Answer
$$28.12$$
Work Step by Step
Since
\begin{align*}
S&=2\pi \int_{a}^{b}f(x) \sqrt{1+[f'(x)]^2}dx\\
&= 2\pi \int_{0}^{4}x^2\sqrt{1+4x^2}dx\\
\end{align*}
Let $2x=\tan \theta \to 2dx= \sec^2\theta d\theta $ and by using
$$
\int \sec ^{n}(x) \mathrm{d} x=\frac{n-2}{n-1} \int \sec ^{n-2}(x) \mathrm{d} x+\frac{\sec ^{n-2}(x) \tan (x)}{n-1}
$$
we get
\begin{align*}
2 \pi \int_{0}^{4} x^{2} \sqrt{1+4 x^{2}} d x&=2 \pi \int_{0}^{\tan ^{-1} x^{-1}} \frac{\tan ^{2} \theta}{4} \sqrt{1+\tan ^{2} \theta} \frac{\sec ^{2} \theta}{2} d \theta\\
&=2 \pi \cdot \frac{1}{8} \int_{0}^{\tan ^{-1} 8} \tan ^{2} \theta \sec \theta \sec ^{2} \theta d \theta \\
&=\frac{\pi}{4} \int_{0}^{\tan ^{-1} \frac{1}{8}} \tan ^{2} \theta \sec ^{3} \theta d \theta \\
&=\frac{\pi}{4} \int_{0}^{\tan ^{-1} 8}\left(\sec ^{2} \theta-1\right) \sec ^{3} \theta d \theta \\
&=\frac{\pi}{4} \int_{0}^{\tan ^{-1} 8}\left(\sec ^{5} \theta-\sec ^{3} \theta\right) d \theta\\
&= \frac{\pi}{4} \cdot\left[\frac{\sec ^{3}\left(\tan ^{-1} 8\right) \tan \left(\tan ^{-1} 8\right)}{4}-\frac{1}{8} \sec \left(\tan ^{-1} 8\right) \tan \left(\tan ^{-1} 8\right)-\left(-\frac{1}{8} \ln \left(\sec \left(\tan ^{-1} 8\right)+\tan \left(\tan ^{-1} 8\right)\right)\right)\right]
\\&=\frac{\pi}{4}\left[\left(2 \sec ^{3}(82.87)-\sec (82.87)-\frac{1}{8} \ln (\sec (88.87))\right)\right]\\
&\approx 28.12
\end{align*}
