CalculusVolume3-OP_mktoy8b-105

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5.13
We just have to integrate the constant function f (x, y) = 1 over the region. Thus, the area A of the bounded
region is ∫
x = 0
x = 2
∫
y = x2
y = 2x
dy dx or ∫
y = 0
x = 4
∫
x = y/2
x = y
dx dy:
A = ∬
D
1dx dy = ∫
x = 0
x = 2
∫
y = x2
y = 2x
1dy dx = ∫
x = 0
x = 2
⎡
⎣y|y = x2
y = 2x⎤
⎦dx = ∫
x = 0
x = 2
⎛
⎝2x − x2⎞
⎠dx = x2 − x3
3 |02 = 4
3.
Find the area of a region bounded above by the curve y = x3 and below by y = 0 over the interval
[0, 3].
We can also use a double integral to find the average value of a function over a general region. The definition is a direct
extension of the earlier formula.
Definition
If f (x, y) is integrable over a plane-bounded region D with positive area A(D), then the average value of the
function is
fave = 1
A(D) ∬
D
f (x, y)dA.
Note that the area is A(D) = ∬
D
1dA.
Example 5.19
Finding an Average Value
Find the average value of the function f (x, y) = 7xy2 on the region bounded by the line x = y and the curve
x = y (Figure 5.25).
Chapter 5 | Multiple Integration 513
5.14
Figure 5.25 The region bounded by x = y and x = y.
Solution
First find the area A(D) where the region D is given by the figure. We have
A(D) = ∬
D
1dA = ∫
y = 0
y = 1
∫
x = y
x = y
1dx dy = ∫
y = 0
y = 1
⎡
⎣x|x = y
x = y⎤
⎦dy = ∫
y = 0
y = 1
( y − y)dy = 2
3y3/2 − y2
2 |01 = 1
6.
Then the average value of the given function over this region is
fave = 1
A(D) ∬
D
f (x, y)dA = 1
A(D) ∫
y = 0
y = 1
∫
x = y
x = y
7xy2 dx dy = 1
1/6 ∫
y = 0
y = 1⎡
⎣
⎢7
2x2 y2|x = y
x = y⎤
⎦
⎥dy
= 6 ∫
y = 0
y = 1
⎡
⎣
7
2y2 ⎛
⎝y − y2⎞
⎠
⎤
⎦dy = 6 ∫
y = 0
y = 1
⎡
⎣
7
2
⎛
⎝y
3 − y4⎞
⎠
⎤
⎦dy = 42
2
⎛
⎝
⎜y4
4 − y5
5
⎞
⎠
⎟|01 = 42
40 = 21
20.
Find the average value of the function f (x, y) = xy over the triangle with vertices
(0, 0), (1, 0) and (1, 3).
Improper Double Integrals
An improper double integral is an integral ∬
D
f dA where either D is an unbounded region or f is an unbounded
function. For example, D = ⎧
⎩
⎨(x, y)||x − y| ≥ 2⎫
⎭
⎬ is an unbounded region, and the function f (x, y) = 1/⎛
⎝1 − x2 − 2y2⎞
⎠ over
the ellipse x2 + 3y2 ≤ 1 is an unbounded function. Hence, both of the following integrals are improper integrals:
i. ∬
D
xy dA where D = ⎧
⎩
⎨(x, y)||x − y| ≥ 2⎫
⎭
⎬;
ii. ∬
D
1
1 − x2 − 2y2dA where D = ⎧
⎩
⎨(x, y)|x2 + 3y2 ≤ 1⎫
⎭
⎬.
In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that f has
only finitely many discontinuities. Not all such improper integrals can be evaluated; however, a form of Fubini’s theorem
does apply for some types of improper integrals.
514 Chapter 5 | Multiple Integration
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Theorem 5.6: Fubini’s Theorem for Improper Integrals
If D is a bounded rectangle or simple region in the plane defined by ⎧
⎩
⎨(x, y): a ≤ x ≤ b, g(x) ≤ y ≤ h(x)⎫
⎭
⎬ and also by
⎧
⎩
⎨(x, y): c ≤ y ≤ d, j(y) ≤ x ≤ k(y)⎫
⎭
⎬ and f is a nonnegative function on D with finitely many discontinuities in the
interior of D, then
∬
D
f dA = ∫
x = a
x = b
∫
y = g(x)
y = h(x)
f (x, y)dy dx = ∫
y = c
y = d
∫
x = j(y)
x = k(y)
f (x, y)dx dy.
It is very important to note that we required that the function be nonnegative on D for the theorem to work. We consider
only the case where the function has finitely many discontinuities inside D.
Example 5.20
Evaluating a Double Improper Integral
Consider the function f (x, y) = ey
y over the region D = ⎧
⎩
⎨(x, y): 0 ≤ x ≤ 1, x ≤ y ≤ x⎫
⎭
⎬.
