1
n
= max( U 1
n
Then it can be shown4 that the joint p.d.f. of X and Y is
n( n − 1)( y − x) n−2, for 0 ≤ x ≤ y ≤ 1
f ( x, y) =
(3.56)
0,
elsewhere.
Thus, the expected value of X is
1
1
E X =
n( n − 1) x( y − x) n−2 d y dx
0
x
1
y=1
=
nx( y − x) n−1
dx
0
y= x
1
=
nx(1 − x) n−1 dx , so integration by parts yields
0
1
1
= − x(1 − x) n −
(1 − x) n+1
n + 1
0
1
E X =
,
n + 1
4See Ch. 6 in HOEL, PORT and STONE.
134
CHAPTER 3. MULTIPLE INTEGRALS
and similarly (see Exercise 3) it can be shown that
1
y
n
EY =
n( n − 1) y( y − x) n−2 dx d y =
.
0
0
n + 1
So, for example, if you were to repeatedly take samples of n = 3 random real numbers from
(0, 1), and each time store the minimum and maximum values in the sample, then the aver-
age of the minimums would approach 1 and the average of the maximums would approach
4
3 as the number of samples grows. It would be relatively simple (see Exercise 4) to write a
4
computer program to test this.
Exercises
B
1. Evaluate the integral ∞ e− x 2 dx using anything you have learned so far.
−∞
2. For σ > 0 and µ > 0, evaluate ∞
1
e−( x− µ)2/2 σ 2 dx.
−∞ σ 2 π
3. Show that EY = n in Example 3.18
n+1
C
4. Write a computer program (in the language of your choice) that verifies the results in
Example 3.18 for the case n = 3 by taking large numbers of samples.
5. Repeat Exercise 4 for the case when n = 4.
6. For continuous random variables X , Y with joint p.d.f. f ( x, y), define the second moments
E( X 2) and E( Y 2) by
∞
∞
∞
∞
E( X 2) =
x 2 f ( x, y) dx d y
and
E( Y 2) =
y 2 f ( x, y) dx d y ,
−∞ −∞
−∞ −∞
and the variances Var( X ) and Var( Y ) by
Var( X ) = E( X 2) − ( EX )2 and Var( Y ) = E( Y 2) − ( EY )2 .
Find Var( X ) and Var( Y ) for X and Y as in Example 3.18.
7. Continuing Exercise 6, the correlation ρ between X and Y is defined as
E( X Y ) − ( EX )( EY )
ρ =
,
Var( X ) Var( Y )
where E( X Y ) = ∞
∞ xy f ( x, y) dx d y. Find
−∞ −∞
ρ for X and Y as in Example 3.18.
(Note: The quantity E( X Y ) − ( EX )( EY ) is called the covariance of X and Y .)
8. In Example 3.17 would the answer change if the interval (0, 100) is used instead of (0, 1)?
Explain.
4 Line and Surface Integrals
4.1 Line Integrals
In single-variable calculus you learned how to integrate a real-valued function f ( x) over an
interval [ a, b] in R1. This integral (usually called a Riemann integral) can be thought of as
an integral over a path in R1, since an interval (or collection of intervals) is really the only
kind of “path” in R1. You may also recall that if f ( x) represented the force applied along the
x-axis to an object at position x in [ a, b], then the work W done in moving that object from
position x = a to x = b was defined as the integral:
b
W =
f ( x) dx
a
In this section, we will see how to define the integral of a function (either real-valued or
vector-valued) of two variables over a general path (i.e. a curve) in R2. This definition will
be motivated by the physical notion of work. We will begin with real-valued functions of two
variables.
In physics, the intuitive idea of work is that
Work = Force × Distance .
Suppose that we want to find the total amount W of work done in moving an object along a
curve C in R2 with a smooth parametrization x = x( t), y = y( t), a ≤ t ≤ b, with a force f ( x, y) which varies with the position ( x, y) of the object and is applied in the direction of motion
along C (see Figure 4.1.1 below).
y
t = t
C
i
∆ s
2
2
i ≈
∆ xi + ∆ yi
t = a ∆ yi
t = ti+1
∆ xi
t = b
x
0
Figure 4.1.1
Curve C : x = x( t), y = y( t) for t in [ a, b]
We will assume for now that the function f ( x, y) is continuous and real-valued, so we only
consider the magnitude of the force. Partition the interval [ a, b] as follows:
a = t 0 < t 1 < t 2 < ··· < tn−1 < tn = b , for some integer n ≥ 2
135
136
CHAPTER 4. LINE AND SURFACE INTEGRALS
As we can see from Figure 4.1.1, over a typical subinterval [ t , t
] the distance
traveled
i
i+1
∆ si
along the curve is approximately
∆ x 2
2, by the Pythagorean Theorem. Thus, if the
i
+ ∆ yi
subinterval is small enough then the work done in moving the object along that piece of the
curve is approximately
Force × Distance ≈ f ( x , y )
2
2 ,
(4.1)
i∗
i∗
∆ xi + ∆ yi
where ( x , y )
, t
], and so
i∗
i∗ = ( x( ti ∗), y( ti ∗)) for some ti ∗ in [ ti
i+1
n−1
W ≈
f ( x , y )
2
2
(4.2)
i∗
i∗
∆ xi + ∆ yi
i=0
is approximately the total amount of work done over the entire curve. But since
∆ x 2
∆ y 2
∆ x 2
2
i
i
,
i
+ ∆ yi =
+
∆ ti
∆ ti
∆ ti
where ∆ t
, then
i = ti+1 − ti
n−1
∆ x 2
∆ y 2
W ≈
f ( x , y )
i
i
.
