Welcome to an example
on how to use a divergence theorem
to evaluate a flux integral
where the surface integral is shown here,
where the vector field F is given
and the surface S is the solid bounded by the cylinder
y-squared plus z-squared equals 25,
and the planes, x equals negative four and x equals three.
Let's first look at this graphically.
The surface S is the surface that bounds the cylinder
between these two yellow planes
which are x equals negative four
and x equals three.
And the vector field F is graphed in purple,
and therefore the value of the flux integral
will give us a net flow across the surface of this cylinder.
Before we set this up though, let's review
the divergence theorem.
The divergence theorem states the total divergence
of a vector field in a solid region V
equals a total flow across the boundary surface S,
where V must be a solid region bounded by S
oriented by a unit normal vector directed outward.
And the partial derivatives of the components
of the vector field F must be continuous on V.
So under these conditions, the flux integral,
often expressed in one of these two forms,
is equal to the triple integral over the region V
of the divergent of the vector field F differential V,
which can also be expressed in this form here.
So this shows the relationship between the triple integral
over a solid region and a surface integral
over a surface.
So going back to our example,
looking at the vector field F in our formula,
P equals three-xy-squared,
Q equals x to the z,
and R equals e to the third.
So let's go ahead and find these partial derivatives.
So the partial of P with respect to x is equal to
the derivative of the three-xy-squared with respect to x.
That'd be three-y-squared.
And the partial of Q with respect to y
is equal the derivative of x times e to the power of z
with respect to y, which would be zero.
And the partial of R with respect to z
is equal to the derivative of z to the third
with respect to z, which would be three-z-squared.
Which means the given surface integral or flux integral...
again, often expressed as a double integral
over the surface S of F dot N
differential S is equal to the triple integral
over the solid region V of
the sum of the partial derivatives,
which would be three-y-squared
plus three-z-squared
differential V.
And now because the surface S is a cylinder,
it does make sense to use integral coordinates.
But we need to be careful here though because notice
how this cylinder is running along the x axis
and is bounded by x equals negative four
and x equals three.
So differential V is not going to be equal to
rdzdrd-theta,
it's going to be equal to rdxdrd-theta.
Again, because the cylinder's running along the x axis
not the z axis.
And that also means y-squared plus z-squared
is equal to r-squared.
So this triple integral is equal to the triple integral
of, well if you factor out three, we're going to have
three times the quantity y-squared plus z-squared
which would be three-r-squared.
And again, differential V is equal to
rdxdrd-theta.
The limits integration for x are going to be from
negative four to three.
We can use the yz trace if we need to to help us find
the limits integration for r and theta,
the limits integration for r from zero to five,
and for theta, it's from zero to two-pi.
Let's go and evaluate this on the next slide.
Of course, we can write the [mumbles] function here
as three-r-cubed.
So we first integrate respect to x treating r as a constant.
So I'd have three-r-cubed-x.
Performing substitution for x,
when x is three,
I'm going to have three-r-cubed times three
minus when r is negative four,
we have three times r-cubed times negative four.
So we have nine-r-cubed plus 12r-cubed,
which gives us 21r-cubed.
And now we integrate with respect to r.
So the antiderivative is going to be 21 times
r to the fourth divided by four
or 21/4ths r to the fourth.
Performing substitution for r,
when r is five, we have 21/4ths
times five to the fourth.
And when r is zero, we have zero.
This simplifies to 13,125/4ths
d-theta.
And now we integrate with respect to theta.
Performing substitution for theta,
we're just going to have 13,125 divided by four
times the quantity two-pi minus zero.
Of course, we have the common factor of two.
So one, two, and two; and two twos and four.
So the exact value of the flux integral
is 13,125pi
divided by two,
which is approximately 20,616.7018.
So going back to the graph, this value represents
the total flow across the boundary of surface S
as well as the total divergence of the vector field
in the solid region V.
I hope you found this helpful.