Considering the answer of the ProfRob https://physics.stackexchange.com/questions/649640/magnetic-flux-of-a-dipole-through-a-sphere/649706#649706, just I was curious how he get:
$$ \oint_A \mathbf{B} \cdot d\mathbf{a} = \frac{\mu_0}{r} \int^{\pi}_{0} \sin\theta\cos\theta \ d\theta = 0 $$ knowing that $A$ it is a spherical closed surface and that in spherical polar coordinates, the magnetic field of a dipole is $$ \mathbf{B} = \frac{\mu_0 m \cos \theta}{2\pi r^3} \hat{\bf r} + \frac{\mu_0 m \sin \theta}{4\pi r^3}\hat{\boldsymbol{\theta}} \tag 1$$ where $m$ is the magnetic dipole moment.
Why, being the magnetic dipole moment a vector $\mathbf{m}$, is it computed like a scalar in the $(1)$?
I know that for a function $f$ in polar coordinates is:
$$\iint_{\text{area surface } A} f(r,\theta) \, dA = \int_a^b \int_0^{r(\theta)} f(r,\theta)\,r\,dr d\theta \tag2 $$
where $dA$ is an infinitesimal area element in polar coordinates that can be written as
$$dA = |J|\,dr d\theta = r\,dr d\theta.$$
where $A$ corresponds to the area enclosed by the curve $r(\theta)$ and the straight lines $\theta =a$ and $\theta =b$.
Hence, why and where is $m$? $$ \oint_A \mathbf{B} \cdot d\mathbf{a} \equiv \color{orange}{\frac{\mu_0}{r} \int^{\pi}_{0} \sin\theta\cos\theta d\theta = 0}$$
$\require{cancel}$
On a sphere, using spherical coordinates, the area element is radial, moreover
$$ {\rm d}{\bf a} = r^2\sin\theta~{\rm d}\theta~{\rm d}\phi~\hat{\bf r} $$
where $0 < \phi < 2\pi$ and $0 < \theta < \pi$
with this in mind
\begin{eqnarray} {\bf B}\cdot {\rm d}{\bf a} &=& \left(\frac{\mu_0 m}{2\pi r^3}\cos\theta~\hat{\bf r} + \frac{\mu_0 m}{4\pi r^3}\sin\theta~\hat{\bf \theta} \right) \cdot \left(r^2\sin\theta~{\rm d}\theta~{\rm d}\phi~\hat{\bf r} \right) \\ &=& \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta{\rm d}\theta~{\rm d}\phi \cancelto{1}{(\hat{\bf r} \cdot \hat{\bf r})} + \frac{\mu_0 m}{4\pi r}\sin^2\theta {\rm d}\theta~{\rm d}\phi \cancelto{0}{(\hat{\bf r}\cdot \hat {\bf \theta})} \\ &=& \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta{\rm d}\theta~{\rm d}\phi \end{eqnarray}
\begin{eqnarray} \oint_A {\rm d}{\bf a} \cdot {\bf B} &=& \int_{0}^{2\pi} {\rm d\phi} \int_{0}^{\pi}{\rm d}\theta~ \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta \\ &=& \left(\int_{0}^{2\pi} {\rm d}\phi\right)\left(\int_{0}^{\pi}{\rm d}\theta~ \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta \right) \\ &=& 2\pi\frac{\mu_0 m}{2\pi r} \int_0^\pi {\rm d}\theta~\sin\theta\cos\theta \\ &=& \frac{\mu_0 m}{r} \int_0^\pi {\rm d}\theta~\sin\theta\cos\theta \\ &=& 0 \end{eqnarray}