e y 2 dx d y
1
0
0
0
π/2
y
∞
∞
5.
cos x sin y dx d y
6.
x ye−( x 2+ y 2) dx d y
0
0
0
0
2
y
1
x 2
7.
1 dx d y
8.
2 d y dx
0
0
0
0
9. Find the volume V of the solid bounded by the three coordinate planes and the plane
x + y + z = 1.
10. Find the volume V of the solid bounded by the three coordinate planes and the plane
3 x + 2 y + 5 z = 6.
B
11. Explain why the double integral
1 d A gives the area of the region R. For simplicity,
R
you can assume that R is a region of the type shown in Figure 3.2.1(a).
C
12. Prove that the volume of a tetrahedron with mutually perpendic-
c
b
ular adjacent sides of lengths a, b, and c, as in Figure 3.2.6, is abc .
6
( Hint: Mimic Example 3.5, and recall from
a
Section 1.5 how three noncollinear points determine a plane. )
13. Show how Exercise 12 can be used to solve Exercise 10.
Figure 3.2.6
110
CHAPTER 3. MULTIPLE INTEGRALS
3.3 Triple Integrals
Our definition of a double integral of a real-valued function f ( x, y) over a region R in R2 can
be extended to define a triple integral of a real-valued function f ( x, y, z) over a solid S in R3.
We simply proceed as before: the solid S can be enclosed in some rectangular parallelepiped,
which is then divided into subparallelepipeds. In each subparallelepiped inside S, with sides
of lengths ∆ x, ∆ y and ∆ z, pick a point ( x , y , z ). Then define the triple integral of f ( x, y, z)
∗
∗
∗
over S, denoted by
f ( x, y, z) dV , by
S
f ( x, y, z) dV = lim
f ( x , y , z )
∗
∗
∗ ∆ x ∆ y ∆ z ,
(3.7)
S
where the limit is over all divisions of the rectangular parallelepiped enclosing S into sub-
parallelepipeds whose largest diagonal is going to 0, and the triple summation is over all the
subparallelepipeds inside S. It can be shown that this limit does not depend on the choice
of the rectangular parallelepiped enclosing S. The symbol dV is often called the volume
element.
Physically, what does the triple integral represent? We saw that a double integral could
be thought of as the volume under a two-dimensional surface. It turns out that the triple
integral simply generalizes this idea: it can be thought of as representing the hypervolume
under a three-dimensional hypersurface w = f ( x, y, z) whose graph lies in R4. In general,
the word “volume” is often used as a general term to signify the same concept for any n-
dimensional object (e.g. length in R1, area in R2). It may be hard to get a grasp on the concept
of the “volume” of a four-dimensional object, but at least we now know how to calculate that
volume!
In the case where S is a rectangular parallelepiped [ x , x ]
, y ]
, z ], that is, S
1
2
× [ y 1 2 × [ z 1 2
=
{( x, y, z) : x
, y
, z
}, the triple integral is a sequence of three iterated
1 ≤ x ≤ x 2
1 ≤ y ≤ y 2
1 ≤ z ≤ z 2
integrals, namely
z 2
y 2
x 2
f ( x, y, z) dV =
f ( x, y, z) dx d y d z ,
(3.8)
z 1
y 1
x 1
S
where the order of integration does not matter. This is the simplest case.
A more complicated case is where S is a solid which is bounded below by a surface z =
g ( x, y), bounded above by a surface z
( x, y), y is bounded between two curves h ( x) and
1
= g 2
1
h ( x), and x varies between a and b. Then
2
b
h 2( x)
g 2( x, y)
f ( x, y, z) dV =
f ( x, y, z) d z d y dx .
(3.9)
a
h 1( x)
g 1( x, y)
S
Notice in this case that the first iterated integral will result in a function of x and y (since its
limits of integration are functions of x and y), which then leaves you with a double integral of
3.3 Triple Integrals
111
a type that we learned how to evaluate in Section 3.2. There are, of course, many variations
on this case (for example, changing the roles of the variables x, y, z), so as you can probably
tell, triple integrals can be quite tricky. At this point, just learning how to evaluate a triple
integral, regardless of what it represents, is the most important thing. We will see some
other ways in which triple integrals are used later in the text.
3
2
1
Example 3.7. Evaluate
( x y + z) dx d y dz.
0
0
0
Solution:
3
2
1
3
2
x=1
( x y + z) dx d y dz =
1 x 2 y
d y d z
2
+ xz
0
0
0
0
0
x=0
3
2
=
1 y
2
+ z d y dz
0
0
3
y=2
=
1 y 2
d z
4
+ yz
0
y=0
3
=
(1 + 2 z) dz
0
3
= z + z 2
= 12
0
1
1− x
2− x− y
Example 3.8. Evaluate
( x + y + z) dz d y dx.
