I have been asked to find the volume of a hyperboloid using cylindrical coordinates:
$x=r\cos(\phi)$
$y=r\sin(\phi)$
$r^2 = x^2 + y ^ 2$
The hyperboloid is defined as $r^2 = z^2 + a^2$ between $z = 0$ and $z = b$.
I set up the following integral: $\int_0^b\int_0^{2\pi}\int_a^{\sqrt{z^2-a^2}} 1 drd\phi dz$
However the solution given has the following integral: $\int_0^b\int_0^{2\pi}\int_0^{\sqrt{z^2-a^2}} 1 drd\phi dz$
Specifically, the inner integral has lower limit $0$ not $a$. This doesn't make sense to me, because if the hyperboloid is drawn as a cross section with $z=0$ we can clearly see the radius is never 0, it starts at $a$ and continues to the upper limit defined above.
Infact, $r = 0$ doesn't satisfy the equation for the hyberboloid ($r^2 = z^2 + a^2$) for a non-zero $a$. The same is done in this question without any explanation.
Why do/can we start integrating at $r = 0$ when $r$ is never $0$ and doesn't even satisfy the hyperboloid equation?