<s>[INST] Integrate $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$ $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$
$$
\begin{align}
S &= \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{1} (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{2} \sqrt{1+u^2}du ~~~~~~\textrm{substitution with $u = y + y^3$, $du = 1 + 3y^2$} \\
&= 2\pi \int_{0}^{2} \sec^3\theta d\theta = \frac{2\pi}{2} \bigg(\sec\theta \tan\theta + \ln|\sec\theta + \tan\theta| \bigg)\\
&= \pi \bigg[ \sec\theta \tan\theta + ln|\sec\theta + \tan\theta|\bigg]_{0}^{2} \\
&= \pi \bigg[ \sec(\sqrt{1+u}) u + ln|\sec(\sqrt{1+u}) + u|\bigg]_{0}^{2} \\
&= \pi \bigg[ \sec(\sqrt{11}) \cdot 11 + ln|\sec(\sqrt{11}) + 11|\bigg] - \pi\bigg[ \sec(\sqrt{1}) \cdot 0 + ln|\sec(\sqrt{1}) + 0|\bigg]\\
&\approx 21.918\pi
\end{align}
$$
However the approximation per Wolfram Alpha and my textbook both agree on a value of 13.5134.
 [/INST] We see that 
$$S/2\pi=\int_0^1\sqrt{1+(1+3x^2)^2}\,xdx+\int_0^1x^2\sqrt{1+(1+3x^2)^2}\,xdx$$
Then set $t=1+3x^2\Rightarrow xdx=\frac16dt$ in both integrals to get
$$S/2\pi=\frac16\int_1^4 \frac{t-1}{3}\sqrt{1+t^2}\,dt+\frac16\int_1^4\sqrt{1+t^2}\,dt\\
=\frac1{18}\int_1^4t\sqrt{1+t^2}\,dt+\frac19\int_1^4\sqrt{1+t^2}\,dt$$
In the first integral, sub $u=t^2$, and in the second integral, use a trig sub. Can you take it from here?
</s>