Why does integrating $\frac{1}{r}$ over a cylinder result in $2\pi z$?

Question

The time average intensity is given by $I(r)=\frac{a E_0^2}{2Z r}$ (...). If we integrate $I(r)$ over a cylinder with radius $r$ and length $z$, we get $I=2\pi z$.

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Obviously, they used cylindrical coordinates: $\int_0^z \int_0^{2\pi} \int_0^r \frac{1}{r} dr d\phi dz$. My question is: why does $\int_0^r \frac{1}{r} dr$ seem to equal $1$ and not $\ln{r}$ ?

Thank you

Answer 1

As @ChaoticGood says, you forgot the Jacobian. In cylindrical coordinates, $dxdydz = r dr d\phi dz$. Maybe you can review https://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Line_and_volume_elements to get a refresher