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I've been reading through FLP Vol. II, and he has proven that as the flux through a closed surface is: $\ \int_{surface} \mathbf{F} \space \mathrm{d}\mathbf{a} $, according to the divergence theorem, the flux through a surface can be defined as: $\ \int_{volume} \nabla \cdot \mathbf{F} \space \mathrm{d}V $, where $\ \mathbf{F} $ is any vector field, and the volume is that which is enclosed by the surface.

Previously he had stated as a word equation that: $\ \text{Flux of } \mathbf{B} \text{ through any closed surface}=0. $ I would therefore assume that $\ \int_{volume} \nabla \cdot \mathbf{B} \space \mathrm{d}V = 0$, however Gauss' law for magnetism states that: $\ \nabla \cdot \mathbf{B} = 0$. Does that mean that $\ \nabla \cdot \mathbf{B} = 0$ and $\ \int_{volume} \nabla \cdot \mathbf{B} \space \mathrm{d}V = 0$ are equivalent statements, or am I making a fundamental error somewhere?

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  • $\begingroup$ They're equivalent alright. This is exactly how you go between the integral and differential formulations of Maxwell's laws. $\endgroup$ – David H Apr 6 '14 at 14:45
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Intuitively, if the volume integral of a function is 0 over any arbitrary volume, the function itself must be 0 at all points in space.

More concretely, consider a function for which $\int_V \, f \, \mathrm{d}x = 0$ for any volume $V$. Then, $\int_{V+dV} \, f \, \mathrm{d}x = 0$ for any infinitesimal addition to V.

$$\int_{V+dV} \, f \, \mathrm{d}x - \int_V \, f \, \mathrm{d}x = \int_{dV} \, f \, \mathrm{d}x = f(\text{at dV}) = 0$$

In your case, $f = \nabla \cdot B$, so $\nabla \cdot B = 0$.

(Note: I was a bit lazy with my notation above, so it's not a formal proof. However, it should still provide the intuitive answer to your question.)

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As Draksis said, the condition is that the integral over any volume has to be zero. If you want a formal proof, here you go:

Let's call $f(\mathbf{x}) = \nabla \cdot \mathbf{B}$, and assume it is continuous. Suppose there's some $\mathbf{x}_0 \in \mathbb{R}^3$ with $f(\mathbf{x}_0) \neq 0$, and let's say that $f(\mathbf{x}_0) > 0$ (the proof for $f < 0$ is identical). Then since $f$ is a continous function, there is some ball $B$ around $\mathbf{x}_0$ where $f$ is positive. Therefore, $\int_B f > 0$, which is a contradiction with the assumption that the integral should be zero for any volume.

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