Welcome to our lesson on
triple integrals using spherical coordinates.
In a spherical coordinate system, a point, P,
in space is represented by an
ordered triple rho, theta, phi,
where rho is the distance between the point and origin,
as we see here, in green,
theta is the angle counterclockwise
from the pole, or positive x-axis in the X Y plane,
here in blue,
and then phi is the angle between the
positive Z access and the point, as we see here in red.
We also need to be familiar with the equations used
to convert between spherical and rectangular coordinates.
You may want to copy these into your notes.
Let's take a look at a triple interval
using spherical coordinates.
If we have a triple integral in rectangular form,
as we see here, this is how we
write it in spherical coordinates.
There are several things that we should recognize here.
First, the function must be expressed
in spherical coordinates, where X is
equal to rho sine phi cosine theta,
Y is equal to rho sine phi sine theta,
and Z is equal to rho cosine phi.
Then, notice that differential V is equal to
rho squared sine phi, D rho, D phi, D theta.
There's an extra factor of rho squared sine phi
when converting to spherical coordinates.
just like any other triple integral,
we can change the order of integration.
Let's go in take a look at an example.
Here, we want to integrate our function
over the solid region,
X squared plus Y squared plus Z squared is equal to four.
We should recognize this as a sphere, with radius two.
This represents our region of integration.
Let's go ahead and set this up.
First, let's convert our function
to a function in spherical coordinates.
Well, X squared plus Y squared plus Z squared
is equal to rho squared.
Then, because we are converting to spherical coordinates,
we have extra factors of rho squared sine phi.
Let's go ahead and keep the order of D rho, D phi, D theta.
Let's start by taking a look at the
limits of integration for rho.
Remember, rho is a distance from the
origin to a point on the surface.
Because the radius is equal to two,
the limits of integration for rho will be from zero to two.
Next, the limits of integration for phi.
Remember, phi is the angle between the positive Z axis,
and the points on the surface.
To get the entire sphere, rho would have to
be from zero all the way to pi radiance.
Then, for theta, the angle between the positive X axis
and the points on the surface,
we have to rotate all the way around
a circle in the X Y plane.
Theta would be from zero, all the way to two pi.
Let's go ahead and rewrite our integrand,
and then we can integrate.
We have rho to the fourth sine phi.
Let's go ahead and evaluate this on the next page.
So treating phi as a constant,
we will have one fifth rho to the fifth sine phi.
Replacing rho with two, we would have 32 fifths sine phi,
and then when rho is zero, this should be zero.
Now, we integrate with respects to phi.
That's going to give us negative 32 fifths cosine phi.
When phi is equal to pi, cosine pi is negative one,
so, positive 32 fifths minus --
Then, when phi is zero, cosine zero is one,
so we will have a negative 32 fifths.
We will have a positive 64 fifths.
So, we will have 64 fifths theta,
and evaluate it at zero and two pi.
When theta is two pi, we'll have two pi.
When theta is zero, we'll have zero.
We'll have 128 pi fifths for this triple integral.
Let's go ahead and take a look at another one.
Here, we want to integrate the function
over the region inside X squared plus Y squared
plus Z squared equals one in the first octant.
This would be a region of integration.
Now, let's set this one up.
Let's take a look at our function first.
X squared plus Y squared plus Z squared
is equal to rho squared.
Rho squared raised to the three half power
would give us rho to the third,
so we have three E rho to the third,
and then we have our extra factors of rho squared sine phi,
D rho, D phi, D theta.
Now, let's take a look at the limits of integration.
First, for rho, the radius of this piece of the sphere
in the first octant is going to be from zero to one,
so rho is from zero to one.
Then, because we are in the first octant,
the angle between the positive z-axis,
and the point on the surface will be
from zero to pi over two.
Then, for theta, the angle between
the points on the surface and the positive X axis
would also be from zero to pi over two.
Now let's integrate this.
We want to first integrate this with respects to rho.
Notice we have an exponent of rho cubed,
and then we have a factor of rho squared.
This is going to require I use substitution,
where U is going to be equal to rho to the third.
So, differential U is equal to three rho squared D rho,
and that actually really good news,
because if you look at our integrand,
we do have a factor of three rho squared D rho.
If we write this in terms of U,
we would just have E to the U, sine phi D U.
This would give us E to the U sine phi.
Well, E to the U is E to the rho to the third sine phi.
Let's go ahead and do this on the next page.
So replacing rho with one, and then with zero.
When rho is one, we have E sine phi,
and then when rho is zero, we'd have E to the zero,
which is one, so minus sine phi.
Let's go ahead and factor out the sine phi.
We should have E minus one times sine phi.
Now, integrate with respects to phi,
that's going to give us a negative cosine phi.
So, when phi is pi over two, cosine pi over two is
going to be zero, so we would have zero minus --
Then, when phi is zero, this will be negative one,
so we will have E minus one, times negative one.
This will just give us E minus one.
Now, we will integrate with respects to theta.
That will just give us D minus one times theta.
When theta is pi over two, I have pi over two,
and if theta was zero we would have zero.
It looks like we have pi over two times
E minus one as the value of our triple integral.
That will have to do it for this video.
We'll take a look at determining volume
using spherical coordinates in the next video.