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.