(4.3)
i∗
i∗
+
∆ ti
i=0
∆ ti
∆ ti
Taking the limit of that sum as the length of the largest subinterval goes to 0, the sum over
all subintervals becomes the integral from t = a to t = b, ∆ xi and ∆ yi become x ′( t) and y′( t),
∆ ti
∆ ti
respectively, and f ( x , y ) becomes f ( x( t), y( t)), so that
i∗
i∗
b
W =
f ( x( t), y( t))
x ′( t)2 + y′( t)2 dt .
(4.4)
a
The integral on the right side of the above equation gives us our idea of how to define,
for any real-valued function f ( x, y), the integral of f ( x, y) along the curve C, called a line integral:
Definition 4.1. For a real-valued function f ( x, y) and a curve C in R2, parametrized by
x = x( t), y = y( t), a ≤ t ≤ b, the line integral of f ( x, y) along C with respect to arc length s is
b
f ( x, y) ds =
f ( x( t), y( t))
x ′( t)2 + y′( t)2 dt .
(4.5)
C
a
The symbol ds is the differential of the arc length function
t
s = s( t) =
x ′( u)2 + y′( u)2 du ,
(4.6)
a
4.1 Line Integrals
137
which you may recognize from Section 1.9 as the length of the curve C over the interval [ a, t],
for all t in [ a, b]. That is,
ds = s ′( t) dt =
x ′( t)2 + y′( t)2 dt ,
(4.7)
by the Fundamental Theorem of Calculus.
For a general real-valued function f ( x, y), what does the line integral
f ( x, y) ds rep-
C
resent? The preceding discussion of ds gives us a clue. You can think of differentials as
infinitesimal lengths. So if you think of f ( x, y) as the height of a picket fence along C, then
f ( x, y) ds can be thought of as approximately the area of a section of that fence over some
infinitesimally small section of the curve, and thus the line integral
f ( x, y) ds is the total
C
area of that picket fence (see Figure 4.1.2).
y
f ( x, y)
C
ds
x
0
Figure 4.1.2
Area of shaded rectangle = height × width ≈ f ( x, y) ds
Example 4.1. Use a line integral to show that the lateral surface area A of a right circular
cylinder of radius r and height h is 2 πrh.
z
Solution: We will use the right circular cylinder with base circle C
given by x 2 + y 2 = r 2 and with height h in the positive z direction
r
(see Figure 4.1.3). Parametrize C as follows:
x
h
= x( t) = r cos t ,
y = y( t) = r sin t , 0 ≤ t ≤ 2 π
= f ( x, y)
y
Let f ( x, y) = h for all ( x, y). Then
0
b
x
C : x 2 + y 2 = r 2
A =
f ( x, y) ds =
f ( x( t), y( t))
x ′( t)2 + y′( t)2 dt
C
a
Figure 4.1.3
2 π
=
h
(− r sin t)2 + ( r cos t)2 dt
0
2 π
= h
r
sin2 t + cos2 t dt
0
2 π
= rh
1 dt = 2 πrh
0
138
CHAPTER 4. LINE AND SURFACE INTEGRALS
Note in Example 4.1 that if we had traversed the circle C twice, i.e. let t vary from 0 to
4 π, then we would have gotten an area of 4 πrh, i.e. twice the desired area, even though the
curve itself is still the same (namely, a circle of radius r). Also, notice that we traversed the
circle in the counter-clockwise direction. If we had gone in the clockwise direction, using the
parametrization
x = x( t) = r cos(2 π − t) ,
y = y( t) = r sin(2 π − t) , 0 ≤ t ≤ 2 π ,
(4.8)
then it is easy to verify (see Exercise 12) that the value of the line integral is unchanged.
In general, it can be shown (see Exercise 15) that reversing the direction in which a curve
C is traversed leaves
f ( x, y) ds unchanged, for any f ( x, y). If a curve C has a parametriza-
C
tion x = x( t), y = y( t), a ≤ t ≤ b, then denote by − C the same curve as C but traversed in the opposite direction. Then − C is parametrized by
x = x( a + b − t) ,
y = y( a + b − t) , a ≤ t ≤ b ,
(4.9)
and we have
f ( x, y) ds =
f ( x, y) ds .
(4.10)
C
− C
Notice that our definition of the line integral was with respect to the arc length parameter
s. We can also define
b
f ( x, y) dx =
f ( x( t), y( t)) x ′( t) dt
(4.11)
C
a
as the line integral of f ( x, y) along C with respect to x, and
b
f ( x, y) d y =
f ( x