<s>[INST] Lateral surface area by hand I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.
I have come up with my own solution, but I'm not sure if it's right.
Before substituting values back in, I got
$$ \frac\pi{16} \left( (1+4x^2)^{3/2} 2x - \frac32 \ln\sqrt{1+4x^2+2x} + x\sqrt{1+4x^2} \right)  \text{ from }0\text{ to }\sqrt2$$
Can someone please verify this? For my integration by parts, I set $u = \sec^3\theta - \sec\theta$ and $dv = \sec^2\theta d\theta$
Thanks so much in advance.
 [/INST] With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$.  Now, let
$$\begin{align}
I_1&=\int\tan^2\theta\sec^3\theta d\theta
\\
&=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta
\\
&=\tan\theta(\sec^3\theta-\sec\theta)-\int3\sec^3\theta\tan^2\theta-\sec\theta\tan^2\theta d\theta
\\
&=\tan\theta(\sec^3\theta-\sec\theta)-3I_1+\int\sec\theta\tan^2\theta d\theta
\end{align}$$
(using the integration by parts as you suggested) so that $I_1=\frac{1}{4}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)$.  Let
$$\begin{align}
I_2&=\int\sec\theta\tan^2\theta d\theta
\\
&=\sec\theta\tan\theta-\int\sec^3\theta d\theta
\\
&=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta d\theta-\int\sec\theta d\theta
\\
&=\sec\theta\tan\theta-I_2+\log|\sec\theta-\tan\theta|
\end{align}$$
(using integration by parts with $u=\tan\theta$ and $dv=\sec\theta\tan\theta d\theta$) so that $I_2=\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)$.  Now,
$$\begin{align}
\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta
&=\frac{\pi}{4}I_1=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\int\sec\theta\tan^2\theta d\theta\right)
\\
&=\frac{\pi}{16}\left(\tan\theta(\sec^3\theta-\sec\theta)+\frac{1}{2}\left(\sec\theta\tan\theta+\log|\sec\theta-\tan\theta|\right)\right)
\\
&=\frac{\pi}{16}\left(\tan\theta\sec^3\theta-\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\log|\sec\theta-\tan\theta|\right)
\\
&=\frac{\pi}{16}\left(2x(1+4x^2)^{3/2}-x\sqrt{1+4x^2}+\frac{1}{2}\log\left|\sqrt{1+4x^2}-2x\right|\right)
\end{align}$$
which is similar to what you have, but not the same.
(Of course, doing this by hand and covering a sheet of paper with scribbles to do the two instances of over-and-back integration by parts, I may well have made a mistake along the way, but my result does appear to check by the method described in aschepler's answer.)
</s>