Reference: "Theoretische Physik" (2015) by Bartelsmann and others, page 391, equation (11.23).
While deriving the first Maxwell equation based on Coulomb's law, the authors are using the following integral identity:
$$
\begin{align}
\oint_{\partial V} \mathbf{d}f \cdot \mathbf{E}(\mathbf{r}) & = k\oint_{\partial V} \mathbf{d}f \cdot \int dV' \rho(\mathbf{r}') \frac{\mathbf{r} – \mathbf{r}'}{|\mathbf{r} – \mathbf{r}'|^3} \\
& = k \int dV' \rho(\mathbf{r}') \oint_{\partial V} \mathbf{d}f \cdot \frac{\mathbf{r} – \mathbf{r}'}{|\mathbf{r} – \mathbf{r}'|^3},
\end{align}
$$
where $\mathbf{d}f$ is a surface element, $\mathbf{E}$ is electric field, $k$ is some constant, $\rho$ is volume specific charge density, $\mathbf{r}$ is location of reference charge, and $\mathbf{r}'$ is the integration variable.
Is it true that they first swapped the boundary and volume integral, and then pulled $\rho$ out of the boundary integral? Is it true that $\rho$ is a function of $\mathbf{r}'$, and can't be pulled out of the integral?
Since I could not follow the argument of the answer given by naturallyInconsistent, I made up an example and evaluated it once with integrals swapped, once left in the original order. Here is the result for $\rho = r$, evaluated at $\mathbf{r} = \mathbf{0}$, and integrated over a unit sphere:
Original order of integration:
$$
\Phi(\mathbf{r}) =
\int_{\partial V}
\left(
\int_V
\rho \frac{\mathbf{r} – \mathbf{r}'}{||\mathbf{r} – \mathbf{r}'||^3}
\text{d}V
\right)
\cdot
\mathbf{n}
\text{d}A.
$$
How not to do it
In what follows, the integration is done wrong. The misconception here is that the electric field $\mathbf{E}$ can be evaluated at a certain point $\mathbf{r}=\mathbf{0}$, and then integrated over the surface. But this would require the electric field to be constant everywhere. Instead, it is a function of position $\mathbf{E}(\mathbf{r})$.
First, compute $\mathbf{E}(\mathbf{0}):$
$$
\mathbf{E}(\mathbf{0}) = 8 \int_{\varphi=0}^{\pi/2} \int_{\theta=0}^{\pi/2} \int_{r=0}^1 \frac{r}{r^3}
\left(
\begin{array}{c}
r \sin(\theta)\cos(\varphi) \\
r \sin(\theta)\sin(\varphi) \\
r \cos(\theta)
\end{array}
\right)
r^2 \sin(\theta)
\text{d}r
\text{d}\theta
\text{d}\varphi
=
\left(
\begin{array}{c}
2 \pi/3 \\
4 \pi/3 \\
\pi
\end{array}
\right)
$$
Then, compute the flux:
$$
\Phi(\mathbf{0}) = 8 \int_{\varphi=0}^{\pi/2} \int_{\theta=0}^{\pi/2} \mathbf{E}(\mathbf{0}) \cdot
\left(
\begin{array}{c}
\sin(\theta) \cos(\varphi) \\
\sin(\theta) sin(\varphi) \\
\cos(\theta)
\end{array}
\right)
\sin(\theta)
\text{d}\theta
\text{d}\varphi
=
2 \pi^2 + 16 \pi
$$
Now if I swap the order of integration, I'm evaluating the expression
$$
\int_V \int_{\partial V}
\left(
\rho \frac{\mathbf{r}'}{||\mathbf{r}'||^3}
\right)
\cdot
\mathbf{n}
\text{d}A
\text{d}V
= \int_V \int_{\partial V}
\left(
\frac{\rho}{||\mathbf{r}'||^2}
\right)
\mathbf{n}
\cdot
\mathbf{n}
\text{d}A
\text{d}V
= 8 \pi^2.
$$
So there is at least one case where the order of integration may not be changed?
Best Answer
The integral on the LHS only depends upon $\vec r$ whereas, after swapping the integrals, the $\vec r\,^\prime$ integral is on the outside, and so the $\rho$ that only depends upon $\vec r\,^\prime$ can be pulled out. The inner integral collects all the dependence upon $\vec r$