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210
CHAPTER 4
...
7: Bounded region in R2
4
...
3
...
The only di¢ culty in evaluating the de…nite integral
Rb
f (x) dx came from the function f and the di¢ culty to …nd an antiderivative
a
for it
...
In the
previous section, we restricted ourselves to rectangular regions
...
A region of R2 can have any shape
...
We …rst de…ne the integral of a function over two
variables over a general region
...
Let us assume we are given a function f (x; y) de…ned on a closed and
bounded region D as shown in …gure 4
...
Suppose we want to integrate f
over D
...
8
...
3)
4
...
DOUBLE INTEGRALS OVER GENERAL REGIONS
211
Figure 4
...
De…nition 330 If
ZZ
F (x; y) dA exists, then we de…ne
R
ZZ
f (x; y) dA =
D
ZZ
F (x; y) dA
R
This de…nition makes sense since, as we can see in …gures 4
...
10, the
portion of the graph of F which is not 0 is identical to the graph of f: The
portion which is 0 will not contribute to the integral
...
In the
ZZ
case, f (x; y) 0,
f (x; y) dA corresponds to the volume of the solid which
D
lies above D and below the graph of z = f (x; y)
...
This is not always a simple
R
task
...
4
...
2
Regions of Type I
When describing a region, one has to give the condition x and y must satisfy
so that a point (x; y) lies in the region
...
More
precisely, we have the following de…nition:
De…nition 331 A plane region D is said to be of type I if it is of the form
D = (x; y) 2 R2 j a
x
b and g1 (x)
y
g2 (x)
212
CHAPTER 4
...
9: Graph of f
Figure 4
...
3
...
11: Region of type I
where g1 and g2 are two continuous functions of x
...
11, 4
...
13
...
3
...
We de…ne F (x; y) as explained in equation 4
...
So, we have
ZZ
ZZ
f (x; y) dA =
F (x; y) dA
D
=
Z
R
b
a
Z
d
F (x; y) dydx by Fubini’ theorem
s
c
To evaluate this iterated integral, we …rst evaluate the inside integral
...
3
...
Also when y
is between g1 (x) and g2 (x), F (x; y) = f (x; y), so we have
Z d
Z g2 (x)
F (x; y) dy =
f (x; y) dy
c
g1 (x)
Therefore, we have the following proposition:
214
CHAPTER 4
...
12: Region of type I
Figure 4
...
3
...
4)
g1 (x)
Remark 333 Notice that the function F (x; y) does not appear in the formula
...
When computing an integral,
it does not come into play, as the next example will show
...
Remark 335 To evaluate a double integral over a general region, the …rst step
is always to write it as an iterated integral
...
The region will determine what the limits of integration of the iterated
integrals are
...
Graphing it will help
...
ZZ
Z 1 Z x2
f (x; y) dA =
f (x; y) dydx
D
0
x2
x
1 and
x2
y
x2
216
CHAPTER 4
...
The region is shown in …gure 337
...
0 -0
...
6 -0
...
2
0
...
4
0
...
8
1
...
2
1
...
6
x
1 and 0
y
1
...
0
x
-1
Region D
We see that we can write D as: D = f(x; y) j 0
hence
ZZ
f (x; y) dA =
0
D
Example 338 Evaluate
Z
ZZ
D
x4
1
Z
sin xg and
sin x
f (x; y) dydx
0
2y dA where D = (x; y) 2 R2 j
1
x
1 and
x2
y
x2
4
...
DOUBLE INTEGRALS OVER GENERAL REGIONS
217
D is clearly a type I region
...
3
...
A region is said to be of type II if y is between two constants, and x is between
two continuous functions of y
...
5)
h1 (y)
y 3 dA if D is the region bounded by the x-
D
axis; the lines x = 1, y = 1 and y = x
...
This
region is shown in …gure 4
...
This is a region of type II, it can be described
by D = (x; y) 2 R2 j 0 y 1 and
1 x y
...