Notice that the function is nonnegative and continuous at all points on D except (0, 0). Use Fubini’s theorem
to evaluate the improper integral.
Solution
First we plot the region D (Figure 5.26); then we express it in another way.
Figure 5.26 The function f is continuous at all points of the
region D except (0, 0).
The other way to express the same region D is
D = ⎧
⎩
⎨(x, y): 0 ≤ y ≤ 1, y2 ≤ x ≤ y⎫
⎭
⎬.
Chapter 5 | Multiple Integration 515
5.15
Thus we can use Fubini’s theorem for improper integrals and evaluate the integral as
∫
y = 0
y = 1
∫
x = y2
x = y
ey
y dx dy.
Therefore, we have
∫
y = 0
y = 1
∫
x = y2
x = y
ey
y dx dy = ∫
y = 0
y = 1
ey
y x|x = y2
x = y dy = ∫
y = 0
y = 1
ey
y
⎛
⎝y − y2⎞
⎠dy = ∫
0
1
⎛
⎝ey − yey⎞
⎠dy = e − 2.
As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the
function f is continuous in an unbounded rectangle R.
Theorem 5.7: Improper Integrals on an Unbounded Region
If R is an unbounded rectangle such as R = ⎧
⎩
⎨(x, y): a ≤ x ≤ ∞, c ≤ y ≤ ∞⎫
⎭
⎬, then when the limit exists, we have
∬
R
f (x, y)dA = lim
(b, d) → (∞, ∞)
∫
a
b ⎛
⎝
⎜∫
c
d
f (x, y)dy
⎞
⎠
⎟dx = lim
(b, d) → (∞, ∞)
∫
c
d ⎛
⎝
⎜∫
a
b
f (x, y)dy
⎞
⎠
⎟dy.
The following example shows how this theorem can be used in certain cases of improper integrals.
Example 5.21
Evaluating a Double Improper Integral
Evaluate the integral ∬
R
xye−x2 − y2
dA where R is the first quadrant of the plane.
Solution
The region R is the first quadrant of the plane, which is unbounded. So
∬
R
xye−x2 − y2
dA = lim
(b, d) → (∞, ∞)
∫
x = 0
x = b⎛
⎝
⎜ ∫
y = 0
y = d
xye−x2 − y2
dy
⎞
⎠
⎟dx = lim
(b, d) → (∞, ∞)
∫
y = 0
y = d⎛
⎝
⎜ ∫
x = 0
x = b
xye−x2 − y2
dy
⎞
⎠
⎟dy
= lim
(b, d) → (∞, ∞)
1
4
⎛
⎝1 − e−b2⎞
⎠
⎛
⎝1 − e−d2⎞
⎠ = 1
4
Thus, ∬
R
xye−x2 − y2
dA is convergent and the value is 1
4.
Evaluate the improper integral ∬
D
y
1 − x2 − y2
dA where D = ⎧
⎩
⎨(x, y)x ≥ 0, y ≥ 0, x2 + y2 ≤ 1⎫
⎭
⎬.
516 Chapter 5 | Multiple Integration
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In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over
general regions. Before we go over an example with a double integral, we need to set a few definitions and become familiar
with some important properties.
Definition
Consider a pair of continuous random variables X and Y , such as the birthdays of two people or the number of
sunny and rainy days in a month. The joint density function f of X and Y satisfies the probability that (X, Y) lies
in a certain region D:
P⎛
⎝(X, Y) ∈ D⎞
⎠ = ∬
D
f (x, y)dA.
Since the probabilities can never be negative and must lie between 0 and 1, the joint density function satisfies the
following inequality and equation:
f (x, y) ≥ 0 and ∬
R2
f (x, y)dA = 1.
Definition
The variables X and Y are said to be independent random variables if their joint density function is the product of
their individual density functions:
f (x, y) = f1 (x) f2 (y).
Example 5.22
Application to Probability
At Sydney’s Restaurant, customers must wait an average of 15 minutes for a table. From the time they are seated
until they have finished their meal requires an additional 40 minutes, on average. What is the probability that a
customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the
meal are independent events?
Solution
Waiting times are mathematically modeled by exponential density functions, with m being the average waiting
time, as
f (t) =
⎧
⎩
⎨
0 if t < 0,
1
me−t/m if t ≥ 0.
If X and Y are random variables for ‘waiting for a table’ and ‘completing the meal,’ then the probability density
functions are, respectively,
f1(x) =
⎧
⎩
⎨
0 if x < 0,
1
15e−x/15 if x ≥ 0. and f2(y) =
⎧
⎩
⎨
0 if y < 0,
1
40e−y/40 if y ≥ 0.
Clearly, the events are independent and hence the joint density function is the product of the individual functions
f (x, y) = f1(x) f2(y) =
⎧
⎩
⎨
0 if x < 0 or y < 0,
1
600e−x/15 e−y/60 if x, y ≥ 0.
Chapter 5 | Multiple Integration 517