# Flux Over a Surface

I am teaching a multivariable calculus course and we are starting to go over surface integrals. I am a math professor with little knowledge of physics. At one point the book discusses fluid flow. It is trying to convince me that the surface integral of a velocity field is the flow of the fluid across the surface. If $\vec{v}$ is my vector field, then we integrate $\vec{v}\cdot \vec{n}$, where $\vec{n}$ is the outward normal. So, we are integrating the work of $\vec{v}$ in the normal direction. Doesn't this tell you how much the fluid wants to leave the surface in the normal direction? Why does it give you flow across the surface?

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Let's first assume that "how much fluid" means "the mass of fluid." Let $\rho(t, \mathbf x)$ denote the volume mass density of the fluid at time $t$ and position $\mathbf x$ in space and let $\mathbf v(t, \mathbf x)$ denote the velocity vector field of the fluid. The vector field $$\mathbf j = \rho\mathbf v$$ gives the mass per unit area, per unit time passing through a given surface often called the mass current density.

To motivate this, imagine some small oriented surface element $d\mathbf a = da\mathbf n$ at some point in space and some short period of time $dt$, then the quantity $$\mathbf v\cdot d\mathbf a\, dt$$ gives the volume of the fluid passing through the area element in the time $dt$. To see this, think of the "tube" of fluid that is "behind" the surface area element and that is about to flow through it. Multiplying, then, by the volume mass density gives the mass of the fluid that passed through the area element in the time $dt$ $$dM = \rho\mathbf v\cdot d\mathbf a\,dt$$ Drawing a picture would really help here. Notice that if we divide both sides by $dt$ then we obtain the rate of mass flow across the surface element $$\frac{dM}{dt} = \rho\mathbf v\cdot d\mathbf a$$

The first part of the argument shows that the if by "how much fluid" you mean "volume of fluid" you just need to eliminate the $\rho$ factor and simply use the velocity vector field. The argument would then show that the rate of volume flow across the surface element is $$\frac{dV}{dt} = \mathbf v \cdot d \mathbf a$$

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