210 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.7: Bounded region in R2
4.3 Double Integrals Over General Regions
4.3.1 Introduction
When dealing with integrals of functions of one variable, we are always inte-grating over an interval. The only diffi culty in evaluating the definite integral∫ baf (x) dx came from the function f and the diffi culty to find an antiderivative
for it. With functions of two or more variables, not only the function can causethe integral to be diffi cult, but also the region over which we integrate. In theprevious section, we restricted ourselves to rectangular regions. However, notevery region is rectangular. A region of R2 can have any shape. Even if thefunction is easy to integrate, if the region is complex enough, the integral willstill be diffi cult to evaluate. We first define the integral of a function over twovariables over a general region. We will then show how this integral can beevaluated over some simple regions which we will call regions of types I and II.
Let us assume we are given a function f (x, y) defined on a closed andbounded region D as shown in figure 4.7. Suppose we want to integrate fover D. Because D is bounded, it can be enclosed in a rectangle R as shown infigure 4.8.
To define the integral of f over D, we first define the function F (x, y) asfollows:
F (x, y) =
f (x, y) if (x, y) ∈ D0 if (x, y) ∈ R−D (4.3)
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 211
Figure 4.8: Bounded region in R2, enclosed in a rectangle.
Definition 330 If∫∫R
F (x, y) dA exists, then we define
∫∫D
f (x, y) dA =
∫∫R
F (x, y) dA
This definition makes sense since, as we can see in figures 4.9 and 4.10, theportion of the graph of F which is not 0 is identical to the graph of f. Theportion which is 0 will not contribute to the integral. In particular, this meansthat for our definition, it does not matter which rectangle R we select. In the
case, f (x, y) ≥ 0,∫∫D
f (x, y) dA corresponds to the volume of the solid which
lies above D and below the graph of z = f (x, y).
We still must be able to compute∫∫R
F (x, y) dA. This is not always a simple
task. But it is for certain regions, which we consider next.
4.3.2 Regions of Type I
When describing a region, one has to give the condition x and y must satisfyso that a point (x, y) lies in the region. A region is said to be of type I if x isbetween two constants, and y is between two continuous functions of x. Moreprecisely, we have the following definition:
Definition 331 A plane region D is said to be of type I if it is of the form
D =
(x, y) ∈ R2 | a ≤ x ≤ b and g1 (x) ≤ y ≤ g2 (x)
212 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.9: Graph of f
Figure 4.10: Graph of F
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 213
Figure 4.11: Region of type I
where g1 and g2 are two continuous functions of x.
Possible regions of type I are shown in figures 4.11, 4.12 and 4.13. To evaluate∫∫D
f (x, y) dA, we enclose D in the rectangle [a, b] × [c, d] as shown in figure
4.3.2.We define F (x, y) as explained in equation 4.3. So, we have∫∫D
f (x, y) dA =
∫∫R
F (x, y) dA
=
∫ b
a
∫ d
c
F (x, y) dydx by Fubini’s theorem
To evaluate this iterated integral, we first evaluate the inside integral. Usingthe properties of the definite integral, we have:∫ d
c
F (x, y) dy =
∫ g1(x)
c
F (x, y) dy +
∫ g2(x)
g1(x)
F (x, y) dy +
∫ d
g2(x)
F (x, y) dy
=
∫ g2(x)
g1(x)
F (x, y) dy
since F (x, y) = 0 for y < g1 (x) and y > g2 (x) (see figure 4.3.2). Also when yis between g1 (x) and g2 (x), F (x, y) = f (x, y), so we have∫ d
c
F (x, y) dy =
∫ g2(x)
g1(x)
f (x, y) dy
Therefore, we have the following proposition:
214 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.12: Region of type I
Figure 4.13: Region of type I
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 215
Proposition 332 If f is continuous on a type I region D such that
D =
(x, y) ∈ R2 | a ≤ x ≤ b and g1 (x) ≤ y ≤ g2 (x)
then ∫∫D
f (x, y) dA =
∫ b
a
∫ g2(x)
g1(x)
f (x, y) dydx (4.4)
Remark 333 Notice that the function F (x, y) does not appear in the formula.It was simply a device used to derive this formula. When computing an integral,it does not come into play, as the next example will show.
Remark 334 When writing∫∫D
f (x, y) dA as an iterated integral, it is impor-
tant to remember that the outer integral must be the one with constant limit ofintegrations.
