Is this way of integrating correct? I came across this integration in a paper and I'm a bit unsure about if it is correct or not.
This is an integral in cylindrical polar co-ordinates. The integration is over the entire volume of space (no bounds). The integration was done as follows:
$$\iiint \,f(z)\,g(r)\, r\,dr\,d\theta\, dz$$
$$= \iint  \,g(r) \, r\, dr\,d\theta\, \int_{-r}^r\,f(z)\,dz$$
$$ = \iint \,g(r)\, F(r)\,r\,dr\,d\theta\,$$
where $F(r) = \int_{-r}^r dz \,f(z)$ 
Now to me this seems just wrong. I would like a second opinion on this, as the paper was published in a reputed journal.
 A: Observe
\begin{align}
&\int^\infty_{-\infty} ds \int d^3\vec{k}\ |\rho_{|\vec{k}|}|^2 e^{-i(\Omega+\omega_{|\vec{k}|}\cosh\beta -k_3\sinh\beta) s} \\
= {} & \int^\infty_{-\infty} ds \int^\infty_0 dK \int_{|\vec{k}|=K} d\sigma\ |\rho_{|\vec{k}|}|^2e^{-i(\Omega+\omega_{|\vec{k}|}\cosh\beta -k_3\sinh\beta) s}\\
= {} & \int^\infty_{-\infty}ds \int^\infty_0 dK\ |\rho_K|^2 e^{-i(\Omega+\omega_{K}\cosh\beta)s} \int_{|\vec{k}|=K} d\sigma\ e^{ik_3\sinh \beta s} \\
= {} & \int^\infty_{-\infty}ds \int^\infty_0 dK\ |\rho_K|^2 e^{-i(\Omega+\omega_{K}\cosh\beta)s} \int_{|\vec{k}|=K} d\sigma\ \frac{\partial_{k_3}e^{ik_3\sinh \beta s}}{i\sinh \beta\ s} \\
= {} & \frac{1}{\sinh \beta} \int^\infty_{-\infty}ds \int^\infty_0 dK\ |\rho_K|^2 \frac{e^{-i(\Omega+\omega_{K}\cosh\beta)s}}{is} \int^K_{-K} dk_3\ \frac{K}{\sqrt{K^2-k_3^2}} \int_{\sqrt{k_1^2+k_2^2}=\sqrt{K^2-k_3^2}}dS\ \partial_{k_3}e^{ik_3\sinh \beta\ s}\\
= {} & \frac{2\pi}{\sinh \beta} \int^\infty_{-\infty}ds \int^\infty_0 dK\ |\rho_K|^2K \frac{e^{-i(\Omega+\omega_{K}\cosh\beta)s}}{is} \int^K_{-K} dk_3\ \partial_{k_3}e^{ik_3\sinh \beta\ s}  
\end{align}
Hence we get 
\begin{align}
&\frac{2\pi}{\sinh \beta} \int^\infty_{-\infty}ds \int^\infty_0 dK\ |\rho_K|^2K \frac{e^{-i(\Omega+\omega_{K}\cosh\beta)s}}{is} \int^K_{-K} dk_3\ \partial_{k_3}e^{ik_3\sinh \beta\ s} \\
&=\ \frac{4\pi}{\sinh\beta} \int^\infty_{-\infty} ds \int^\infty_0 dK\ |\rho_K|^2K e^{-i(\Omega+\omega_{K}\cosh\beta)s} \frac{\sin(K\sinh\beta s)}{s}.
\end{align}
I hope this helps.
Additional Info: Consider the sphere of radius $K$ (fixed radius), say $x^2+y^2+z^2 = K^2$, then
\begin{align}
\text{ Surface Area of the Sphere} =\int_{x^2+y^2+z^2=K^2}d\sigma = \int^K_{-K} \frac{K}{\sqrt{K^2-z^2}}dz\int_{x^2+y^2=K^2-z^2} dS
\end{align}
where $dS$ is the measure on the circle. Now, for each fixed $z$, we see that
\begin{align}
\int_{x^2+y^2= K^2-z^2} dS = \text{ Circumference of a circle of radius } \sqrt{K^2-z^2}
\end{align}
which is just $2\pi \sqrt{K^2-z^2}$. Thus, it follows 
\begin{align}
\int_{x^2+y^2+z^2=K^2}d\sigma = 2\pi \int^K_{-K} \frac{K}{\sqrt{K^2-z^2}}dz\ \sqrt{K^2-z^2}  = 2\pi \int^K_{-K} = 4\pi K^2.
\end{align}
Note: I have used the Co-area formula for smooth manifold.
A: I think it must have been intended to be read like this:
$$\iiint f(z)\,g(r)\, r\,dr\,d\theta\, dz = \iint \left( g(r) \, r\, dr\,d\theta\, \int_{-r}^r f(z)\,dz \right)$$
Thus the integral $\displaystyle \int_{-r}^r$ is inside the integral with respect to $r$, even though it's written to the right of $dr$.
