Welcome to a lesson
on triple integrals using cylindrical coordinates.
Let's first review cylindrical coordinates.
In a cylindrical coordinate system,
a point P in space is represented by an ordered triple,
r comma theta comma z.
Our r comma theta is the polar representation
of a point in the x-y plane.
So r is the directed distance,
this distance here, from the origin, in the x-y plane.
The angle theta, this angle here,
is angled counter-clockwise from the polar axis,
or the positive x-axis in the x-y plane.
And then finally z is the directed distance,
up or down, this distance here,
from r comma theta to the point in space.
We also need to be familiar with these equations,
that we use to convert from cylindrical
to rectangular coordinates, as well as these equations,
that we use to convert from rectangular
to cylindrical coordinates.
So if you don't have these in your notes,
you may want to include them.
Now let's talk about converting a triple integral
in rectangular form to cylindrical form.
The first thing to notice is the integrand function,
f of x comma y comma z,
must be written as a function of r, theta, and z
by substituting r cosine theta for x,
r sine theta for y,
and z remains the same.
And then differential V is equal
to r dz, dr d theta.
So notice how I'm converting from rectangular
to cylindrical form.
We will have an extra factor of r in the integrand function.
So if the solid region of integration
can be easily defined using cylindrical coordinates,
the result is the triple integral, in cylindrical form,
is often much easier to evaluate.
And when converting from rectangular
to cylindrical coordinates, the xy-trace is often helpful
in order to find the limit of integration for r and theta.
Let's look at our first example.
I want to integrate f of x comma y comma z
over the solid region V, bounded by z equals zero
and z equals four minus the square root
of the quantity x squared plus y squared.
Let's first take a look at the region V in space.
The region of integration is the region shown here,
bounded by the cone and the yellow plane.
If we look down the x-y plane,
we can see the xy-trace,
which is the circle here, centered at the origin.
Of course, we can also look up on the x-y plane,
and view the xy-trace here.
So going back to our work, let's begin setting this up
as a triple integral using cylindrical coordinates.
So we know we'll have a triple integral.
Integrand function in rectangular form
is f of x comma y comma z equals x squared plus y squared,
which must be expressed as a function of r, theta and z,
and because x squared plus y squared is equal to r squared,
f of r cosine theta comma r sine theta comma z
is just r squared,.
And then differential V is equal
to r dz,
dr, d theta.
And now let's work on determining limits integration
for z r theta.
Well, z is bounded by z equals zero,
and z equals four minus the square root
of the quantity x squared plus y squared.
But we must express limits integration
as functions of r and theta.
So lower limit of integration for z is going to be zero,
but then for the upper limit of integration,
because x squared plus y squared is equal to r squared,
and the square root of r squared is equal to r,
the upper limit of integration for z
is going to be four minus r.
And now we'll use the xy-trace
to help us determine the limits integration for r and theta.
To find the xy-trace by hand,
we would set z equal to zero,
so using this equation here, if we set z equal to zero,
we get, add the square root of both sides,
and then square both sides of the equation,
which will give us x squared plus y squared equals 16,
which is this circle here, centered at the origin,
with a radius of four.
And therefore to trace out this circle,
using cylindrical coordinates,
r would go from zero to four,
which are the limits integration for r.
And then for theta, to go around the circle,
the limits integration for theta
would be from zero to two pi.
Let's go ahead and rewrite
the integrand function as r cubed.
Now we integrate with respect to z,
treating r cubed as a constant.
So the antiderivative is, r cubed z,
the limits integration for z are
from z equals zero to z equals four minus r.
So now we need to find big F of b minus big F of a.
So when z equals four minus r, we have
r cubed times the quantity four minus r,
minus when z is zero, we have r cubed times zero.
So notice how the new integrand function
is just going to be four r cubed minus r to the fourth.
And now integrate with respect to r,
so we'd have four times r to the fourth
divided by four, that's just r to the fourth,
minus r to the fifth divided by five,
or minus one fifth r to the fifth.
So when r is equal to four,
we have four to the fourth minus 1/5
times four to the fifth,
minus, when r is zero, both terms are zero.
So simplifying,
four to the fourth minus one fifth times four to the fifth,
is equal to 256/5.
And now we integrate with respect to theta,
which gives us 256 divided by five theta.
Now we need to find big F of b
minus big F of a one more time,
which should be 256 divided by five times pi minus zero,
which gives us 512 pi divided by five.
So this is the value of our triple integral.
Let's look at another example.
Here, we're asked to integrate f of x comma y 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.
In this case, because the integrand function
is just the constant function one,
if this solid region V is above the x-y plane,
this would give us the volume of the region V.
Let's begin by looking at the region V in space.
The region V is the region below the light blue parabaloid,
and above z equals zero, graphed in yellow,
between these two green cylinders.
If we look down on the x-y plane,
again, we can see the xy-trace,
which should be the region bounded by these two circles.
And again, this case, because the integrand function
is f of x comma y comma z equals one,
the triple integral is going to give us
the volume of this solid region V.
So beginning to set this up,
f of x comma y comma z equals one
is going to be the same function in cylindrical coordinates,
so we have one, and then differential V,
again, is equal to r dz,
dr, d theta.
And again, z is bounded by z equals zero,
and the paraboloid given by z equals four
minus the quantity x squared plus y squared,
but again, the limits integration must be expressed
in terms of r and theta, so limits integration for z
are going to be from zero to four minus r squared.
We have r squared here because x squared
plus y squared is equal to r squared.
And now we'll look at the xy-trace
to help us determine limits integration for r and theta.
The xy-trace can also be determined
by using these two equations here,
and by setting z equal to zero in this equation.
In this case, if we set z equal to zero here,
it gives us the same equation we have here,
x squared plus y squared equals four.
So the xy-trace,
so the x-y trace is the region bounded
between these two circles,
where the smaller circle has a radius of one,
and the outer circle has a radius of two.
Which means, the limits integration for r
are going to be from r equals one to r equals two,
and then to trace out the region
between these two circles, theta has to go
from zero all the way around the circle to two pi radians.
So limits integration for theta are from zero to two pi.
And now we integrate respect to z.
So the antiderivative is going to be
one r z, or just r z.
Limits integration are from z equals zero
to z equals four minus r squared.
Now I need to find big F of b minus big F of a,
so we'd have r times the quantity four minus r squared,
minus r times zero.
So simplifying, the new integrand function
is just four r minus r to the third.
Let's continue on the next slide.
And now we integrate respect to r,
so the antiderivative is four times r squared,
divided by two, or two r squared,
minus r to the fourth divided by four,
minus 1/4 r to the fourth.
Now I need to find big F of b minus big F of a.
So first, r is equal to two,
so we'd have two times two squared
minus 1/4 times two to the fourth,
minus, when r is one we have two times one squared,
minus 1/4 times one to the fourth.
So simplifying, here we have eight minus four,
minus the quantity two minus 1/4,
which simplifies to 9/4.
Integrating respect to theta, we have 9/4 theta,
so we have 9/4 times the quantity two pi minus zero,
which gives us nine pi divided by two.
And again, in this case, because the integrand function
is just a constant function one,
and the solid region V is above the x-y plane,
the triple integral is equal
to the volume of the region V.
I hope you found this helpful.