Welcome to our lesson
on Changing the Order of Integration using Triple Integrals.
Here, we want to express the triple integral
for the volume bounded by
four z, equals x squared plus y squared,
and z equals four.
We want to express the triple integral
using the following three orders of integration.
The first thing we should remember
is that we can determine the volume
of a solid region using a triple integral,
evaluated over the three-dimensional region v,
integrated with respects to v.
And then, we replace differential v
with our different orders of integration
with respects to x, y, and z.
Before we start,
let's go and take a look at this in three dimensions.
So, we're trying to determine the volume bounded by this
blue paraboloid in this yellow plane.
So, we're trying to determine
the volume of the region in here.
A couple things I want to point out,
first, notice that the volume is symmetrical
in the first four octants.
So, what we can do,
is determine the volume in the first octant,
and then, multiply that by four.
So, we're going to go ahead and do that.
And the last thing I want to mention is
that the xy-trace,
which would be where z is equal to zero,
would only have one point at the origin.
So, instead of determining the xy-trace,
we're going to need to determine the z equal four trace,
or the slice cut by this yellow plane,
and you can see, it's going to be a circle,
centered at zero, zero.
So, what we're going to do now is
determine the xy-trace, xz-trace, and yz-trace.
And then, we'll determine the limits of integration
for these three orders of integration.
So, as I just mentioned, for the xy-trace,
instead of letting z equal to zero, we'll let z equal four.
Then, looking at this equation here,
we would have 16 equals, x squared plus y squared.
So, we have a circle centered at zero, zero,
with a radius of four.
For the xz-trace, we'll set y equal zero.
So, we'd have four z equals x squared.
We could solve this for z.
We'd have z equals, x squared over four.
So, if this is the x-axis or this is the z-axis,
we have this parabola that opens up.
We also know that z is equal to four.
And then, for the yz-trace, we set x equal to zero.
So, we'd have four z equals y squared.
And then, solving for z,
we'd have z equals, y squared divided by four.
Again, we have this parabola here in blue that opens up.
And we also know that z is equal to four.
So, we're calling this the y-axis, and this the z-axis.
So, because we're going to need to determine the volume
in the first octant and then multiply it by four,
we're only concerned
about the regions in the first quadrant, so here,
here,
and here.
So, let's go ahead and set up this first triple integral,
integrating with respect to z, y, and then x.
Remember, we're going to determine the
volume in the first octant and then multiply it by four.
The limits of integration with respects to z
must be expressed in terms of y and x.
Well, the values of z are bounded this paraboloid
and z equals four.
So, if we solve this for z,
we would have x squared, plus y squared, divided by four.
That'll be our lower limit of integration for z.
And the upper limit of integration will be z equals four.
We now need to express y, as a function of x.
Let's go and take a look xy-trace.
Well, y would start at zero and then go up to the circle.
So, if we solve this equation here for y,
we would subtract x squared on both sides,
and then take the square root.
So, the limits of integration for a y would be from zero
all the way to the square root of 16 minus x squared.
And then, for x,
we could look at either xy-trace or the xz-trace.
X starts at zero, then goes out to four here,
as well as here.
So, here's the first triple integral,
integrating with respect to z, and then y, and then x.
Now, let's consider the order of integration dydxdz.
So, the limits of integration for y
must be expressed as a function of x and z.
Looking at the yz-trace, we can see that y starts at zero
and goes out to this parabola, which a piece of the
four z equals x squared plus y squared.
And since we have to express y as a function of x and z,
we can start y at zero,
but we do have to use this equation up here
to express y as a function of x and z.
So, to solve this for y,
we would subtract x squared on both sides,
and then take the square root of both sides.
So, the upper limit for y would be the square root
four z, minus x squared.
And now, for the limits of integration for x,
which must be expressed as a function of z,
we'll look at the xz-trace, as we see here.
So, x starts at zero and then goes out to this parabola.
And the equation the parabola is four z equals x squared.
So, to express this as function of z,
we'll have to solve this equation for x.
So, if we square root both sides of this equation,
we would have
x equals the square root of four z, or two square root z.
So, x starts at zero,
and it goes up to two square root z.
And then, for z, we can see from the xz and the yz-trace,
the lower limit would be zero,
and the upper limit would be four, here, and here.
Now, let's take a look at the order of dxdzdy.
So, limits of integration for x must be expressed
as a function of y and z.
Looking at these two traces here,
we can see that x starts at zero, and goes out to the curve.
And these curves are a slice of the paraboloid.
So, x would go from zero,
all the way to this equation here is solve for x
So, we'll have the square root of four z,
minus y squared.
And then, for the limits of z, expressed as a function of y,
we'll look at the yz-trace, which is here.
So, z starts on this curve here,
and goes up to four.
So, the upper limit will be four,
and the lower limit will be the equation of this graph,
expressed as a function of y.
So, if we solve this equation for z,
we would divide both sides by four.
We'd have y squared over four.
And then, for the limits of integration for y,
we can see from both the xy-trace and the yz-trace
that y starts at zero, and goes up to four, here,
as well as here.
So, now we've completed three different triple integrals
that would represent the same volume
of that three-dimensional region.
And just so we can see them all together, here they are.
You may want to try working some of these out
to see if you get the same answer,
but I believe this middle one
becomes very challenging to integrate.
And so, often it is helpful
to think about the order of integration
when setting up a triple integral
because it will affect how difficult it is to integrate.
I hope you found this helpful.