Now that we know how to integrate over a two-dimensional
region we need to move on to integrating over a three-dimensional region. We used a double integral to integrate over a
two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a
three dimensional region. The notation
for the general triple integrals is,
Let’s start simple by integrating over the box,
Note that when using this notation we list the x’s first, the y’s second and the z’s
third.
The triple integral in this case is,
Note that we integrated with respect to x first, then y, and
finally z here, but in fact there is
no reason to the integrals in this order.
There are 6 different possible orders to do the integral in and which
order you do the integral in will depend upon the function and the order that
you feel will be the easiest. We will
get the same answer regardless of the order however.
Let’s do a quick example of this type of triple integral.
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Example 1 Evaluate
the following integral.
 ,
Solution
Just to make the point that order doesn’t matter let’s use
a different order from that listed above.
We’ll do the integral in the following order.
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Before moving on to more general regions let’s get a nice
geometric interpretation about the triple integral out of the way so we can use
it in some of the examples to follow.
Fact
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The volume of the three-dimensional region E is given by the integral,
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Let’s now move on the more general three-dimensional
regions. We have three different
possibilities for a general region. Here
is a sketch of the first possibility.
In this case we define the region E as follows,
where is the notation that means that the point lies in the region D from the xy-plane. In this case we will evaluate the triple
integral as follows,
where the double integral can be evaluated in any of the
methods that we saw in the previous couple of sections. In other words, we can integrate first with
respect to x, we can integrate first
with respect to y, or we can use
polar coordinates as needed.
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Example 2 Evaluate
where E
is the region under the plane that lies in the first octant.
Solution
We should first define octant. Just as the two-dimensional coordinates
system can be divided into four quadrants the three-dimensional coordinate
system can be divided into eight octants.
The first octant is the octant in which all three of the coordinates
are positive.
Here is a sketch of the plane in the first octant.
 
We now need to determine the region D in the xy-plane. We can get a visualization of the region by
pretending to look straight down on the object from above. What we see will be the region D in the xy-plane. So D will be the triangle with vertices
at ,
,
and . Here is a sketch of D.
Now we need the limits of integration. Since we are under the plane and in the
first octant (so we’re above the plane ) we have the following limits for z.
We can integrate the double integral over D using either of the following two
sets of inequalities.
Since neither really holds an advantage over the other
we’ll use the first one. The integral
is then,
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Let’s now move onto the second possible three-dimensional
region we may run into for triple integrals.
Here is a sketch of this region.
For this possibility we define the region E as follows,
So, the region D
will be a region in the yz-plane. Here is how we will evaluate these integrals.
As with the first possibility we will have two options for
doing the double integral in the yz-plane
as well as the option of using polar coordinates if needed.
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Example 3 Determine
the volume of the region that lies behind the plane and in front of the region in the yz-plane that is bounded by and .
Solution
In this case we’ve been given D and so we won’t have to really work to find that. Here is a sketch of the region D as well as a quick sketch of the
plane and the curves defining D
projected out past the plane so we can get an idea of what the region we’re
dealing with looks like.
 
Now, the graph of the region above is all okay, but it
doesn’t really show us what the region is.
So, here is a sketch of the region itself.
 
Here are the limits for each of the variables.
The volume is then,
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We now need to look at the third (and final) possible three-dimensional
region we may run into for triple integrals.
Here is a sketch of this region.
In this final case E
is defined as,
and here the region D
will be a region in the xz-plane. Here is how we will evaluate these integrals.
where we will can use either of the two possible orders for
integrating D in the xz-plane or we can use polar coordinates
if needed.
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Example 4 Evaluate
where E
is the solid bounded by and the plane .
Solution
Here is a sketch of the solid E.
 
The region D in
the xz-plane can be found by
“standing” in front of this solid and we can see that D will be a disk in the xz-plane. This disk will come from the front of the
solid and we can determine the equation of the disk by setting the elliptic
paraboloid and the plane equal.
This region, as well as the integrand, both seems to
suggest that we should use something like polar coordinates. However we are in the xz-plane and we’ve only seen polar coordinates in the xy-plane. This is not a problem. We can always “translate” them over to the xz-plane with the following
definition.
Since the region doesn’t have y’s we will let z take
the place of y in all the
formulas. Note that these definitions
also lead to the formula,
With this in hand we can arrive at the limits of the
variables that we’ll need for this integral.
The integral is then,
Now, since we are going to do the double integral in polar
coordinates let’s get everything converted over to polar coordinates. The integrand is,
The integral is then,
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