When describing surface integrals in vector fields, it's common to use the notation $$\iint_S \vec{F} \cdot \text{d} \vec{S}$$ as a shorthand for $$\iint_S \vec{F} \cdot \vec{n}\, \text{d}S$$ This seems to be analogous to the notation for line integrals: $$\int_C \vec{F} \cdot \text{d}\vec{r} = \int_C \vec{F} \cdot \vec{T} \,\text{d}s$$ However, I don't understand why it makes sense to shorten $\vec{n} \,\text{d}S$ to $\text{d}\vec{S}$.

With line integrals, it makes sense to have $\vec{T} \,\text{d}s = \text{d}\vec{r}$ because $\text{d}\vec{r}$ notationally represents an infinitesimal movement along the curve described by $\vec{r}$. But with surface integrals, it doesn't seem like $\text{d}\vec{S}$ represents movement along the surface because the vector is normal to the surface.

At first I assumed that it was just a weird notation that resulted from the analogy to line integrals. However, I found a resource from MIT (http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v9.pdf) that called the $\text{d}\vec{S}$ notation "suggestive". Of what is that notation suggestive?

  • 4
    $\begingroup$ If you want to attach a vector to a surface, your only choice is in the normal direction. Think of a plane. No direction in the plane is a preferential direction, but the perpendicular to the plane is unique (up to a sign) $\endgroup$
    – Andrei
    Jul 10 '20 at 3:24

Generally a surface integral is answering the informal question of, "How much "stuff" is flowing through the surface?" That's why it makes sense to dot with the normal vector to the surface, because just as $\text{d}\vec{r}$ represents movement along the curve, $\text{d}\vec{S}$ represents movement through the surface. See flux, or the divergence theorem.

  • $\begingroup$ The thing that bothers me about this notation is that $\text{d}\vec{S}$ isn't related to some original value $\vec{S}$ in the way that, for example, $\text{d}\vec{r}$ is related to a curve $\vec{r}$ or $\text{d}A$ is related to area $A$. However, I think this explanation is still helpful. $\endgroup$
    – Mu Prime
    Jul 10 '20 at 6:19
  • $\begingroup$ @MuPrime But d$\vec S$ is the differential area element a surface $S$ (with position vector also $\vec r,$ lying in 3-d space). In fact, roughly speaking, the magnitude of d$\vec S$ approaches that of d$A$ as the surface becomes flatter. $\endgroup$
    – ryang
    Dec 18 '21 at 6:00

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