Find the volume above the x-y plane and below the surface $f(\theta, r) = \frac{5}{r+4}-\frac{5}{8}$.
I do know how to find the answer using a double integration with respect to $r $ and $\theta$, but the issue im having is just some slight confusion regarding the small details.
My attempt:
In my reasoning, $f( \theta, r)$ is positive and defined for $-4 < r <4$, thus, the integration of the function with respect to $r$ will be from $-4$ to $4$.
Correct answer:
$f(\theta, r)$ is only positive for $0 < r < 4$, thus the integration of the function with respect to $r$ will be from $0$ to $4$.
I don't really understand why this is the case. Is it because that a positive $f(\theta, r)$ and a negative $r$ will cause the resulting coordinate to be below the xy plane? Does that mean a negative $f(\theta, r)$ and a negative $r$ results in a coordinate above the xy plane?
When working with polar coordinates it is assumed that $r \geq 0$. This is because $\theta$ is in $[0, 2\pi )$ (or $[-\pi, \pi)$). If you were to use a negative $r$ then it would change $\theta$'s interval.