I have the following question:
Determine the total flux of the velocity field
$$\vec{v} = \begin{pmatrix} 1 \\ x + y \end{pmatrix}$$
out of the disk $D$, a circle with radius 2 and center at the origin
The answer is:
the Boundary of this disk $D \in \mathbb{R}^{2}$ may be parametrized by
$$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = \begin{pmatrix} 2 \cos t \\ 2 \sin t \end{pmatrix} \quad \text{for} \quad 0\leq t \leq 2\pi$$
Thus we have $$ \begin{pmatrix} -2 \sin t \\ 2 \cos t \end{pmatrix}dt$$ and $$ds = ||\vec{ds}|| = 2dt \tag{1}\label{eq1}$$
The outer unit normal vector a time (or angle) $t$ is given by
$$\vec{n}(t) = \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} \tag{2}\label{eq2}$$
Which mathematical axiom / theorem do we have to invoke to be able to state (1) and (2)
The flux of a closed curved $\boldsymbol{\gamma}:[a,b] \rightarrow \mathbb{R}^2$ is given by the formula: $$\oint_{\gamma} \left < \mathbf{v}, \mathbf{\hat{n}}\right> ds \;\;\;(1)$$
Where $\mathbf{v}$ is the velocity field, $\mathbf{\hat{n}}$ is the unit normal of $\boldsymbol{\gamma}$ and $<.,.>$ is the dot product.
As the comments suggested $ds$ is the infinitesimal arc-length of the curve, which is given by: $$ds = |\dot{\boldsymbol{\gamma}}| dt$$
And $\mathbf{\hat{n}}$ can be thought as the left/right rotation of the unit tangent vector of the curve: $$\mathbf{\hat{n}} = \mathbf{R}\mathbf{\hat{t}}$$
Where $\mathbf{\hat{t}}$ is the unit tangent vector, and $\mathbf{R}$ is the rotation matrix, that rotates a vector to the right.
So, $$\mathbf{\hat{t}} = \frac{\boldsymbol{\dot\gamma}}{|\boldsymbol{\dot\gamma}|}$$ and $$\mathbf{R} = \begin{bmatrix} 0& 1\\ -1&0 \end{bmatrix}$$
Therefore (1) becomes: $$\oint_{\gamma} \left < \mathbf{v}, \mathbf{\mathbf{R}\mathbf{\hat{t}}}\right> |\dot{\boldsymbol{\gamma}}| dt$$ $$\oint_{\gamma} \left < \mathbf{v}, \mathbf{\mathbf{R}\frac{\boldsymbol{\dot\gamma}}{|\boldsymbol{\dot\gamma}|}}\right> |\dot{\boldsymbol{\gamma}}| dt$$
Taking the scalar inside the dot product:
$$\oint_{\gamma} \left < \mathbf{v}, \mathbf{\mathbf{R}\boldsymbol{\dot\gamma}}\right> |\dot{\boldsymbol{\gamma}}|/ |\boldsymbol{\dot\gamma}| dt$$
$$\oint_{\gamma} \left < \mathbf{v}, \mathbf{\mathbf{R}\boldsymbol{\dot\gamma}}\right> dt$$
Using the info above, I can answer your questions:
Note that $\boldsymbol{\gamma}(t) = \begin{bmatrix} 2\cos(t)\\ 2\sin(t) \end{bmatrix}$ then:
$$ds = |\dot{\boldsymbol{\gamma}}| dt = \sqrt{\left(2\cos(t)\right)^2 + \left(2\sin(t)\right)^2} dt = \sqrt{4\cos^2(t) + 4\sin^2(t)}dt = 2dt$$
And $\mathbf{n}(t)$: $$\mathbf{n}(t) = \mathbf{R}\mathbf{t} = \mathbf{R}\dot{\boldsymbol{\gamma}}(t) =\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \begin{bmatrix} -2\sin(t)\\ 2\cos(t) \end{bmatrix} = \begin{bmatrix} 2\cos(t)\\ 2\sin(t) \end{bmatrix}$$
Hope this answer your question, in some way :)