Negative volume of solid of revolution around x axis. There was a question on my exams yesterday about the volume of revolotion of the solid generated when the circle $x^2+(y-3)^2=1$ rotates around x axis. That moment I thought it would be best if I used parametric equations. So I took $x(t)=\cos t$, $y(t)=3+\sin t$, $t\in[0,2\pi]$. Now, I know the volume is $$V=\pi\int_{t_1}^{t_2}y^2 x' dt$$
and that's the formula I used. But, the result that came up was negative: $-6\pi^2$. But, if you take "clockwise" parametric equations like $x(t)=\sin t,\,y(t)=3+\cos t,\,t\in[0,2\pi]$, then you get $6\pi^2$.
So I checked on the internet if I forgot the formula had an absolute $x'$ or something but no. 
So my question is: is there a physical explanation for this or is there something else going on here?
My guess was that it has to do with the direction of rotation maybe?

The upper part of the circle contributes more to the volume of revolution than the lower half does. So, if you look at the relative signs of $dx$ and $dt$ for the upper half-circle in the your parametrisation, you will see which one will give a positive volume.

