further details on the calculation, or do you have any other questions?

Here are some related questions to explore:
1. How is the formula for surface area of revolution derived?
2. How would the surface area change if the curve were revolved around the x-axis instead?
3. What is the effect of changing the range of \( y \)-values on the surface area?
4. Can this problem be solved using a parametric approach?
5. How does the surface area formula differ for revolution around lines other than the y-axis?

**Tip:** When dealing with integrals involving square roots, sometimes it's helpful to check if a substitution can simplify the expression.

Find the area of the surface generated by revolving the curve x = y^3 / 2, 0 ≤ y ≤ 3, about the y-axis.

Learn how to find the surface area of the surface generated by revolving the curve x = y^3 / 2 from 0 ≤ y ≤ 3 around the y-axis. This solution uses the surface area of revolution formula and involves integration with square roots, suitable for students studying advanced calculus.

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