What's a formula for the flux across a graph in 2 dimensions using the upward pointing normal?
I know a flux extends to 2 dimensions by the formula, $\int_ γ\vec{F}\cdot\vec{n}ds$. And we can define a function f(x), with a path defined by that function by $(t, f(t))$. I believe the formula should be similar to the integral written above.
Yes, the main question is how to get $\vec{n}$
If we let $\vec{r}=t\mathbf{i}+f(t)\mathbf{j}$ then we can construct the unit tangent vector $\vec{T}=\frac{\mathbf{r}'(t)}{\left|\mathbf{r}'(t)\right|}$
$$\vec{T}=\frac{\mathbf{i}+f'(t)\mathbf{j}}{\sqrt{1+f'(t)^2}}$$
We are interested in flux so we wish to find the normal vector $\vec{n}$ with the relation
$$\vec{T}\cdot\vec{n}=0$$
So
$$\vec{n}=\frac{-f'(t)\mathbf{i}+\mathbf{j}}{\sqrt{1+f'(t)^2}}$$
This is the same normal vector as
$$\vec{n}=\frac{\mathbf{T}'(t)}{\left|\mathbf{T}'(t)\right|}\space\space\space\text{(show this)}$$
So given that $\vec{r}(t)$ traces $f(x)$, $\vec{F}(x,y)$ is the vector field, and $\gamma$ is the segment of arc length to be considered, then
$$\Phi(t)=\int_\gamma\left(\vec{F}(x,y)\cdot\vec{n}\right)ds=\int_{t_0}^t\left(\vec{F}(\vec{r}(t))\cdot\vec{n}\right)\left|\mathbf{r}'(t)\right|dt$$
Where $\vec{F}(\vec{r}(t))\equiv\vec{F}(x(t),y(t))$