using the divergence theorem

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The divergence theorem only applies for closed surfaces S. However, we can sometimes work out a flux integral on a surface that is not closed by being a little sneaky. NOTE that this is NOT always an efficient way of proceeding. However, it sometimes is, and this is a nice example of both the divergence theorem and a flux integral, so we'll go through it as is.

figure showing the surface
figure 1: the surface
figure showing the surface with a bottom
figure 2: the surface with a bottom added

Find the flux of the vector field F = x y i + y z j + x z k through the surface z = 4 - x2 - y2, for z >= 3.

The surface is shown in the figure to the right. Because this is not a closed surface, we can't use the divergence theorem to evaluate the flux integral. However, if we had a closed surface, for example the second figure to the right (which includes a bottom surface, the yellow section of a plane) we could. We'll consider this in the following.

The divergence theorem says

int int_S F.dS = int int int_E div F dV

where the surface S is the surface we want plus the bottom (yellow) surface. So we can find the flux integral we want by finding the right-hand side of the divergence theorem and then subtracting off the flux integral over the bottom surface. This gives us nice practice both applying the divergence theorem and finding a surface integral, so we'll do it.

The divergence theorem part of the integral: Here div F = y + z + x. Note that here we're evaluating the divergence over the entire enclosed volume, so we can't evaluate it on the surface. Doing the integral in cylindrical coordinates, we get

int int int_E div F dV = int int int_E x + y + z dV = (solution for integral over the unit circle and 3<=z<=4-r^2) = 5pi/3

The flux through the bottom boundary: Note that here we have a very easy parameterization of the surface, r = <x, y, 3>. The normal vector N = <0, 0, -1> (because we want an outward normal), and dS = dx dy. Thus on the surface F = F = x y i + y z j + 3 x k, and the surface integral becomes

int int_S F.dS = int int_S F.N dS = int int_D -3x dA = 0

Putting it together: here, things dropped out nicely. Using the divergence theorem, we get the value of the flux through the top and bottom surface together to be 5 pi / 3, and the flux calculation for the bottom surface gives zero, so that the flux just through the top surface is also 5 pi / 3.

(How were the figures here generated? In Maple, with this maple worksheet.)


using the divergence theorem
Last modified: Mon Apr 19 12:06:56 EDT 2004
©2004 Gavin LaRose, UM Math Dept.