22.4. SOME APPLICATIONS OF THE DIVERGENCE THEOREM 409

Proof: The divergence theorem holds for balls because they are cylindrical in everydirection. Therefore,

1v(B(x,δ ))

∫∂B(x,δ )

F ·ndA =1

v(B(x,δ ))

∫B(x,δ )

divF (y) dV.

Therefore, since divF (x) is a constant,∣∣∣∣divF (x)− 1v(B(x,δ ))

∫∂B(x,δ )

F ·ndA∣∣∣∣

=

∣∣∣∣divF (x)− 1v(B(x,δ ))

∫B(x,δ )

divF (y) dV∣∣∣∣

=

∣∣∣∣ 1v(B(x,δ ))

∫B(x,δ )

(divF (x)−divF (y)) dV∣∣∣∣

≤ 1v(B(x,δ ))

∫B(x,δ )

|divF (x)−divF (y)| dV

≤ 1v(B(x,δ ))

∫B(x,δ )

ε

2dV < ε

whenever ε is small enough, due to the continuity of divF . Since ε is arbitrary, this shows(22.4). ■

How is this definition independent of coordinates? It only involves geometrical notionsof volume and dot product. This is why. Imagine rotating the coordinate axes, keepingall distances the same and expressing everything in terms of the new coordinates. Thedivergence would still have the same value because of this theorem.

22.4 Some Applications Of The Divergence Theorem

22.4.1 Hydrostatic PressureImagine a fluid which does not move which is acted on by an acceleration g. Of course theacceleration is usually the acceleration of gravity. Also let the density of the fluid be ρ , afunction of position. What can be said about the pressure p in the fluid? Let B(x,ε) be asmall ball centered at the point x. Then the force the fluid exerts on this ball would equal

−∫

∂B(x,ε)pndA.

Here n is the unit exterior normal at a small piece of ∂B(x,ε) having area dA. By thedivergence theorem, (see Problem 1 on Page 426) this integral equals

−∫

B(x,ε)∇pdV.

Also the force acting on this small ball of fluid is∫B(x,ε)

ρgdV.