Remark 335 To evaluate a double integral over a general region, the first stepis always to write it as an iterated integral. How this is done depends on theregion. The region will determine what the limits of integration of the iteratedintegrals are. Hence, it is extremely important to really understand what theregion looks like. Graphing it will help.
Example 336 Write∫∫D
f (x, y) dA as an iterated integral where D =
(x, y) ∈ R2 | 0 ≤ x ≤ 1 and − x2 ≤ y ≤ x2
Here, the region is given explicitly, we know the limits for both x and y, so wecan write the iterated integral.∫∫
D
f (x, y) dA =
∫ 1
0
∫ x2
−x2f (x, y) dydx
216 CHAPTER 4. MULTIPLE INTEGRALS
Example 337 Write∫∫D
f (x, y) dA as an iterated integral where D is the re-
gion bounded by the x-axis, y = sinx, the lines x = 0 and x = 1.The region is shown in figure 337.
1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
1
1
2
x
y
Region D
We see that we can write D as: D = (x, y) | 0 ≤ x ≤ 1 and 0 ≤ y ≤ sinx andhence
∫∫D
f (x, y) dA =
∫ 1
0
∫ sin x
0
f (x, y) dydx
Example 338 Evaluate∫∫D
(x4 − 2y
)dA where D =
(x, y) ∈ R2 | −1 ≤ x ≤ 1 and − x2 ≤ y ≤ x2
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 217
D is clearly a type I region.∫∫D
(x4 − 2y
)dA =
∫ 1
−1
∫ x2
−x2
(x4 − 2y
)dydx
=
∫ 1
−1
[x4y − y2
∣∣x2−x2
]dx
=
∫ 1
−1
(x6 − x4 + x6 + x4
)dx
=
∫ 1
−1
2x6dx
=2
7x7
∣∣∣∣1−1
=4
7
4.3.3 Regions of Type II
The condition a point (x, y) must satisfy to be in a type II region is as follows.A region is said to be of type II if y is between two constants, and x is betweentwo continuous functions of y. More precisely, we have the following definition:
Definition 339 A plane region D is said to be of type II if it is of the form
D =
(x, y) ∈ R2 | c ≤ y ≤ d and h1 (y) ≤ x ≤ h2 (y)
Using the same method as for type I regions, we can show that
Proposition 340 If f is continuous on a type II region D such that
D =
(x, y) ∈ R2 | c ≤ y ≤ d and h1 (y) ≤ x ≤ h2 (y)
then ∫∫D
f (x, y) dA =
∫ d
c
∫ h2(y)
h1(y)
f (x, y) dxdy (4.5)
Example 341 Evaluate∫∫D
(xy − y3
)dA if D is the region bounded by the x-
axis, the lines x = −1, y = 1 and y = x.It often helps to sketch the region so we can see what type of region it is. Thisregion is shown in figure 4.14. This is a region of type II, it can be describedby D =
(x, y) ∈ R2 | 0 ≤ y ≤ 1 and − 1 ≤ x ≤ y
. In type II regions, x is the
variable between two functions of y. Make sure these are functions of y, not x.
218 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.14: Region bouded by the x−axis, and the lines y = −1, x = −1, andy = x.
In this case, the right side of the region is delimited by the line y = x, whichwhen we write as a function of y becomes x = y. Therefore, we have∫∫
D
(xy − y3
)dA =
∫ 1
0
∫ y
−1
(xy − y3
)dxdy
=
∫ 1
0
[x2
2y − xy3
∣∣∣∣y−1
]dy
=
∫ 1
0
(y3
2− y4 −
(y2
+ y3))
dy
=
∫ 1
0
(−y4 − y3
2− y
2
)dy
=
(−y
5
5− y4
8− y2
4
)∣∣∣∣10
= −23
40
4.3.4 Properties
We list without proof several properties the double integral satisfies. Theseproperties are very similar to the properties of the definite integral of functionsof one variable.