0
0
0
Solution:
1
1− x
2− x− y
1
1− x
z=2− x− y
( x + y + z) dz d y dx =
( x + y) z + 1 z 2
d y dx
2
0
0
0
0
0
z=0
1
1− x
=
( x + y)(2 − x − y) + 1 (2
2
− x − y)2 d y dx
0
0
1
1− x
=
2 − 1 x 2
y 2 d y dx
2
− xy − 12
0
0
1
y=1− x
=
2 y − 1 x 2 y
x y 2
y 3
dx
2
− xy − 12
− 16
0
y=0
1
=
11
x 3 dx
6 − 2 x + 1
6
0
1
= 11 x
x 4
6
− x 2 + 1
24
= 7
0
8
112
CHAPTER 3. MULTIPLE INTEGRALS
Note that the volume V of a solid in R3 is given by
V =
1 dV .
(3.10)
S
Since the function being integrated is the constant 1, then the above triple integral reduces
to a double integral of the types that we considered in the previous section if the solid is
bounded above by some surface z = f ( x, y) and bounded below by the xy-plane z = 0. There
are many other possibilities. For example, the solid could be bounded below and above by
surfaces z = g ( x, y) and z
( x, y), respectively, with y bounded between two curves h ( x)
1
= g 2
1
and h ( x), and x varies between a and b. Then
2
b
h 2( x)
g 2( x, y)
b
h 2( x)
V =
1 dV =
1 d z d y dx =
( g ( x, y)
( x, y)
2
− g 1
) d y dx
a
h 1( x)
g 1( x, y)
a
h 1( x)
S
just like in equation (3.9). See Exercise 10 for an example.
Exercises
A
For Exercises 1-8, evaluate the given triple integral.
3
2
1
1
x
y
1.
x yz dx d y d z
2.
x yz d z d y dx
0
0
0
0
0
0
π
x
x y
1
z
y
3.
x 2 sin z d z d y dx
4.
ze y 2 dx d y d z
0
0
0
0
0
0
e
y
1/ y
2
y 2
z 2
5.
x 2 z dx d z d y
6.
yz dx d z d y
1
0
0
1
0
0
2
4
3
1
1− x
1− x− y
7.
1 dx d y d z
8.
1 d z d y dx
1
2
0
0
0
0
9. Let M be a constant. Show that
z 2
y 2
x 2 M dx d y dz
)( y
)( x
).
z
= M( z 2 − z 1
2 − y 1
2 − x 1
1
y 1
x 1
B
10. Find the volume V of the solid S bounded by the three coordinate planes, bounded above
by the plane x + y + z = 2, and bounded below by the plane z = x + y.
C
b
z
y
b
11. Show that
f ( x) dx d y d z =
( b− x)2 f ( x) dx. ( Hint: Think of how changing the
2
a
a
a
a
order of integration in the triple integral changes the limits of integration. )
3.4 Numerical Approximation of Multiple Integrals
113
3.4 Numerical Approximation of Multiple Integrals
As you have seen, calculating multiple integrals is tricky even for simple functions and
regions. For complicated functions, it may not be possible to evaluate one of the iterated in-
tegrals in a simple closed form. Luckily there are numerical methods for approximating the
value of a multiple integral. The method we will discuss is called the Monte Carlo method.
The idea behind it is based on the concept of the average value of a function, which you
learned in single-variable calculus. Recall that for a continuous function f ( x), the average
value ¯
f of f over an interval [ a, b] is defined as
b
¯
1
f =
f ( x) dx .
(3.11)
b − a a
The quantity b − a is the length of the interval [ a, b], which can be thought of as the
“volume” of the interval. Applying the same reasoning to functions of two or three variables,
we define the average value of f ( x, y) over a region R to be
¯
1
f =
f ( x, y) d A ,
(3.12)
A( R)
R
where A( R) is the area of the region R, and we define the average value of f ( x, y, z) over a solid S to be
¯
1
f =
f ( x, y, z) dV ,
(3.13)
V ( S)
S
where V ( S) is the volume of the solid S. Thus, for example, we have
f ( x, y) d A = A( R) ¯ f .
(3.14)
R
The average value of f ( x, y) over R can be thought of as representing the sum of all the
values of f divided by the number of points in R. Unfortunately there are an infinite number
(in fact, uncountably many) points in any region, i.e. they can not be listed in a discrete
sequence. But what if we took a very large number N of random points in the region R
(which can be generated by a computer) and then took the average of the values of f for
those points, and used that average as the value of ¯
f ? This is exactly what the Monte Carlo
method does. So in formula (3.14) the approximation we get is
f 2 − ( ¯ f)2
f ( x, y) d A ≈ A( R) ¯ f ± A( R)
,
(3.15)
N
R
where
N
f ( x , y )
N ( f ( x , y ))2
¯
i
i
i
i
f =
i=1
and
f 2 =
i=1
,
(3.16)
N
N
114
CHAPTER 3. MULTIPLE INT