Multivariate Calculus - Bruce E. Shapiro

LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM 199
the surface integral is
F · dA = F(x + dx, y + dy/2, z + dz/2)dydz · (1, 0, 0)
S
+ F(x, y + dy/2, z + dz/2)dydz · (−1, 0, 0)
+ F(x + dx/2, y + dy, z + dz/2)dxdz · (0, 1, 0)
+ F(x + dx/2, y, z + dz/2)dxdz · (0, −1, 0)
+ F(x + dx/2, y + dy/2, z + dz)dxdy · (0, 0, 1)
+ F(x + dz/2, y + dy/2, z)dxdy · (0, 0, −1)
= [M(x + dx, y + dy/2, z + dz/2) − M(x, y + dy/2, z + dz/2)]dydz
+ [N(x + dx/2, y + dy, z + dz/2) − N(x + dx/2, dy, z + dz/2)]dxdz
+ [P (x + dx/2, y + dy/2, z + dz) − P (x + dx/2, y + dy/2, z)]dxdy
Hence
1
V
S
F · dA =
1
F · dA
dxdydz
S
M(x + dx, y + dy/2, z + dz/2) − M(x, y + dy/2, z + dz/2)
=
dx
N(x + dx/2, y + dy, z + dz/2) − N(x + dx/2, dy, z + dz/2)
+
dy
P (x + dx/2, y + dy/2, z + dz) − P (x + dx/2, y + dy/2, z)
+
dz
Taking the limit as V → 0,
1
lim F · dA = M x + N y + P z = ∇ · F
V →0 V
S
Theorem 24.2 Gauss’ Divergence Theorem. Let F = (M, N, P ) be a vector
field with M, N, and P continuously differentiable on a solid S whose boundary is
∂S. Then
F · n dA = ∇ · F dV (24.2)
where n is an outward pointing unit normal vector.
∂S
Proof. Break the volume down into infinitesimal boxes of volume
∆V = ∆x × ∆y × ∆z
Then we can expand the volume integral as a Riemann Sum
∇ · F dV =
V
S
n∑
∆V i ∇ · F i (24.3)
Math 250, Fall 2006 Revised December 6, 2006.
i=1