Proposition 342 Assuming that the following integrals exist and that C is aconstant, we have:
1.∫∫D
(f (x, y)± g (x, y)) dA =
∫∫D
f (x, y) dA±∫∫D
g (x, y) dA
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 219
2.∫∫D
Cf (x, y) dA = C
∫∫D
f (x, y) dA
3. If f (x, y) ≥ g (x, y) for all (x, y) in D, then∫∫D
f (x, y) dA ≥∫∫D
g (x, y) dA
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∫∫
D
f (x, y) dA =
∫∫D1
f (x, y) dA+
∫∫D2
f (x, y) dA
This property corresponds to the property for functions of one variable whichsays:
∫ baf (x) dx =
∫ caf (x) dx+
∫ bcf (x) dx. It can sometimes be used to evalu-
ate integrals over regions which are neither of type I, nor of type II. If we havesuch 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. Each integralis then an integral of type I or II, which we know how to evaluate.
Proposition 344∫∫D
1dA = A (D) where A (D) denotes the area of D.
This property says that if we integrate the constant function f (x, y) = 1over a region D, we get the area of D. This makes sense. The volume of a solida section D between two parallel planes is (area of the section)×height. In thiscase, the height is simply 1. This is also very important as it gives us a way ofcomputing the area of general regions. We illustrate it with an example.
Example 345 Find the area of the region D in the first quadrant bounded byy = x and y = x2.The region is shown in figure 4.15. We can treat this as a type 1 or a type 2region. We show the solution both ways.
• Type 1 region. The area of the region is∫∫D
dA where D is the region
described in the problem, in green on figure 4.15. To evaluate this integral,we must rewrite it as an iterated integral treating D as a type 1 region. Weimagine covering the region with vertical strips as shown in figure 4.16.We would use a strip for every x such that 0 ≤ x ≤ 1 and for each x, the
220 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.15: Region between y = x and y = x2
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 221
Figure 4.16: Type 1 region
range of y values along the strip would be: x2 ≤ y ≤ x. Therefore∫∫D
dA =
∫ 1
0
∫ x
x2dydx
=
∫ 1
0
y|xx2 dx
=
∫ 1
0
(x− x2
)dx
=
(x2
2− x3
3
)∣∣∣∣10
=1
2− 1
3
=1
6
• Type 2 region. This is similar to above, but we imagine covering the regionwith horizontal strips as shown in figure 4.17. we would use such stripsfor 0 ≤ y ≤ 1. For each y, the range of x values along the strip would be
222 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.17: Type 2 region
y ≤ x ≤ √y. Therefore∫∫D
dA =
∫ 1
0
∫ √yy
dxdy
=
∫ 1
0
x|√y
y dy
=
∫ 1
0
(√y − y) dy
=
2
3y
3
2 − y2
2
∣∣∣∣∣∣1
0
=2
3− 1
2
=1
6
Combining the above properties, we have:
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 223
Figure 4.18: Region bounded by y = x2 and y = 4√x
Proposition 346 If m ≤ f (x, y) ≤M for all (x, y) in D, then
mA (D) ≤∫∫D
f (x, y) dA ≤MA (D)
Definition 347 (Average Value) We define the average value of a functionf (x, y) over a region D to be
Average value of f over D =1
area of D
∫∫D
f (x, y) dA
4.3.5 More Practice Problems
In many cases, a region can be considered of type I or II. One then has to decidewhether to treat it as a type I region or a type II. It may be that either way willwork. However, in some cases, one way may lead to an integral which is verydiffi cult, while the other way leads to an easier integral. In the example thatfollows, we look at cases to illustrate these various options. The more problemyou practice on, the better you will become at identifying the type of a region,and how to perform the integration.
Example 348 Evaluate∫∫
Ω
(√x− y2
)dA where Ω is the region bounded by
y = x2 and y = x14 .
Once again, it helps to draw the region. It is shown in figure 4.18
224 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.19: Region bounded by y = x2 and y = 4√x
To completely define the region, one need to find the points where the two regionsmeet. For this, we solve
y = x2
y = 4√x
Combining the two gives
x2 = x14
If we raise both sides to the fourth power, we get
x8 = x
x8 − x = 0
x(x7 − 1
)= 0
x = 0 or x = 1
When x = 0, we get y = 0. When x = 1, we get y = 1. We see that this region
can be seen as a type I region: Ω =
(x, y) | 0 ≤ x ≤ 1 and x2 ≤ y ≤ x 14
. Note
that y is between two functions of x. Ω can also be seen as a type II region.(seefigure 4.19). In this case, Ω =
(x, y) | 0 ≤ y ≤ 1 and y4 ≤ x ≤ √y
. Note that
x is between two functions of y. We obtained them as follows. From figure 4.19,we see that the region is bounded on the left by y = x
14 which can be written as
x = y4. It is bounded on the right by y = x2 which can be written as x =√y.
