Gauss' Theorem

Consider the following volume enclosed by a surface we will call S S .

Now we will embed S S in a vector field:

We will cut the the object into two volumes that are enclosed by surfaces we will call S 1 S 1 and S 2 S 2 .

Again we embed it in the same vector field. It is clear that flux through S 1 S 1 + S 2 S 2 is equal to flux through S . S . This is because the flux through one side of the plane is exactly opposite to the flux through the other side of the plane: So we see that ∮ S F ⃗ ⋅ d a ⃗ = ∮ S 1 F ⃗ ⋅ d a 1 ⃗ + ∮ S 2 F ⃗ ⋅ d a 2 ⃗ . ∮ S F ⃗ ⋅ d a ⃗ = ∮ S 1 F ⃗ ⋅ d a 1 ⃗ + ∮ S 2 F ⃗ ⋅ d a 2 ⃗ . We could subdivide the surface as much as we want and so for n n subdivisions the integral becomes:

∮ S F ⃗ ⋅ d a ⃗ = ∑ i = 1 n ∮ S i F ⃗ ⋅ d a i ⃗ . ∮ S F ⃗ ⋅ d a ⃗ = ∑ i = 1 n ∮ S i F ⃗ ⋅ d a i ⃗ . What is ∮ S i F ⃗ ⋅ d a i ⃗ ∮ S i F ⃗ ⋅ d a i ⃗ .? We can subdivide the volume into a bunch of little cubes:

To first order (which is all that matters since we will take the limit of a small volume) the field at a point at the bottom of the box is F z + Δ x 2 ∂ F z ∂ x + Δ y 2 ∂ F z ∂ y F z + Δ x 2 ∂ F z ∂ x + Δ y 2 ∂ F z ∂ y where we have assumed the middle of the bottom of the box is the point ( x + Δ x 2 , y + Δ y 2 , z ) ( x + Δ x 2 , y + Δ y 2 , z ) . Through the top of the box ( x + Δ x 2 , y + Δ y 2 , z + Δ z ) ( x + Δ x 2 , y + Δ y 2 , z + Δ z ) you get F z + Δ x 2 ∂ F z ∂ x + Δ y 2 ∂ F z ∂ y + Δ z ∂ F z ∂ z F z + Δ x 2 ∂ F z ∂ x + Δ y 2 ∂ F z ∂ y + Δ z ∂ F z ∂ z Through the top and bottom surfaces you get Flux Top - Flux bottom

Which is Δ x Δ y Δ z ∂ F z ∂ z = Δ V ∂ F z ∂ z Δ x Δ y Δ z ∂ F z ∂ z = Δ V ∂ F z ∂ z

Likewise you get the same result in the other dimensionsHence ∮ S i F ⃗ ⋅ d a i ⃗ = Δ V i [ ∂ F x ∂ x + ∂ F y ∂ y + ∂ F z ∂ z ] ∮ S i F ⃗ ⋅ d a i ⃗ = Δ V i [ ∂ F x ∂ x + ∂ F y ∂ y + ∂ F z ∂ z ]

or ∮ S i F ⃗ ⋅ d a i ⃗ = ∇ ⃗ ⋅ F ⃗ Δ V i ∮ S i F ⃗ ⋅ d a i ⃗ = ∇ ⃗ ⋅ F ⃗ Δ V i ∮ S F ⃗ ⋅ d a ⃗ = ∑ i = 1 n ∮ S i F ⃗ ⋅ d a i ⃗ = ∑ i = 1 n ∇ ⃗ ⋅ F ⃗ Δ V i ∮ S F ⃗ ⋅ d a ⃗ = ∑ i = 1 n ∮ S i F ⃗ ⋅ d a i ⃗ = ∑ i = 1 n ∇ ⃗ ⋅ F ⃗ Δ V i

So in the limit that Δ V i → 0 Δ V i → 0 and n → ∞ n → ∞ ∮ S F ⃗ ⋅ d a ⃗ = ∮ V ∇ ⃗ ⋅ F ⃗ ⅆ V ∮ S F ⃗ ⋅ d a ⃗ = ∮ V ∇ ⃗ ⋅ F ⃗ ⅆ V

This result is intimately connected to the fundamental definition of the divergence which is ∇ ⃗ ⋅ F ⃗ ≡ lim V → 0 1 V ∮ S F ⃗ ⋅ d a ⃗ ∇ ⃗ ⋅ F ⃗ ≡ lim V → 0 1 V ∮ S F ⃗ ⋅ d a ⃗ where the integral is taken over the surface enclosing the volume V V . The divergence is the flux out of a volume, per unit volume, in the limit of an infinitely small volume. By our proof of Gauss' theorem, we have shown that the del operator acting on a vector field captures this definition.