because in cylindrical co-ordinates the volume is \int_a^b\int_0^{2\pi}\int_0^{f(z)} r dr d\varphi dz = 2\pi \int_a^b \frac{1}{2} (f(z))^2 dz = \pi \int_a^b (f(z))^2dz where f(z) is understood to be non-negative, i.e. if f(z) has a zero one has to stop integration at it and continue with the next positive part of the boundary description becoming f(z) and so on.