Welcome to an example of how to use
the divergence theorem to evaluate a flux integral
using spherical coordinates.
So we're going to calculate the flux integral,
often expressed in this form here as well
where the vector field F is given, and the surface S
is the surface of the solid bounded by the hemispheres
given by z equals the square root of the quantity
four minus x squared minus y squared.
Z equals the quantity one minus x squared
minus y squared, and the plane z equals zero.
So let's first take a look at this graphically.
The surface S is the surface that bounds the solid
between these two hemispheres above the xy plane
graphed here in yellow.
And also a graph of vector field F in purple.
So the value of the flux integral is equal
to the net flow or total flow across the surface
again, that bounds the solid between these two hemispheres
and above the xy plane.
Before we set this up, let's review the divergence theorem.
The divergence theorem states that the total divergence
of a vector field in the solid region V equals a total
flow across the boundary surface S.
So V must be a solid region bounded by the surface S
oriented by a unit normal vector directed outward.
And the partial derivatives of the vector field F
must be continuous over the solid region V.
So we can evaluate the flux integral shown here
on the left but writing it as a triple integral
over the solid region V of the divergence
of the vector field F, differential V
which can also be expressed in this form here.
So this shows a relationship between a triple integral
over a solid region in the surface integral over a surface.
Going back to our example, using the formula here
on the far right, let's first find these partial derivatives
where P is equal to four x cubed z,
Q is equal to four y cubed z, and R equals z to the fourth.
So the partial of P with respect to x
equals a derivative of four x to the third
with respect to x which would be 12 x squared z.
The partial of Q with respect to y
equals the derivative of four y cubed z
with respect to y, which would be 12 y squared z.
And the partial of R with respect to z
equals the derivative of three z to the fourth
with respect to z which would be 12 z to the third.
Which means, the given flux integral
often expressed in this form here as well,
the integral over the surface S of F dot N
differential S, where this differential S represents
a small change in vector area.
And this differential S represents a small change
in surface area.
So applying the divergence theorem,
this is equal to the triple integral
over the solid v of the sum of these partial derivatives
which we already found here.
So we have 12 x squared z, plus 12 y squared z,
plus 12 z cubed differential v.
And now from here, because the solid region V
is bounded by two hemispheres, we'll use
spherical coordinates to evaluate this triple integral.
But as a quick review, in spherical coordinates,
x equals rho sine phi cosine theta,
y equals rho sine phi sine theta,
and z equals rho cosine phi.
X squared times y squared plus z squared equals rho squared.
Looking at this integram function here,
notice how if we factored out 12z, we would have
12z times the quantity x squared plus y squared,
plus z squared which means in spherical coordinates
we would have the triple integral.
We'll come back to the limits of integration
of 12z again, where z equals rho cosine phi.
So we have 12 rho cosine phi.
And then x squared plus y squared plus z squared
equals rho squared.
Then for spherical coordinates, remember differential V
is equal to rho squared sine phi, d rho, d phi, d theta.
Now let's work on determining the limits of integration
for rho, phi, and theta.
Remember, spherical coordinates, rho's the distance
between the point and the origin.
Theta is the angle counter-clockwise from the pole,
or positive x-axis in the xy plane.
And phi is the angle between the positive z-axis
and that point of the information that's shown here.
So limits of integration for rho would be from one to two,
because we already said the inner hemisphere is one.
The radius of the other hemisphere is two.
Then for phi, remember, that's the angle down
from the positive d-axis.
And because we're only concerned about the solid
above the xy plane, rho would be from zero
to pi over two.
Then Theta would be from zero to two pi
all the way around the circle in the xy plane.
So from the positive x-axis all the way around the circle.
Now let's work on evaluating this on the next slide.
Let's go ahead and factor out the 12, so we have 12
times the triple integral.
Then we have rho times rho squared times rho squared.
That's four to the fifth.
Cosine phi times phi,
d rho, d phi, d theta.
So we first integrate with respect to rho,
treating phi as a constant.
We'd have rho to the sixth divided by six,
times cosine phi sine phi.
Or one-sixth rho to the sixth cosine phi sine phi.
Let's go ahead and factor out the one-sixth
and then perform substitutions for rho.
So 12 times one-sixth would be two.
Then we'd have two to the sixth cosine phi sine phi
minus one to the sixth cosine phi sine phi.
Here we have 64 cosine phi sine phi,
minus one cosine phi sine phi, which would be 63 cosine
phi sine phi.
We factor out the 63.
Two times 63 equals 126.
So 126 times the double integral of cosine phi sine phi,
d phi, d theta.
Integrating with respect to phi will perform
the u substitution.
We'll just let u be equal to sine phi.
And therefore, differential u equals cosine phi d phi.
So all of this you can think of as just udu.
In terms of u, the anti-derivative would be one-half
u squared.
Which means in terms of phi we'd have one-half
sine squared phi.
Let's go ahead and factor out the one half.
126 times one-half is 63.
So we'd have sine squared pi over two
minus sine squared zero.
Sine pi over two is one, one squared is one.
Sine zero is zero, so sine squared zero would be zero.
So we end up with just 63 times the integral
from zero to two pi, d theta.
If we want one d theta integrating with respect to theta
we have 63, and we have theta.
Let's finish this on the next slide.
So we just have 63 times the quantity two pi minus zero.
The exact value of flux integral is 126 pi.
Do it as one approximation is approximately
395.8407.
Going back to our graph one last time,
we just found the total flow across the boundary
of the surface S which bounds the solid
between these to hemispheres above the xy plane
is 126 pi which, based upon the divergence theorem
is also the total divergence of the vector field
in the solid region V.
I hope you found this helpful.