The following Lemniscate revolves around the $x$-axis with given values for the boundaries:
$$r^2=9\cos(2\theta)$$ $$ \theta = 0, \pi/4$$
The resulting integral for the volume is also given
$$V_x = \int_{\pi/4}^ 0(r\sin(\theta)^2 (r\cos(\theta))' \, d\theta$$
I was wondering how $\pi/4$ ends up as the bottom boundary in the volume integral. Where as if you would calculate the surface area for the lemniscate it would end up in the top boundary. I have seen boundaries getting switched around several times now and I don't quite understand why. Can anyone give me an easy to understand explanation?
Thanks in advance