Finding volume of region rotated around y-axis. here's a picture of the problem:

I see now that (obviously) the volume can't be negative, but if I use the absolute value of it, is it the correct volume?
Here's my math on how I got the volume:
$$V=\int2\pi x(6sin(2x^2))dx$$
Use u-sub:
$$u=2x^2\;\; {du\over4}=xdx$$
So,
$$V=12\pi (-cos(2x^2)*{1\over4}) \;\; Where\;\; a=6 \;\; and\;\; b=2$$
After using a and b, I get the current volume in the bow with the red "X" by it.
Thanks for any help! These are just non-graded homework problems, but I think I'll learn more if someone points out what I'm doing wrong. Thank you.
EDIT: nvm, I found out the answer is $6\pi$. I'm still not sure where I went wrong.

Here is my approach:
$$
\begin{align}
V&
=2\pi\int_0^{\sqrt{\pi/b}}x~y~dx\\
&=2\pi a\int_0^{\sqrt{\pi/b}}x~\sin(bx^2)~dx\\
&=\frac{\pi a}{2}\int_0^{\pi}\sin t~dt,\quad t=bx^2\\
&=-\frac{\pi a}{b}\cos t\biggr|_0^{\pi}\\
&=\frac{\pi a}{b}
\end{align}
$$
In the present case we get $V=3\pi$. I have verified this results numerically.