Make sure these are functions of y, not x
...
MULTIPLE INTEGRALS
Figure 4
...
axis, and the lines y =
1, x =
1, and
In this case, the right side of the region is delimited by the line y = x, which
when we write as a function of y becomes x = y
...
3
...
These
properties are very similar to the properties of the de…nite integral of functions
of one variable
...
(f (x; y) g (x; y)) dA =
f (x; y) dA
g (x; y) dA
D
D
D
4
...
DOUBLE INTEGRALS OVER GENERAL REGIONS
2
...
If f (x; y)
ZZ
219
f (x; y) dA
D
g (x; y) for all (x; y) in D, then
ZZ
f (x; y) dA
D
ZZ
g (x; y) dA
D
Proposition 343 If D, D1 and D2 are three regions in R2 such that D =
D1 [ D2 and D1 and D2 do not overlap then
ZZ
f (x; y) dA =
ZZ
f (x; y) dA +
D1
D
ZZ
f (x; y) dA
D2
This property corresponds to the property for functions of one variable which
Rb
Rc
Rb
says: a f (x) dx = a f (x) dx + c f (x) dx
...
If we have
such a region and we can divide it into two (or more regions) of type I or II ,
then we can break the integral into integrals over the subregions
...
Proposition 344
ZZ
1dA = A (D) where A (D) denotes the area of D
...
This makes sense
...
In this
case, the height is simply 1
...
We illustrate it with an example
...
The region is shown in …gure 4
...
We can treat this as a type 1 or a type 2
region
...
Type 1 region
...
15
...
We
imagine covering the region with vertical strips as shown in …gure 4
...
We would use a strip for every x such that 0 x 1 and for each x, the
220
CHAPTER 4
...
15: Region between y = x and y = x2
4
...
DOUBLE INTEGRALS OVER GENERAL REGIONS
221
Figure 4
...
Therefore
1
x
x2 dx
0
=
=
=
1
2
1
6
x3
3
x2
2
1
0
1
3
Type 2 region
...
17
...
For each y, the range of x values along the strip would be
222
CHAPTER 4
...
17: Type 2 region
y
x
p
y
...
3
...
18: Region bounded by y = x2 and y =
Proposition 346 If m
f (x; y)
mA (D)
ZZ
223
p
4
x
M for all (x; y) in D, then
f (x; y) dA
M A (D)
D
De…nition 347 (Average Value) We de…ne the average value of a function
f (x; y) over a region D to be
1
Average value of f over D =
area of D
ZZ
f (x; y) dA
D
4
...
5
More Practice Problems
In many cases, a region can be considered of type I or II
...
It may be that either way will
work
...
In the example that
follows, we look at cases to illustrate these various options
...
Example 348 Evaluate
1
ZZ
p
x
y 2 dA where
is the region bounded by
y = x2 and y = x 4
...
It is shown in …gure 4
...
MULTIPLE INTEGRALS
Figure 4
...
For this, we solve
y = p2
x
y= 4x
Combining the two gives
1
x2 = x 4
If we raise both sides to the fourth power, we get
x8
x
8
x x7
x
1
x
= x
=
0
=
0
=
0 or x = 1
When x = 0, we get y = 0
...
We see that this region
o
n
1
can be seen as a type I region: = (x; y) j 0 x 1 and x2 y x 4
...
can also be seen as a type II region
...
19)
...
Note that
x is between two functions of y
...
From …gure 4
...
It is bounded on the right by y = x2 which can be written as x = y
...
4
...
DOUBLE INTEGRALS OVER GENERAL REGIONS
Method 1 Treating
ZZ
p
x
as a type I region
...
y
2
dA =
Z
1
Z
1
0
=
Z
Z
Z
1
1
=
ZZ
cos
2 3
y4
3
2 3
y4
3
y2
xy 2
y4
5
y2 +
3
5 dy
2 6
y
3
5
y2
8 7
2 7
y4
y2
21
7
8
2
1
+
21 7 21
=
Example 349 Evaluate
y 2 dxdy
x
1
4 2x3
2
3
0
=
p
2
0
=
1
y2
y4
0
=
y = 0 and y = x
...
