Welcome to part three
of triple integrals and volume.
I thought we should take a look at one more example.
Here we're going to determine the volume
of a solid bounded by X squared plus Y squared equals nine
above Z equals zero and below X plus Z equals four.
So by now we should know that the volume
can be determined by this triple integral.
Where V is the bounded three dimensional region
when we integrate with respects to V,
replacing the dv with, or any other order
of those differentials.
So let's start by taking a look at
the XY, XZ, and YZ traces.
So to determine the XY trace, we'll set Z equal to zero.
So here we'll have the circle centered at the origin
with radius three.
And the region in this plane is inside the circle.
For the XZ trace you'll set Y equal to zero.
Well if Y is zero we'll have X squared equals nine.
So X equals plus or minus thee.
So this is the X axis, this is the Z axis.
So we have a vertical line here
and a vertical line here.
Then we have this line X plus Z equals four
where we have both an X and a Y intercept equal to four.
So this trace would be the trapezoidal region
as we see here.
Because we also know Z is equal to zero which would be here,
and then lastly for the YZ trace we'll set X equal to zero.
So if X is zero we'll have Y equals plus or minus three.
We'll let this equal the Y axis and this be Z axis.
So we'll have two vertical lines here and here.
Then if X is zero here we have Z equals four,
a horizontal line through here.
So the region in the YZ plane would be this rectangle here.
Let's see if we can determine the best order of integration
as well as the limits of integration.
Limits of integration for X and Y will be affected
by the graph of the circle,
and the limits of integration for Z
will be affected by this line here.
We don't want to use this trace here
to determine the limits of integration for X,
'cause moving from left to right,
here it's bounded by this vertical line
and this slanted line,
and here it's bounded by this vertical line
and another vertical line.
So that would make things a lot more complicated.
Let's go ahead and integrate with respects to Y first.
Which means the limits of integration for Y
must be expressed as a function of X and Z.
This equation of the circle was
X squared plus Y squared equals nine.
If we solve this equation for Y,
we'll have Y equals plus or minus the square root
of nine minus X squared.
So the lower limit of integration will be negative
square root of nine minus X squared.
The upper limit will be positive square root of nine
minus X squared.
And then to avoid having to integrate this result
with respect to X because of the square roots,
let's go ahead and integrate with respects to Z second
and then X last.
So the limits of integration for Z,
looking at this trace here would be from zero to this line.
So we'll start at zero,
and the equation of this line here was the equation
X plus Z equals four.
Subbing this for Z would give us four minus X,
and then for X it will be from negative three
to positive three, as we see here.
As well as, as we see here.
Before we start, let's go ahead and take a look
at this three dimensional region in space.
The region we're looking for is underneath
this slanted yellow plane inside this blue cylinder
above the XY plane.
So we're looking for the volume in this region here.
Let's go ahead and start this on the next slide
where we have more room.
When we integrate with respects to Y,
we'll just get Y.
So when we sub the upper [mumbles] we'll have
the square root of nine minus X squared,
and then minus a negative square root of nine
minus X squared.
That will give us two square roots of nine minus X squared.
Now when you got respects to Z, we're treating X
as a constant so we're just going to have
two Z times the square root of nine minus X squared.
So now we're going to place Z with four minus X
and then with zero.
Again let's go ahead and do that on the next slide.
So we'll have two times the quantity of four minus X
times the square root of nine minus X squared,
and then when Z is zero this will be zero.
Now this is going to be eight minus two X.
Let's break this up into two different [mumbles] integrals.
So we'd have negative two X times the square root
of nine minus X squared plus.
Then we'd have eight times the square root of nine
minus X squared.
Now the reason I wanted to do that is this is going to
require U-substitution,
this is going to require trig substitution.
So here we'll let U equal nine minus X squared.
So differential U is going to be negative two X DX.
That's a perfect match here.
The integral of the square root of U,
or U to the one half DU.
So that's going to give us U to the three halves
divided by three halves or two thirds U to the three halves.
Plus, now this integral is going to require trig substitution.
So if you let X equal three sine theta,
then DX would be equal to three cosine theta.
Let's go ahead and model theta.
So sine theta would be X over three.
So this would be the square root of nine minus X squared.
Looks like I forgot the D theta here.
Let's go ahead and perform the substitution now.
We'll have eight times, this will be the square root
of nine minus nine sine square theta.
That'll simplify to three cosine theta.
And then DX is equal to three cosine theta D theta.
Let's go ahead and simplify this.
Eight times three time three, it's going to be 72,
and we'll have cosine squared theta D theta.
Let's go to the next slide.
Let's go ahead and apply the power reducing formula here
for a cosine squared theta.
It's going to give us one plus cosine two theta divided by two.
Let's go ahead and simplify this.
This will give us integral of 36 D theta
plus the integral of 36 cosine two theta D theta,
and now we're getting away with this respect to theta,
but now this is going to require another U-substitution.
Let's just call it V.
This is going to be 36 theta plus
if we let V equal two theta
that's going to give us 36 times,
this will give us an extra factor of one half.
So we'll half times one half sine two theta.
Let's go ahead and perform a substitution here.
Let's replace sine two theta with
two sine theta cosine theta.
We can see when we do that,
that this one half and this two are going to simplify out.
Let's go ahead and rewrite this in terms of X.
Then remember the limits of integration
were from negative three to positive three.
Let's see how we're going to write theta.
Looking at this equation here,
sine theta is equal to X over three.
So theta would be equal to arc sine of X over three.
And then here we're going to have 36 times
sine theta cosine theta.
Well sine theta would be X over three,
and cosine theta would be the square root
of nine minus X squared over three.
Let's go ahead and leave it like this,
and then evaluate this at our upper
and lower limits of integration.
So when X is three, this is going to be zero.
Here we're going to have 36 times arc sine of three over three.
The arc sine of one is going to be pie over two.
So we'll have 36 times pie over two,
and then again, when X is three, this will be zero,
and then when X is negative three, this is going to be zero.
This is going to be 36 times arc sine of negative one.
That's going to give us negative pie over two.
So we'll have 36 times negative pie over two,
and then again this is going to be zero
when X is negative three.
So we have 18 pie minus negative 18 pie.
Which gives us a total of 36 pie cubic units
for this volume.
So this triple integral had a lot of good
integration review topics.
I hope you found it helpful.