Welcome to a lesson triple integrals
and volume using cylindrical coordinates.
The goal is to determine the volume of a solid
using cylindrical coordinates.
In a cylindrical coordinate system,
a point P in space is represented by an ordered triple,
r, theta, and z, as we see here below.
r, theta is the polar representation
of a point in the xy-plane, where r here
is the directed distance from the origin
in the xy-plane, and theta is the angle
counter-clockwise from the polar axis
or the positive x-axis.
And z is the directed distance up or down
from r, theta in the xy-plane.
And then also, we need to be familiar with the formulas
used to convert between rectangular and cylindrical
coordinates, as we see here and here.
So to determine the volume of a solid
using cylindrical coordinates, we're going to have
a triple integral over the region V,
integrated with respects to V.
Where differential V is equal to r, d z,
d r, d theta.
Remember, have an extra factor of r,
and then the limits of integration will be expressed
using r and theta.
And this is very similar to what we did
in the previous video, when we introduced
triple integrals using cylindrical coordinates.
Let's go and take a look at that real quickly.
The only difference is for the volume,
f of x, y, z, would be equal to one.
So our procedure for evaluating the triple integral
is going to be the same as we did in the first video,
it's just that now the limits of integration
will define the solid region, which will then give us
the volume of the solid using this triple integral.
Let's go and take a look at our first example.
Here we want to use a triple integral
and cylindrical coordinates to determine the volume
of the solid bounded by z equals zero,
and z equals nine minus the quantity
x squared plus y squared.
Let's take a look at this in 3D first.
So here's our graph.
We have a paraboloid in blue,
and then we have this yellow plane.
And our goal is to determine the volume
inside the paraboloid above the yellow plane.
So we'll start by taking a look at the x, y,
x, z, and y, z traces.
So for the x, y trace, we'll set z equal to zero.
So if z was zero, we could move the quantity
x squared plus y squared to the left side,
and we'd have x squared plus y squared
equals nine, so we have a circle
centered at the origin, with a radius of three.
And then for the x, z trace,
if we set y equal to zero, we'd have z equals
nine minus x squared, well,
that's a parabola that opens down,
with a z-intercept of nine, and x-intercepts
of positive three and negative three.
So it'd look like this.
And then the y, z trace is similar.
If we let x equal zero, we'd have z equals
nine minus y squared.
So this is the y-axis and this is the z-axis,
the trace would look just like the x, z trace.
So our volume is going to be equal to
the triple integral, where differential V
is going to be equal to r, d z, d r, d theta.
So we'll start by determining
the limits of integration for z.
And for the x, z and the y, z trace,
I might as well go ahead and graph z equals zero,
here and here, so we can see that z starts
at zero, and then for the upper limit
of integration with respect to z,
which must be expressed in terms of r and theta,
would be determined by these two parabolas here,
which were slices of z equals nine minus x squared
plus y squared, and since x squared plus y squared
is equal to r squared, we'd have nine minus r squared
for our upper limit of integration for z.
And then to determine r, we can look at the x, y trace.
r is going to start at zero, and then go out to three.
And then for theta, we need to go all the way around
this circle, so it'd be from zero all the way
to two pi.
So this triple integral will give us the volume
of this bounded three-dimensional region.
Let's go ahead and evaluate this on the next page.
Well, first integrate with respect to z,
treating r as a constant.
That's going to give us r, z.
When z is equal to nine minus r squared
we'll have nine r, minus r cubed.
And then when z is zero, this will be zero.
Now integrate in respects to r, we're going to have
nine times r squared over two,
or 9/2 r squared, minus r to the fourth
over four, 1/4 r to the fourth.
So when r is three, we're going to have
nine times nine is 81/2, minus three to the fourth
is 81, so 81/4.
And then when r is zero, these will both be zero.
81/2 minus 81/4 would equal 81/4.
Integrate with respects to theta.
So we'll have 81/4 times two pi
over one and then minus zero.
This simplifies here.
So this would be 81 pi over two cubic units,
for the volume of this solid region.
Let's take a look at one more example.
We want to use a triple integral and cylindrical
coordinates to determine the volume of the solid
bounded by z equals four minus the square root
of x squared plus y squared, x squared plus
y squared equals one, and z equals zero.
Let's go and take a look at this graphically.
So the goal is to determine the volume
inside the cylinder, above the green plane,
and below the blue surface.
So let's start by taking a look at the traces.
Let's take a look at the x, y, x, z, and y, z traces.
So for the x, y trace, we'll let z equal to zero.
So if z is equal to zero here, we'd have
zero equals four minus the square root
of x squared plus y squared.
So if we move this square root to the other side,
and then squared both sides, we would have
x squared plus y squared equals 16.
But we're not actually going to use this trace,
because take a look at the other trace,
x squared plus y squared equals one,
we're looking for the volume that's inside
this smaller cylinder, so we're only going to sketch
the circle with radius one, centered at the origin.
So here's our x, y trace that we need.
For the x, z trace, we'll set y equal to zero.
Be a little careful here.
We'll have z equals four minus the square root
of x squared, which is simplified to the absolute
value of x, and then here we'd have
x squared equals one.
And here we have z equals zero.
So here's z equals zero.
For z equals four minus the absolute value of x,
remember, absolute value graphs make a V,
so this will be a V with a z-intercept of four.
Something like this.
Then x is equal to plus or minus one.
So the x, z trace is this region in here.
Here we had inside the circle.
And then the y, z trace is going to be similar
to the x, z trace, setting x equal to zero
we'll have z equals four minus the absolute value of y.
Here we'll have y squared equals one.
And again, z is equal to zero.
So here's z equals zero.
Here is y equals negative one,
here's y equals positive one, and we have
this V-shaped graph from the absolute value.
So our region is in here.
A lot of times, when setting this up,
we only need the x, y trace, but I think
getting in the habit of making all three traces
is often helpful to get a better understanding.
So our volume is going to be equal to the triple integral
where we have r, d z, d r, d theta.
So for the limits of integration for z,
it starts at zero, and then it goes up to these V-shaped
graphs, that are generated by this cone
z equals four minus the square root of x squared
plus y squared.
Remember, we have to express z as a function of r and theta.
So the upper limit for z is going to be four
minus, this will be the square root
of r squared, or just four minus r.
And then for r and theta, we can use the x, y trace.
r will be from zero to one.
And then theta will be from zero
all the way around to two pi.
So let's go ahead and evaluate this on the next page.
So we're going to have r times z.
Now we'll replace z with four minus r and then with zero.
Replacing z with four minus r,
it'll give us four r minus r squared,
and then when z is zero, this will be zero.
Integrating respects to r, we're going to have
four times r squared over two,
or two r squared minus 1/3 r cubed.
Replacing r with one, we're going to have
two minus 1/3, and then when r is zero,
these will both be zero.
Two minus 1/3 is 5/3.
So we'll have 5/3 theta.
Two pi, I'll put it over one minus zero.
We'll have 10 pi over three.
So the volume of the solid,
inside the cylinder, above the green plane,
below the blue surface is equal to 10 pi over three
cubic units.
And that'll do it.
Thank you for watching.