MULTIPLE INTEGRALS
Figure 4
...
20
...
Method 1 Treating as a type I region
...
The region is determined by = f(x; y)g0 x 1 and 0
In this case:
ZZ
cos
x2
dA =
2
=
Z
1
0
Z
Z
x
x2
dydx
2
cos
0
1
x2
2
y cos
0
=
Z
1
x cos
0
x
dx
0
x2
dx
2
2
x
We …nish with a substitution
...
Also,
when x = 0, u = 0 and when x = 1, u = 2
...
The region is determined by
=
y
xg
...
3
...
So
x2
dA =
2
Z
1
0
Z
1
cos
y
x2
dxdy
2
We cannot evaluate the inside integral as we do not know an antiderivative
R1
x2
with respect to x of y cos 2 dx
...
The region is shown in …gure 350
...
The solid is also bounded by the planes
z = 0 and z = y + a
...
The region D can be
written as
n
o
p
p
D = (x; y) j b x b and
b2 x 2 y
b2 x2
Thus,
ZZ
D
ydA =
Z
b
b
Z
p
b2 x2
p
ydydx
b2 x2
You will recall that integrating an odd function over aRsymmetric integral gives
a
0, in other words, if f (x) is an odd function, then
f (x) dx = 0
...
MULTIPLE INTEGRALS
R pb 2
p
x2
b2 x2
ydy = 0
...
3
...
As we know, there are two kinds of iterated integrals
...
The other kind is the reverse
...
If R =
s
ZZ
RbRd
RdRb
[a; b] [c; d]
...
R
When the region of integration is not a rectangle, we may not have a choice
...
That example also suggests that given an iterated integral, to evaluate it
might require switching the order of integration
...
It involves changing the type of region we
are integrating over
...
R1R1
x2
Example 351 Switch the order of integration 0 y cos 2 dxdy
...
Then,
we rewrite it as a region of the other type and write the integral in terms of
that region
...
Thus, this is a type II region
...
20
...
From …gure 4
...
Therefore,
cos 2 dA = 0 0 cos 2 dydx
...
3
...
3
...
Since not every region in the plane is a rectangle, we generalized what we
had learned in the previous section to more general regions, closed and
bounded regions
...
D
It is very important to understand that for multiple integrals, the region of
integration, D, plays a very important role
...
Every multiple integral should
D
start with drawing D, the region of integration, and writing a concise
mathematical description of it
...
Type I regions and type II
regions
...
Type I regions: D = f(x; y) j a x b and g1 (x) y
case
2
ZZ
Zb gZ(x)
f (x; y) dydx
f (x; y) dA =
D
g2 (x)g
...
Type II regions: D = f(x; y) j c y d and h1 (y) x
case
ZZ
Zd h2 (x)
Z
f (x; y) dA =
f (x; y) dxdy
D
h2 (y)g
...
Given
ZZ
f (x; y) dA, students must be able to determine if D is a type
D
I or II region and write the double integral as the corresponding iterated
integral
...
MULTIPLE INTEGRALS
2
Zb gZ(x)
Zd h2 (x)
Z
f (x; y) dydx or
f (x; y) dxdy,
Given an iterated integral, that is one of
a g1 (x)
c h1 (x)
students must be able to …nd the region D such that the iterated integral
ZZ
is the same as
f (x; y) dA
...
Sometimes
it is necessary to do so
...
– Write D as a region of the other type
...
If the graph of z = f (x; y) is above the xy-plane then
ZZ
f (x; y) dA is
D
the volume of the solid with cross section D, between the xy-plane and
z = f (x; y)
...
D
4
...
8
Assignment
1
...
2 in your book
...
Do odd # 1, 3, 5, 9, 11 at the end of 12
...