How might we find the volume of the solid whose surface is $\rho = \sin{\phi}^{1/3}$?
Of course, the obvious way to proceed is to write the triple integral $$\int_V dV$$ taking of course $dV = \rho^2 \sin{\phi} d \rho d \phi d \theta$, noting that $\phi$ goes from 0 to $2\pi$, etc. However, this integral is zero.
$\int^{2\pi}_0 \int^\pi_0 \int^{sin^{1/3}(\phi)}_0 \rho^2 sin (\phi) d\rho d\phi d\theta$
This shouldn't be zero.