Finding volume of region rotated around y-axis.

Question

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.

Answer 1

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.