Flux of a vector field that is a partial derivative

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Let $X = \partial_x$ and $Y = x \partial_y$ be vectorfields on $M = \mathbb{R}^2$

I want to calculate the flux of $X$ and $Y$. I am confused on how to approach this problem.
The formula to calculate the flux of a vector field is: $$\Phi = \int_S F \cdot dS$$ However, If I use $F = \partial_x$ we obtain: $$\int_S \partial_x \cdot dS$$ which doesn't make a lot of sense to me

Edit: $S$ refers to a surface, meaning that I am using a surface integral

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This is ALMOST not an answer, but a follow-up comment.

In the language of vector calculus: when $S$ is a region in the plane, the flux $\Phi$ of a vector field $F$ is calculated across the boundary $\partial S$ of $S$: $$ \Phi = \int_{\partial S} F\cdot n\, ds,$$ where $n$ is the outward normal, and $ds$ is the length element (i.e., the integral is one dimensional).

But, by the divergence/Gauss theorem, one can also calculate the flux $\Phi$ by the formula $$ \Phi= \int_S \mathop{\text{div}} F\, dA,$$ where the integral is now an area integral (i.e., a double integral), and $dA$ is an area element, and $\mathop{\text{div}} F$ is the divergence of $F$ (which is often also denoted $\nabla\cdot F$).

Typically one needs to know $S$ (or $\partial S$). However! In your case, the divergence theorem is pretty relevant (i.e., hint: calculate the divergence!) - I am assuming you know it.

Hope this helps.