**Statement: “**The volume integral of the divergence of a vector field A taken over any volume Vbounded by a closed surfaceS is equal to the surface integral of A over the surfaceS.”

Or

”

The volume integral of the divergence of a vector field over the volume enclosed by surface S isequal to the flux of that vector field taken over that surface S.”

Hence, this theorem is used to convert volume integral into surface integral.

Mathematically,

ʃʃʃ_{V} div A dv = ʃ ʃʃ_{V} (∆ .A) d v = ʃ ʃ_{S } A .ds

**Proof :** Let a volume V ) e enclosed a surface S of any arbitrary shape.

Let a small volume element PQRT T’P’Q’R’ of volume dV lies within surface S as shown in Figure 7.13. As we know that flux diverging per unit volume per second is given by div Ai therefore, for volume element dV the flux diverging will be div AdV. If V be the volume enclosed by the surface S, then the total flux diverging through volume V will be equal to the volume integral.

ʃʃʃ_{V} div A dV

Now consider a surface element dS and n a unit vector normal to dS. Let θ be the angle between A and iz at dS, then AdS will give the flux through the surface element dS. Where dS = n dS = area vector along n . Therefore, the total flux passing through the surface S may be obtained by the integral.

ʃʃ_{S} A .n d _{S} = ʃ ʃ S A .dS

But the total flux through the entire surface S [given by Equation (2)] must be equal to the total flux diverging from the volume V enclosed by surfaceS [given by Equation (1)] and therefore

ʃʃʃ_{V} div A D_{v} = ʃ ʃ S A . Ds