We compute the integral both ways.
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 225
Method 1 Treating Ω as a type I region.
∫∫Ω
(√x− y2
)dA =
∫ 1
0
∫ x14
x2
(√x− y2
)dydx
=
∫ 1
0
y√x− y3
3
∣∣∣∣x14
x2
dx=
∫ 1
0
((x34 − x
34
3
)−(x52 − x6
3
))dx
=
∫ 1
0
(2
3x34 − x 5
2 +x6
3
)dx
=
(8
21x74 − 2
7x72 +
x7
21
)∣∣∣∣10
=8
21− 2
7+
1
21
=1
7
Method 2 Treating Ω as a type II region.
∫∫Ω
(√x− y2
)dA =
∫ 1
0
∫ y12
y4
(√x− y2
)dxdy
=
∫ 1
0
(2
3x32 − xy2
)∣∣∣∣y12
y4
dy=
∫ 1
0
((2
3y34 − y 5
2
)−(
2
3y6 − y6
))dy
=
∫ 1
0
(2
3y34 − y 5
2 +y6
3
)dy
=
((8
21y74 − 2
7y72 − y7
21
))∣∣∣∣10
=
(8
21− 2
7+
1
21
)=
1
7
Example 349 Evaluate∫∫
Ω
cos πx2
2 dA where D is the region bounded by x = 1,
y = 0 and y = x.
226 CHAPTER 4. MULTIPLE INTEGRALS
Figure 4.20: Region bounded by x = 0, x = 1, y = 0 and y = x
The region is shown in figure 4.20. Once again, we see that this region can betreated as a type I or type II region.
Method 1 Treating Ω as a type I region. The corner points of the region are(0, 0), (1, 0) and (1, 1). The region is determined by Ω = (x, y)0 ≤ x ≤ 1 and 0 ≤ y ≤ x.In this case: ∫∫
Ω
cosπx2
2dA =
∫ 1
0
∫ x
0
cosπx2
2dydx
=
∫ 1
0
[y cos
πx2
2
∣∣∣∣x0
]dx
=
∫ 1
0
x cosπx2
2dx
We finish with a substitution. If we let u = πx2
2 then du = πxdx. Also,when x = 0, u = 0 and when x = 1, u = π
2 . So, we have∫∫Ω
cosπx2
2dA =
1
π
∫ π2
0
cosudu
=1
π(sinu)|
π20
=1
π
Method 2 Treating Ω as a type II region. The region is determined by Ω =
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 227
(x, y) | 0 ≤ y ≤ 1 and y ≤ x ≤ 1. So∫∫Ω
cosπx2
2dA =
∫ 1
0
∫ 1
y
cosπx2
2dxdy
We cannot evaluate the inside integral as we do not know an antiderivativewith respect to x of
∫ 1
ycos πx
2
2 dx.
Example 350 Calculate the volume of the solid within the cylinder x2 + y2 =b2, between the planes y + z = a and z = 0, given that a ≥ b > 0.The region is shown in figure 350. The region D in R2 by which the solid isbounded is D =
(x, y) | x2 + y2 ≤ b2
. The solid is also bounded by the planes
z = 0 and z = −y + a. Therefore, its volume is given by the integral∫∫D
(a− y) dA =
∫∫D
adA−∫∫D
ydA
The second integral is easier to evaluate than it seems. The region D can bewritten as
D =
(x, y) | −b ≤ x ≤ b and −√b2 − x2 ≤ y ≤
√b2 − x2
Thus, ∫∫
D
ydA =
∫ b
−b
∫ √b2−x2−√b2−x2
ydydx
You will recall that integrating an odd function over a symmetric integral gives0, in other words, if f (x) is an odd function, then
∫ a−a f (x) dx = 0. Thus,
228 CHAPTER 4. MULTIPLE INTEGRALS
∫√b2−x2−√b2−x2 ydy = 0. It follows that
∫∫D
(a− y) dA =
∫∫D
adA
= a
∫∫D
dA
= aA (D) by one of the properties above
= πab2
4.3.6 Switching the Order of Integration
Evaluating a double integral over a region amounts to evaluating one or moreiterated integrals. As we know, there are two kinds of iterated integrals. Onein which the dx integral is the inner integral and the dy integral is the outerintegral. The other kind is the reverse. When the region of integration is arectangle, Fubini’s theorem tells us that the two integrals are the same. If R =
[a, b] × [c, d]. Then∫∫R
f (x, y) dA =∫ ba
∫ dcf (x, y) dydx =
∫ dc
∫ baf (x, y) dxdy.
