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.