gauss's divergence theorem

Vector CalculusGauss’s Divergence Theorem

The theorem of Gauss, like Green’s theorem, relate an integral over a set S to another integral over the boundary of S.To emphasize the similarity in these theorems, we introduce the notation ∂S to stand for the boundary of S.

Thus, one form of Green’s Theorem can be written as

∮∂S F · n ds = ∫∫ div F dAs

It says that the flux of F across the boundary ∂S of a closed bounded plane tegion S is equal to the double integrak of div F over that region. Gauss’s Theorem (also called the Divergence Theorem) lifts this result up one dimension.

Let S be a closed bounded solid in three-space that is completely enclosed by a piecewise smooth surface ∂S (Figure 1).

Gauss’s Theorem

Let F = Mi + Nj +Pk be a vector field such M,N, and P have continuos first-order partial derivates on a solid S with boundary ∂S. If n denotes the outer unit normal to ∂S, then

∫∫ F · n ds = ∫∫∫div F dV

∫∫

S∂S

In other words, the flux of F across the boundary of a closed region in three-space is the triple integral of its divergence over that region.

It is useful both for some applications and for the proud to state the conclusion to Gauss’s Theorem in its Cartesian (nonvector form). We may write

n = cos α i + cos β j + cos γ k

where α, β and γ are the direction angles for n.

Thus

F · n = M cosα i + N cosβ j + P cosγ k

and so Gauss’s formula becomes

∫∫(Mcosα i + Ncosβ j + Pcosγ k) dS = ∫∫∫( + + ) dV∂S S

We first consider the case where the region S is x-simple, y-simple, and z-simple. It will be sufficient to show that

∫∫ M cos α dS = ∫∫∫ dV∫∫ N cos β dS = ∫∫∫ dV∫∫ P cos γ dS = ∫∫∫ dV

Proof of Gauss’s Theorem

∂S

∂S

∂S

S

S

S

Since these demonstrations are similar, we show only the third.Since S is z-simple, it can be described by the inequalities f1(x,y) ≤ z ≤ f2(x,y).

As in Figure 2, ∂S consist of three parts: S1, corresponding to z = f1(x,y); S2, corresponding to z = f2(x,y); and the lateral surface S3, which may be empty. On S3, cosγ = cos90°= 0, so we can ignore its contribution

We can deduce that

∫∫ P cosγ dS = ∫∫ P(x, y, f 2(x, y))dx dy

The result to which we just referred assumes that the normal n points upward. Hence, when we apply it to S1, where n is a lower normal (Figure 2), we must reverse the sign.

S2 R

∫∫ P cosγ dS = -∫∫ P(x, y, f 1(x, y))dx dy

It follows that

∫∫ P cosγ dS = ∫∫ [P(x, y, f 2(x, y))- P(x, y, f 1(x, y))]dx dy

= ∫∫ [∫ dz]dx dy

= ∫∫∫ dV

S1 R

∂S R

f2(x,y)

f1(x,y)R

S

The result just proved extends easily to regions that are finite unions of the type considered. We omit the details