Professor Dave again, let’s talk about the Divergence Theorem.
Green’s Theorem gave us an additional method for calculating
a line integral around a closed curve. Now we will introduce an additional method of
calculating the surface integral for a “closed” surface S. By this we mean that we have some
solid region in space, and its boundary will be our surface. Recall that the surface integral
for this will be expressed as the surface integral over S, of F dot n dS, where F is
some vector field permeating space, and n is the normal vector of the surface. The divergence
theorem, also called Gauss’s Theorem, states that this surface integral is equal to the
triple integral over the solid region of the divergence of F dV. Here, dV is our differential
volume, which in cartesian x, y, z coordinates will simply be dx dy dz.
Let’s take a look at an example to see how to apply this theorem. Consider the surface
of this triangular prism which occupies x from 0 to 1 and y from 0 to 2. It has a flat
base in the xy plane, and the top surface is given by the plane z = 1 – x. Now let’s
say there’s a vector field in this space which is given by F = . If we
wanted to calculate the surface integral of this field over the surface S without using
the theorem, we would have to split each of the five flat surfaces of the prism into their
own surface integrals and calculate each individually. Luckily, the Divergence Theorem reduces our
work to one triple integral over the region of the prism. The divergence theorem states
that the surface integral will be equal to the triple integral of the divergence of F
over the region, so first let’s find the divergence of F. Recall that the divergence
of a vector field is del dot F where del equals . So for this case the divergence
will be del dot F equals y + z + x. Now we have the divergence, but before integration
we must get our bounds sorted out. We get these from the region of our prism. One simple
way to express these bounds is integrating z first, letting it run from 0 to 1 – x.
Now we don’t have any more complicated bounds, so we can integrate y from 0 to 2 and x from
0 to 1. So our integral ends up being the integral from 0 to 1, the integral from 0
to 2, the integral from 0 to 1 – x, of y + z + x dzdydx. Integrating z first, we get
yz + z^2/2 + xz, where z is evaluated from 0 to 1 – x. Factoring out a z and evaluating
we get (1 – x)[y + (1 – x)/2 + x] or simplifying the x terms, (1 – x)(y + x/2 + 1/2). Next
we must integrate with respect to y, which gives us (1 – x)(y^2/2 + xy/2 + y/2) where
y is evaluated from 0 to 2. Evaluating this we get (1 – x)(2 + x + 1) or just (1 – x)(x
+ 3). Now if we multiply this out, we get the integral of (x + 3 – x^2 – 3x) dx.
Simplifying gives us 3 – 2x – x^2. Integrating, we get 3x – x^2 – x^3/3, where x runs
from 0 to 1. And finally, evaluating gives us 3 – 1 – 1/3, simplifying to 5/3. This
is the value of our surface integral, which we used the Divergence Theorem to calculate.
The divergence theorem is another useful tool for calculations, and it has a variety of
applications. In fact, the theorem is used in Gauss’s law to relate the flux of an
electric field, which would be the surface integral part of the theorem, to the amount
of charge contained within the volume, which would be the triple integral part of the theorem.
And with this, we’ve wrapped up another section of mathematics, but before we go on
to more advanced topics, let’s check comprehension.