When the region of integration is not a rectangle, we may not have a choice.We may be forced to use one of the iterated integrals as suggests example349. That example also suggests that given an iterated integral, to evaluate itmight require switching the order of integration. This cannot be done by justreversing the order of the integrals. It involves changing the type of region weare integrating over. We illustrate this with an example.
Example 351 Switch the order of integration∫ 1
0
∫ 1
ycos πx
2
2 dxdy.First, we need to identify the region, decide what type of region it is. Then,we rewrite it as a region of the other type and write the integral in terms ofthat region. From the limits of integration, we see that the region is D =
(x, y) : 0 ≤ y ≤ 1 and y ≤ x ≤ 1. Thus, this is a type II region. Thus∫ 1
0
∫ 1
ycos πx
2
2 dxdy =∫∫D
cos πx2
2 dA where D is as indicated and is shown in figure 4.20. To rewrite
as a type I region, we need to express x between two constants and y betweentwo functions. From figure 4.20, we see that D can also be written as D =
(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ x. Therefore,∫∫D
cos πx2
2 dA =∫ 1
0
∫ x0
cos πx2
2 dydx.
Hence ∫ 1
0
∫ 1
y
cosπx2
2dxdy =
∫ 1
0
∫ x
0
cosπx2
2dydx
4.3. DOUBLE INTEGRALS OVER GENERAL REGIONS 229
4.3.7 Summary
We list important points covered in this section.
• Since not every region in the plane is a rectangle, we generalized what wehad learned in the previous section to more general regions, closed andbounded regions. If we call D such a region, then we learned to compute∫∫D
f (x, y) dA.
• It is very important to understand that for multiple integrals, the region ofintegration,D, plays a very important role. I cannot emphasize enough theimportance of making sure one knows the region of integration well before
attempting to compute∫∫D
f (x, y) dA. Every multiple integral should
start with drawing D, the region of integration, and writing a concisemathematical description of it.
• We actually consider two types of regions. Type I regions and type IIregions. Unlike rectangular regions, the order of integration cannot beswitched easily, at least not without some work.
• Type I regions: D = (x, y) | a ≤ x ≤ b and g1 (x) ≤ y ≤ g2 (x). In thiscase ∫∫
D
f (x, y) dA =
b∫a
g2(x)∫g1(x)
f (x, y) dydx
Note that the limits of integration of the outer integral are always con-stants.
• Type II regions: D = (x, y) | c ≤ y ≤ d and h1 (y) ≤ x ≤ h2 (y). In thiscase ∫∫
D
f (x, y) dA =
d∫c
h2(x)∫h1(x)
f (x, y) dxdy
Note that the limits of integration of the outer integral are always con-stants.
• Given∫∫D
f (x, y) dA, students must be able to determine if D is a type
I or II region and write the double integral as the corresponding iteratedintegral.
230 CHAPTER 4. MULTIPLE INTEGRALS
• Given an iterated integral, that is one ofb∫a
g2(x)∫g1(x)
f (x, y) dydx or
d∫c
h2(x)∫h1(x)
f (x, y) dxdy,
students must be able to find the region D such that the iterated integral
is the same as∫∫D
f (x, y) dA.
• When D, the region of integration, can be written as both a type I and atype II region, it is possible to switch the order of integration. Sometimesit is necessary to do so. To reverse the order of integration of an iteratedintegral, perform the following steps:
—Given an iterated integral, find the region D that it corresponds toand identify the type of region D is.
—Write D as a region of the other type.
—Write the iterated integral corresponding to the newly written region.
• If the graph of z = f (x, y) is above the xy-plane then∫∫D
f (x, y) dA is
the volume of the solid with cross section D, between the xy-plane andz = f (x, y).
•∫∫D
dA gives the area of the region D.
4.3.8 Assignment
1. Do odd # 1 - 27, 29, 35, 37 at the end of 12.2 in your book.
2. Do odd # 1, 3, 5, 9, 11 at the end of 12.3 in your book.
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