Multivariate Calculus - Bruce E. Shapiro

LECTURE 24. FLUX INTEGRALS & GAUSS’ DIVERGENCE THEOREM 197
Example 24.2 Find the flux integral of
through a disk of radius 2 on the plane
in the upward direction.
F = (2, 0, 0)
x + y + z = 2
Solution. A normal vector to the plane that is oriented upward is
N = (1, 1, 1)
which has magnitude √ 3 and hence a unit normal vector is
n = 1 √
3
(1, 1, 1)
The area of the disk is A = π × 2 2 = 4π. Hence an area vector is
Therefore the flux is
A = nA = 4π √
3
(1, 1, 1)
flux = F · A = 4π √
3
(2, 0, 0) · (1, 1, 1) = 8π √
3
Example 24.3 Find the outward flux integral of the vector field F = (2 − x, 0, 0)
over the face of the cube with one corner at the origin, entirely in the first octant,
with each side of length 3.
Solution. Since the vector field is entirely parallel to the x-axis, of the six surfaces
of the cube, four of them have normals that are perpendicular to the vector field.
Since the dot product of these vectors with the vector field is zero, they produce no
contribution to the flux. The remaining two surfaces are those perpendicular to the
x axis, one at x = 3 and the other at x = 0. Each of these two surfaces has an area
of A = 9. At the x = 3 surface, n = (1, 0, 0) and F = (−1, 0, 0), hence
flux = F · nA = (1, 0, 0) · (−1, 0, 0) × 9 = −9
At the x = 0 surface, n = (−1, 0, 0) and F = (2, 0, 0), hence
flux = F · nA = (−1, 0, 0) · (2, 0, 0) × 9 = −18
So the total flux is −9 − 18 = −27.
We have previously defined the divergence as
∇F = M x + N y + P z
when F = (M, N, P ). We can use flux integrals to give an alternate definition of the
divergences, which gives a more physical description of why it is called “divergence.”
In fact, the limit in the following theorem was the original definition of the divergence
and it was not until sometime later that the derivative formula was derived.
Math 250, Fall 2006 Revised December 6, 2006.