To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = (x^3 + 1)^{\frac{1}{2}}\), we have:

\(y^2 = x^3 + 1\)

Thus, the volume is:

\(V = \pi \int_{0}^{2} (x^3 + 1) \, dx\)

Integrating, we get:

\(\pi \left[ \frac{x^4}{4} + x \right]_{0}^{2}\)

Evaluating the definite integral:

\(\pi \left( \left( \frac{2^4}{4} + 2 \right) - \left( \frac{0^4}{4} + 0 \right) \right)\)

\(= \pi \left( 4 + 2 \right)\)

\(= 6\pi\)

Thus, the volume is \(6\pi\) or approximately 18.8.