Multivariate Calculus - Bruce E. Shapiro

198 LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM
Theorem 24.1 Let F = (M, N, P ) be a differentiable vector field defined within
some closed volume V with surface A. Then the divergence is given by
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∇ · F = M x + N y + P z = lim F · dA (24.1)
V →0 V
Example 24.4 Find the divergence of the vector field F = (x, y, z).
Solution: Algebraic Method
∇ · F = M x + N y + P z = 1 + 1 + 1 = 3
Solution: Geometric Method Define a volume V of dimensions dx × dy × dz that is
centered at the point (x, y, z). We can define the following data for this cube.
Side Outward Normal dA dA = nA
x + dx/2 (1,0,0) dydz (1, 0, 0)dydz
x − dx/2 (-1, 0, 0) dydz (−1, 0, 0)dydz
y + dy/2 (0, 1, 0) dxdz (0, 1, 0)dxdz
y − dy/2 (0, -1, 0) dxdz (0, −1, 0)dxdz
z + dz/2 (0, 0, 1) dxdy (0, 0, 1)dxdy
z − dz/2 (0, 0, -1) dxdy (0, 0, −1)dxdy
Assuming the center of the cube is at (x, y, z) then we also have the following by
evaluating the vector field at the center of each face.
Side F F · dA
x + dx/2 (x + dx/2, y, z) (x + dx/2)dydz
x − dx/2 (x − dx/2, y, z) −(x − dx/2)dydz
y + dy/2 (x, y + dy/2, z) (y + dy/2)dxdz
y − dy/2 (x, y − dy/2, z) −(y − dy/2)dxdz
z + dz/2 (x, y, z + dz/2) (z + dz/2)dxdy
z − dz/2 (x, y, z − dz/2) −(z − dz/2)dxdy
Adding up all the fluxes, we find that
F · dA = 3dxdydz
Hence the divergence is
S
∬
s
∇F = lim
F · dA 3dxdydz
= lim
V →0 V V →0 dxdydz = 3
which is the same value we obtained analytically.
Proof of Theorem 24.1. The proof is similar to the example. Consider a box centered
at the point (x + dx/2, y + dy/2, z + dz/2) of dimensions V = dx × dy × dx. Letting
F(x, y, z) = ( M(x, y, z), N(x, y, z), P (x, y, z) )
Revised December 6, 2006. Math 250, Fall 2006
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