Gauss's law for magnetism is stated as followed with the beautiful closed surface double integral (by wikipidia):

$$ \mathop{\vcenter{ \huge\unicode{x222F}\, }}_{S} \mathbf{B} \cdot \text d\mathbf{A} = 0 $$

As I understand, the idea is to say that if we sum (continuous sum since integral) all the scalar products between the vector field $\mathbf{B}$ (i.e., magnetic field) and surface elements $\text d\mathbf{A}$ defined by their surface normals, we get $0$?

Given the above is correct, why using the double integral (I assume the circle is for ''closed surface'') ? But why use a double integral, whereas in other fomulas like for the magnetic flux, they use a simple integral although it is still a continuous summation over a surface - unless I'm mistaken - ?

i.e., :

$\Phi_B = \oint_S \mathbf{B} \cdot \text d\mathbf{S}$


1 Answer 1


A double integral in this context means you are integrating over a surface. The integral here is a double integral because a surface is parametrized by two parameters.

I think your confusion lies in the notation. A surface integral is sometimes denoted with two integral symbols, but not always. So the integral in the definition of magnetic flux is no different from the integral that appears in Gauss' law for magnetism, which in fact says that the magnetic flux through any closed surface is zero.

  • $\begingroup$ Ha ok, so you are saying that the 2 notations are in fact equivalent to the same thing : $\unicode{x222F} B \cdot dS$ = $\oint B \cdot dS$ ? So it is just a matter of taste to choose which one you prefer ? $\endgroup$
    – SheppLogan
    Sep 4, 2019 at 15:27
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    $\begingroup$ Yes. Magnetic flux can also be defined for a surface that is not closed, in which case you don't put the circle on the integral symbol, but it is still a double integral. It is largely a matter of preference. In particular, if you are working with integrals over $n$-dimensional manifolds, where $n$ can be greater than 3, you tend to denote everything with a single integral symbol. $\endgroup$
    – Puk
    Sep 4, 2019 at 15:29
  • $\begingroup$ Ok, but so in the 2D case i guess if you have a non-closed surface and you therefore have to remove the circle, then you should put a double integral and not a simple one? Like $\Phi_B = \iint_S B \cdot dS$ and not $\Phi_B = \int_S B \cdot dS$ ? $\endgroup$
    – SheppLogan
    Sep 4, 2019 at 15:42
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    $\begingroup$ Yes, it is acceptable. $\endgroup$
    – Puk
    Sep 4, 2019 at 15:48
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    $\begingroup$ @Machupicchu It is typical when teaching multi-dimensional calculus to insist on multiple-integration signs to denote the dimensionality of the work. But very often sophisticated readers will to be able to deduce the dimension from other clues, making the terser notional functional. If in doubt that your intent will be clear to your readers, use the multiple integration signs ($\iint$, etc) for clarity. $\endgroup$ Sep 4, 2019 at 16:57

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