Can someone help/correct me in my thinking when working calculating the flux through a surface. Given is the following example:
When a vector field is described by: $$ \vec{E} = xz\,\hat{\imath} - yz\,\hat{\jmath} + (x^2 + y^2)\,\hat{k}$$
And a circular disk through which we want to calculate the flux is described by: $$ D \rightarrow\; z=0,\, x^2+y^2 \leq a^2 $$
I am aware that in order to calculate the flux through the surface we need a vector perpendicular to the surface of the disk to evaluate the integral. Since $z=0$, a unit normal vector is given by $\hat{n} = 0\,\hat{\imath} + 0\,\hat{\jmath} + 1\,\hat{k}$ this could be used. Right?
We are taught that another way to determine a vector perpendicular to the surface $D$ is by parametrizing the surface $D$, as below for example, and then determining $\frac{\partial r}{\partial \rho} \times \frac{\partial r}{\partial \theta}$. $$ D\rightarrow\; r(\rho, \theta) = \rho \cos(\theta)\,\hat{\imath} + \rho\sin(\theta)\,\hat{\jmath} + 0\,\hat{k} $$ with $\rho \in [0, a] \land \theta \in [0, 2\pi]$.
Is it correct that the flux is then given by (question mark indicating I'm not sure about my understanding, and I might have gone wrong somewhere way earlier): $$ \iint_D \vec{B} \cdot d \vec{S} \;\overset{?}{=}\, \iint_D \vec{B}(\vec{r}(\rho, \theta)) \cdot \biggl( \frac{\partial r}{\partial \rho} \times \frac{\partial r}{\partial \theta} \biggr) \, dA $$
The length of this normal vector does matter right. So how has this been taken into account for in the integral with the cross product? Is it because you substitute the parameterization of the surface into the vector field?