Welcome to a second video on triple integrals and volume.
In the previous video, we discussed how we can determine
the volume of a solid region V
by using this triple integral here,
replacing differential V with
dxdydz,
or any of the six possible orders of integration.
So let's go ahead and take a look at another example.
We were to determine the volume
of the solid in the first octant
bounded by x plus two y plus z equals six.
Let's take a look at this in space first.
The idea of this problem is to determine the volume
in the first octant bounded by this blue plane.
So the volume we're looking for
would be the volume underneath this plane
bounded by these three yellow planes.
So our volume is going to be equal to the triple integral
where the limits of the integration will be determined
by the boundaries of the three dimensional region,
and we'll integrate with respects to V,
replacing dv with
dxdydz,
or some other order of integration.
So there's two main things we have to do here
before we can integrate.
We have to determine the order of integration
and then also the limits of integration.
The best way to do this is to look at the traces
in the xy plane, xz plane and yz plane.
It's also important to recognize that there is
often more than one possible order of integration,
that some times one order will be easier than another.
To determine the xy plane, we'll set z equal to zero,
so we'd have x plus 2y equals six.
So the x-intercept would be six,
and the y-intercept would be three.
And because we're in the first octant,
we know that both x and y are positive,
so the xy trace would be this triangular region here.
Now for the xz trace, we'll set y equal to zero
so we'll have x plus z equals six,
so both the x- and the z-intercept are equal to six.
Again, both x and z are positive,
so this is the region of the xz trace.
And lastly, for the yz trace, x equal to zero,
so we'll have 2y plus z equals six.
So the y-intercept would be positive three
and the z-intercept would be positive six.
So the yz trace would be this triangular region here.
But looking at these traces, for the most part
we can use any order of integration.
The one thing I notice is, looking at the original plane,
it's going to be easier if we solve this
for either x or z.
Let's go ahead and solve it for x,
and therefore we'll integrate it with respects to x first.
This means that the limits of integration for x
must be functions of y and z,
so if we solve this equation for x,
we'll have x equals six minus 2y
minus z.
That'll be the upper limit of integration.
And the lower limit of integration will be at zero
because we're in the first octant.
Now we need to decide-- [mumbles]
Now we need to decide if we want to integrate
with respects to z first and then y,
or y and then z, so we'll have either dzdy or dydz.
Looking at the yz trace, it's going to be easier
if we solve this equation for z,
so we'll have z as a function of y,
so let's go ahead and look at respects to z second,
and then y third.
So we must express limits of integration
for z as a function of y,
so z will go from zero all the way to solving this for z,
we'd have six minus 2y.
And then lastly, for y,
we can look at either the xy trace or the yz trace.
Notice that in both cases, y goes from zero to three.
And now we're set to go.
We'll first integrate with respects to x.
This would be x.
So we'll replace x with six minus 2y minus z,
and then with zero,
and that'll just give us six minus 2y minus z.
We'll integrate this with respects to z.
And then with respects to y.
Let's go to the next slide.
So now we're going to treat x and y as constants
as we integrate with respects to z.
So we'll have 6z minus z squared over two
or one half z squared.
This'll be minus 2yz.
We'll evaluate this at zero and six minus 2y.
Let's be careful here.
We just integrated with respects to z
so we're going to replace z
with six minus 2y and zero.
So we'll have six times six minus 2y
minus one half times six minus 2y squared
minus 2y times six minus 2y.
Now we have a bunch of algebra to do.
So we'll have 36 minus 12y.
We have to square this so we'll have 36 times a half,
that's going to be minus 18.
And then we'll have minus 24y
times a negative one half, that'll be plus 12y.
Then we'll have a positive 4y squared
times negative one half.
That'll be minus 2y squared.
Then we'll have minus 12y plus 4y squared.
Let's go ahead and combine our like terms.
Let's say our constant term will be 18.
Our y term will be negative 12y or minus 12y.
And then the y squared term will be positive 2y squared.
Let's take this to the next slide.
So we'll have 18y minus 12 times y squared over two
plus two times y to the third over three.
So we'll have 18y
minus 6y squared plus 2/3
y cubed.
So now we'll substitute in three
and then substitute in zero.
So when y is three, we'll have 18 times three, that's 54.
Minus here we'll have three squared times six.
So minus 54.
And then plus 2/3 times three cubed,
that'll be 27 times 2/3, that'll give us 18.
And then when y is zero, everything will be zero.
So we have a volume of 18 cubic units.
Let's go back and take a look at that graph.
Getting the volume bounded by these yellow planes,
and this blue plane
would be 18 cubic units.
And that'll do it for this video.
We'll take a look at another example in the next video.