Negative volume of solid of revolution around x axis.

Question

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?

Answer 1

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.