Welcome to our lesson
on Triple Integral Using Cylindrical Coordinates.
The goal is to set up and evaluate triple integrals
using cylindrical coordinates.
In the cylindrical coordinate system,
a point P in space, as we see here modeled below,
is represented by an ordered triple r-theta-z
where r, theta is the polar representation
of a point in the xy-plane.
Where r is the directed distance
from the origin in the xy-plane as we see here.
Theta is the angle counter clockwise from the polar axis
or positive x-axis in the xy-plane as we see here.
And z is the directed distance up or down
from r-theta to the point P.
We also need to be familiar with the formulas
used to convert between cylindrical
and rectangular coordinates as we see here and here.
So, we don't really have those in your notes,
so you may want to write those down.
Now, let's talk about converting a triple integral
in rectangular form to cylindrical form.
The first thing we should notice here is that
f of x, y, z must be written as
f of r-theta, r-sine-theta, z.
So, to convert the rectangular equation
to our cylindrical equation,
we replace x with r-cosine-theta,
y with r-sine-theta, and z remains z.
And just as when we were dealing with polar coordinates,
differential V gives us an extra factor of r
and then d-z, d-r, d-theta.
So, if the soild region of integration can be easily defined
by using cylindrical coordinates,
the result of a triple integral
is often much easier to evaluate.
And when converting from rectangular
to cylindrical coordinates,
the x-y, x-z, and y-z traces are often helpful.
Let's go and take a look at our first example.
We want to integrate f of x, y, z over the solid region V
bounded by z equals zero
and z equals four minus the square root
of x squared plus y squared.
Let's look at this graphically.
So, our region of integration is the solid region
inside this cone above the yellow plane.
So, z equals four minus the square root
of x squared plus y squared is this cone here
and z equals zero is this plane here.
If we wanted to sketch the x-y trace,
we'd take this equation here and replace z with zero.
So, if we did that, we could move this square root
to the other side and then, square both sides.
That would give us x squared plus y squared equals 16.
So, we'd have this circle here,
centered at the origin with a radius of four.
So, this would be four, four,
negative four, and negative four.
Let's see if we can use this information to setup
our triple integral using cylindrical coordinates.
Well, first, our function f of x, y, z
is equal to x squared plus y squared.
Well, x squared plus y squared is equal to r squared.
So, we'd have r squared.
Then, remember, differential V
gives us an extra factor of r,
d-z, d-r, d-theta.
So, for the limits of integration for z,
we can see that z starts at zero
and goes up to the surface.
So, the lower limit is going to be zero.
The upper limit is going to be the surface here
and since r squared is equal to x squared plus y squared,
we'd have four minus the square root of r squared
or just four minus r.
And then, to determine the limits of integration for r,
we can take a look at the x-y trace.
We can see that the radius r starts at zero
and goes out to positive four.
And then, for theta, because we want this entire cone,
we would start at zero
and rotate all the way around to two pi.
So, lower limit for theta would be zero
and the upper limit would be two pi.
So, our integrand is going to be r cubed
and then, we have d-z, d-r, d-theta.
Let's go ahead and evaluate this on the next slide.
Integrating with respects to z, r is treated as a constant.
So, we'd have r cubed, z.
Now, we'll replace z with four minus r and then, zero.
So, we're going to have r cubed times four minus r
and then when z is zero, it'd be zero.
Let's go ahead and distribute here.
We'll have four r cubed minus r to the fourth.
And, now, we'll integrate with respects to r.
So, we're going to have four times r to the fourth over four.
That's going to give us r to the fourth.
We'll have minus r to the fifth over five
or 1/5, r to the fifth.
So, when we sub in r equals two,
we're going to have 16
minus, this'll be 32/5,
and then, again, when r is zero, these'll both be zero.
16 minus 32/5, that'll be 48/5.
Now, we integrate with respects to theta.
So, we have 48/5 times two pi.
Put this over one, that's going to give us 96 pi fifths.
Let's go and take a look at another one.
Here, we want to integrate f of x, y, z equals one
over the solid region V bounded by z equals zero,
x squared plus y squared equals one,
x squared plus y squared equals four,
and z equals four minus the quantity
x squared plus y squared.
Let's start by taking a look at a graph of this.
So, our region of integration is between the two cylinders,
below the blue surface, and above the yellow plane.
So, let's go ahead and set this up
using the cylindrical coordinates.
Before we do that, let's go ahead and label a few things.
Before we do that, let's go ahead
and label a couple things here.
This surface here in blue is z equals four
minus the quantity x squared plus y squared.
But, take a look at the x-y trace,
the smaller circle here is
x squared plus y squared equals one.
And this larger circle is
x squared plus y squared equals four.
So, this time f of x, y is equal to one,
so differential of these just going to be
r, d-z, d-r, d-theta.
And, now, for limits of integration for z
expressed as a function of r and theta,
going back to this graph here,
we know now we can replace
x squared plus y squared with r squared.
So, z is going to start at zero, this yellow plane,
and then, go up to the surface
which will be four minus r squared
using the cylindrical coordinates.
And then, for r, looking at the x-y trace,
we're integrating over the solid region
between these two cylinders,
so r is going to be from one on the red circle,
to two on the black circle.
And then, for theta, we want to go
all the way around the circle.
So, it's going to be from zero all the way to two pi again.
And we're going to see in the next video
when f of x, y, z is equal to one as we see here,
this triple integral will actually give us the volume
defined by these limits of integration.
Let's go ahead and first integrate with respects to z.
So, because r is treated as a constant, we'll have r-z.
So, we're going to replace z
with four minus r squared and then, with zero.
So, we're going to have r times four minus r squared,
that'll four r minus r cubed.
And then, when z is zero, this'll be zero.
Let's go ahead and finish this on the next page.
Integrating with respects to r,
that's going to give us four times r squared over two
or two r squared minus r to the fourth over four
or 1/4, r to the fourth.
Then, we replace r with two.
Here, we're going to have eight minus,
this'll be two to the fourth,
that's going to give us 16 divided by 4, that's 4 minus,
and then, when r is one, we'll have two minus 1/4.
That's going to be four minus one and 3/4.
That's going to give us 9/4.
So, we'll have 9/4
times two pi minus zero.
This'll be 18/4 pi or 9/2 pi.
And, again, because f of x, y, z was equal to one,
this actually does give us the volume
defined by the limits of integration.
In the next video, we'll discuss more about
determining volume using triple integrals.
I hope this was helpful.