Answer
$$\pi \left[\frac{9 \sqrt{5}}{4} -\frac{1}{8} \ln (2+\sqrt{5})\right]\approx 15.23$$
Work Step by Step
Given $$y=\frac{1}{2} x^{2}, \quad[0,2]$$
\begin{align*}
y'&= x \\
1+y'^2&= 1+x^{2}
\end{align*}
Then the surface area is given by
\begin{align*}
S&=2\pi \int_{a}^{b} y\sqrt{1+y'^2}dx\\
& =2 \pi \int_{0}^{2} \frac{1}{2} x^2 \sqrt{1+x^{2} } d x\\
&= \pi \int_{0}^{2}x^2 \sqrt{1+x^{2} } d x
\end{align*}
Use the rule
$$\int u^{2} \sqrt{a^{2}+u^{2}} d u=\frac{u}{8}\left(a^{2}+2 u^{2}\right) \sqrt{a^{2}+u^{2}}-\frac{a^{4}}{8} \ln (u+\sqrt{a^{2}+u^{2}})+C$$
Then
\begin{align*}
S &= \pi \int_{0}^{2}x^2 \sqrt{1+x^{2} } d x \\
&=\pi \left[ \frac{x}{8}\left(1+2 x^{2}\right) \sqrt{1+x^{2}}-\frac{1}{8} \ln (x+\sqrt{x+x^{2}})\right]\bigg|_{0}^{2}\\
&= \pi \left[\frac{9 \sqrt{5}}{4} -\frac{1}{8} \ln (2+\sqrt{5})\right]
\approx 15.23
\end{align*}
