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Triple Integrals

Triple Integrals
Now that we know how to integrate over a two-dimensional region we need to move on to
integrating over a three-dimensional region
...
The notation for the general triple integrals is,

∫∫∫ f ( x, y, z ) dV
E

Let’s start simple by integrating over the box,

B = [ a,b ] × [ c, d ] × [ r, s ]

Note that when using this notation we list the x’s first, the y’s second and the z’s third
...
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
...

Let’s do a quick example of this type of triple integral
...

∫∫∫ 8 xyz dV , B = [ 2,3] × [1, 2] × [0,1]
B

Solution
Just to make the point that order doesn’t matter let’s use a different order from that listed above
...

2

3

1

2

0

∫∫∫ 8xyz dV = ∫ ∫ ∫ 8 xyz dz dx dy
1

B

=∫

2

=∫

2

1

1
2

∫
∫

3
2
3
2

4 xyz

2 1
0

dx dy

4 xy dx dy

= ∫ 2 x y dy
1

2

3

2

2

= ∫ 10 y dy = 15
1

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
...
We have three different
possibilities for a general region
...


In this case we define the region E as follows,

E = {( x, y, z ) | ( x, y ) ∈ D, u1 ( x, y ) ≤ z ≤ u2 ( x, y )}

where ( x, y ) ∈ D is the notation that means that the point ( x, y ) lies in the region D from the
xy-plane
...
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
...

Solution
We should first define octant
...
The
first octant is the octant in which all three of the coordinates are positive
...


We now need to determine the region D in the xy-plane
...
What we see will be the region D
in the xy-plane
...
Here is a
sketch of D
...
Since we are under the plane and in the first octant (so
we’re above the plane z = 0 ) we have the following limits for z
...


0≤ x≤3
2
0≤ y ≤− x+2
3

3
0≤ x ≤ − y+3
2
0≤ y≤2

Since neither really holds an advantage over the other we’ll use the first one
...
Here is a sketch of this region
...
Here is how we will evaluate these integrals
...


Example 3 Determine the volume of the region that lies behind the plane x + y + z = 8 and in
front of the region in the yz-plane that is bounded by z =

3
2

y and z = 34 y
...
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
...

So, here is a sketch of the region itself
...


0≤ y≤4
3
3
y≤z≤
y
4
2
0 ≤ x ≤ 8− y − z

The volume is then,

V = ∫∫∫
E

8− y − z
⌠⌠
⎡
dV = ⎮⎮ ∫
dx ⎤ dA
⎥⎦
⌡⌡ ⎢⎣ 0

D
4

=⌠
⎮
⌡0

∫

3 y 2

3y 4

8 − y − z dz dy

4

3 y
⌠
2
1
⎛
⎞
2
= ⎮ ⎜ 8 z − yz − z ⎟
dy
2 ⎠ 3y
⎮ ⎝
4
⌡0
4

1
2

3
57
3
33 2
⌠
2
y− y +
y dy
= ⎮ 12 y −
8
2
32
⌡0
4

⎛
57 2 3
11 3 ⎞
49
= ⎜8y −
y − y +
y ⎟ =
16
5
32 ⎠ 0 5
⎝
3
2

5
2

We now need to look at the third (and final) possible three-dimensional region we may run into
for triple integrals
...


In this final case E is defined as,

E = {( x, y, z ) | ( x, z ) ∈ D, u1 ( x, z ) ≤ y ≤ u2 ( x, z )}

and here the region D will be a region in the xz-plane
...


∫∫∫
E

u2 ( x , z )
⎡
⎤
⌠⌠
f ( x, y, z ) dV = ⎮⎮ ∫
f ( x, y, z ) dy dA
⎢
⎥⎦
u
x
z
,
⌡⌡ ⎣ 1 ( )
D

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
...


∫∫∫

3 x + 3z dV where E is the solid bounded by y = 2x + 2z and
2

2

2

2

E

Solution
Here is a sketch of the solid E
...
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
...
However we are in the xz-plane and we’ve only seen polar coordinates in the
xy-plane
...
We can always “translate” them over to the xz-plane with the
following definition
...
Note that
these definitions also lead to the formula,

x2 + z 2 = r 2
With this in hand we can arrive at the limits of the variables that we’ll need for this integral
...
The integrand is,

3( x + z
2

2

) (8 − ( 2 x

2

+ 2z

2

)) =
=
=

The integral is then,

∫∫∫
E

( 8 − 2r )
3 r ( 8 − 2r )
3 ( 8r − 2r )

3r

2

2

2

3

3x 2 + 3 z 2 dV = ∫∫ 3 ( 8r − 2r 3 ) dA
D

2π

= 3⌠
⌡0

∫

2
0

3
−
8
r
2
r
(
) r dr dθ

2π

⌠ ⎛8 3 2 5⎞
= 3 ⎮ ⎜ r − r ⎟ dθ
5 ⎠0
⌡0 ⎝ 3
2π

128
⌠
= 3⎮
dθ
⌡ 0 15
256 3 π
=
15

2



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