Welcome to a lesson on
triple integrals and volume using
spherical coordinates.
As we saw in the first video,
in the spherical coordinate system, a point P
in space is represented by
an ordered triple ro, theta, and phi,
where ro is the distance between the point
and the origin, here in green,
theta is the angle counterclockwise from the pole
or positive x-axis in the xy-plane, here in blue,
and phi is the angle between the positive
z-axis and the point as we see here in red.
Then we also have to be familiar with
the equations used to convert between
rectangular and spherical coordinates.
Let's take a look at how we're going to
determine the volume of the solid using
triple integrals in spherical coordinates.
Just as we saw in rectangular coordinates
and cylindrical coordinates, if we use
the function f of xyz equals one for our integrand,
and the limits of integration define our solid,
this triple integral will give us the volume
of that solid, but remember in spherical coordinates
differential V gives us ro squared sine phi,
d ro d phi d theta, or it's also commonly
expressed as d ro d theta d phi.
So if the solid region of integration
can be easily defined using spherical coordinates,
the result is often much easier to evaluate.
Let's take a look at our first example.
We want to determine the volume outside
x squared plus y squared plus z squared equals one
which is a sphere of radius one
and inside x squared plus y squared
plus z squared equals four, which is a sphere
with the radius two.
So let's set this up.
Our integrand is going to be ro squared sine phi
because we're using spherical coordinates.
Let's use the order d ro d phi d theta.
So ro will be the distance from the points
on the inner sphere to the outer sphere.
Well this has a radius of one, this is a radius of two
so the limits of integration for ro will be
from one to two.
Remember, phi is the angle between
the positive z-axis and the points on the surface,
so that will be from zero all the way to pi radians.
Then theta, the angle of rotation from
the positive x-axis will be all the way
around the sphere, or from zero all the way to two pi.
So now whenever we have the specs of ro first,
it's going to give us 1/3 ro to the third
sine phi and now we'll evaluate this
at the limits of integration.
Let's go ahead and do that on the next page.
So when ro is equal to two, we'll have
8/3 sine phi minus, and then when ro is one
we'll have 1/3 sine phi.
Well, 8/3 sine phi minus 1/3 sine phi
is going to give us 7/3 sine phi.
Integrated in respects to phi we're going to
have negative 7/3 cosine phi.
Cosine pi is going to give us negative one.
Negative one times negative 7/3 will be 7/3
minus cosine zero would be one.
So we have minus a negative 7/3
so we'll have 14/3 and now we'll integrate
with respects to theta.
We'll have 14/3 theta
so the end result is going to be 28 pi thirds.
Again this would be the volume that is
outside the small sphere but inside the big sphere.
Let's look at another one.
Here we want the volume of the region outside the cone
phi equals pi over three and inside the sphere
ro equals six cosine phi.
Let's take a look at the graph of
these two surfaces.
Here it is.
Again, we're looking for the volume of
the region that's outside this cone
and inside the sphere, so this yellow region
down here is outside the blue cone.
Before we start, let's take a look
at our graph here.
This is the z-axis here.
The cone has the equation phi equals pi over three
so the angle from here to the blue surface
would be pi over three radians
so there's a little slice here that we can't see.
We're trying to find the volume of this region here
that's inside the sphere, outside the cone.
So the volume will be the triple integral
where differential V is ro squared sine phi.
Let's go ahead and keep the order d ro d phi d theta.
The first limits of integration with respects to ro
and they actually gave us the equation
that ro is equal to six cosine phi
so ro is going to be from zero
all the way to six cosine phi.
Now for the limits of integration for phi,
remember phi is the angle between the positive z-axis
and the point on the surface.
Well from here to here is pi over three.
We want the region that's outside the cone
and on the surface so the lower limit of integration
will be pi over three and the upper limit will be
pi over two all the way to the xy plane.
Then for theta we want to go all the way around the circle
so that will be from zero to pi over two.
This one was little bit different
because they gave us equations in spherical form
but we still had to interpret the results
to set up this triple integral.
Integrated with respects to ro we'll have
one third ro cubed sine phi.
Replacing ro with six cosine phi,
well six cubed times 1/3 is going to
give us 72 and then we'll have cosine cubed phi
and we still have our sine phi
and then when ro is zero this will all be zero.
Now we need to integrate this with respects to phi.
Let's do that on the next page.
This is going to require u substitution.
Let's go ahead and let u equal cosine phi,
so differential u is going to be equal to
negative sine phi d phi.
Let's rewrite this in terms of u.
This would be e to the third, sine phi d phi
is going to give us a negative du
so the result will be negative 72 u to the fourth over four
or 1/4 u to the fourth.
That's going to be cosine to the fourth phi.
This is going to be negative 18.
So when phi is equal to pi over two,
cosine pi over two is zero so we'll have
zero minus and then cosine pi over three
is equal to 1/2 so we'll have 1/2 to the fourth.
1/2 to the fourth power is going to give us 1/16
so negative 18 times negative 1/16,
that's going to give us positive 18/16 or 9/8
and now integrating with respects to theta
we'll have 9/8 theta.
So we'll have 18 pi over eight
which gives us nine pi over four.
Again for the volume that's outside the cone
and inside the cylinder in this region here.
That'll do it for this video.
I hope you found these examples helpful.
Thank you for watching.