Given a function $f(x,y):\mathbb R\times \mathbb R\to \mathbb C$ and a real parameter $\theta$, one can use Leibniz's integral rule to solve \begin{equation}\label{eq}\tag{1}\frac{d}{d\theta}\int_{a(\theta)}^{b(\theta)}dx \ f(x,y)= b'(\theta)f(b(\theta),y) - a'(\theta)f(a(\theta),y). \end{equation} Notice that $f$ does not depend itself on $\theta$, and therefore there is no integral term in the RHS.
I am interested in the generalization of $(1)$ in the two-dimensional case, where $f(\mathbf x, \mathbf y): \mathbb R^2\times \mathbb R^2 \to \mathbb C$ is integrated over a region $S(\theta)$ whose geometry depends on $\theta$ (for example, this might be the radius of a circle, or the edge of a square). As far as I understand, Leibniz's rule in three dimension is related to Reynold's transport theorem, according to which given a function $f(\mathbf x, \theta)$ over a region $\Omega(\theta)$ with moving boundary $\partial\Omega(\theta)$ it holds that \begin{equation}\label{2}\tag{2} \frac{d}{d\theta}\int_{\Omega(\theta)}f \ dV=\int_{\Omega(\theta)}\frac{\partial f}{\partial \theta}\ dV + \int_{\partial\Omega(\theta)}(\mathbf v_b\cdot \mathbf n)f\ dA \end{equation} where $\mathbf v_b$ is the velocity of the area element and $\mathbf n$ the normal vector.
However, my situation is a bit different. First of all, since $f$ does not depend on the parameter $\theta$, the first integral in the RHS is zero. Secondly, since I care about the 2D case, I should be able to rewrite $(2)$ as something like $$\tag{3}\frac{d}{d\theta}\int_{S(\theta)}f \ dA= \int_{\partial S(\theta)}(\mathbf v_b\cdot \mathbf n)f \ dl.$$ Is $(3)$ valid under my hypotheses, and if so, is there any way to make the scalar product $\mathbf v_b \cdot \mathbf n$ more explicit without fixing a specific geometry?