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MA1511 Engineering Calculus

Chapter 2 Multiple Integrals
2.1 Double Integrals over Rectangular Domains
This chapter is devoted to definite integrals of functions in two or more variables, known as multiple
integrals. We will focus primarily on double integrals for functions in two variables.
We first recall that the definite integral of a non-negative function 𝑓(𝑥)
𝑏

∫ 𝑓(𝑥)𝑑𝑥
𝑎

represents the area between the graph of 𝑓, the 𝑥 −axis and the lines 𝑥 = 𝑎 and 𝑥 = 𝑏, as shown in
the diagram below.

Analogously, given a non-negative function 𝑓(𝑥, 𝑦) defined over a rectangular domain

𝐷 = {(𝑥, 𝑦): 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑},
the volume of the solid bounded between the surface and the rectangle 𝐷 is given by a double integral
denoted by

∬ 𝑓(𝑥, 𝑦) 𝑑𝐴.
𝐷

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The value of the above double integral is either one of the following iterated integrals
𝑏 𝑑 𝑑 𝑏

𝐼 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦𝑑𝑥 , 𝐽 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦,

𝑎 𝑐 𝑐 𝑎

which can be calculated by first integrating the inner integral with respect to the “inner variable” . To
illustrate this idea, we consider integral 𝐼:
𝑏 𝑑

𝐼 = ∫ (∫ 𝑓(𝑥, 𝑦) 𝑑𝑦) 𝑑𝑥
𝑎 𝑐

The inner integral within the parentheses is to be calculated first. To do this, we treat 𝑥 as a constant
and integrate 𝑓(𝑥, 𝑦) with respect to 𝑦. This will gives us a function of 𝑥, say 𝑔(𝑥). Hence,
b d b


a c
f ( x, y ) dydx   g ( x)dx
a

For example,
3 1 3 1

∫ ∫ 𝑥𝑦 + 𝑥 2 𝑑𝑦𝑑𝑥 = ∫ (∫ 𝑥𝑦 + 𝑥 2 𝑑𝑦) 𝑑𝑥
−3 0 −3 0

3 𝑦=1
𝑥𝑦 2
= ∫[ + 𝑥 2 𝑦] 𝑑𝑥
2 𝑦=0
−3
=𝑔(𝑥)
3
⏞𝑥
= ∫ ( + 𝑥 2 ) 𝑑𝑥
2
−3
𝑥=3
𝑥2 𝑥3
=[ + ] = 18
4 3 𝑥=−3

d b b
Similarly, to find   f ( x, y) dx dy
c a
, we first calculate the “inner integral”  f ( x, y) dx
a

with respect to 𝑥, holding 𝑦 constant. This will give us a function of 𝑦, say ℎ(𝑦). Hence,
b d d


a c
f ( x, y ) dydx   h( y )dy
c

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b d d b
It turns out that the two iterated integrals ,   f ( x, y) dydx
a c
and   f ( x, y) dx dy
c a
are numerically

equal if 𝑓 satisfies certain condition1.

Result 2.1A
Let f be a continuous function on the rectangular domain {(𝑥, 𝑦): 𝑎 ≤ 𝑥 ≤ 𝑏 , 𝑐 ≤ 𝑦 ≤ 𝑑 }.
Then
b d d b


a c
f ( x, y ) dydx    f ( x, y) dx dy
c a

Remarks
1. In subsequent sections, we will consider double integral

∬ 𝑓(𝑥, 𝑦) 𝑑𝐴.
𝐷

for more general domains of integration.

2. When 𝑓(𝑥, 𝑦) = 1,   f ( x, y) dA   1 dA 
D D
area of the region 𝐷.

This is obvious if we recall that the double integral gives the volume, 𝑉 of the solid bounded
between the surface 𝑧 = 𝑓(𝑥, 𝑦) = 1 (which is a horizontal plane of height 1 unit from the
𝑥𝑦 −plane and the region 𝐷. We have

∬ 𝑓(𝑥, 𝑦) 𝑑𝐴 = volume of solid = Area of D × height = Area of D

𝐷
as required.

More generally, for any constant 𝐶,

∬ 𝐶 𝑑𝐴 = 𝐶 ∬ 1 𝑑𝐴 = 𝐶 × area of 𝐷
𝐷 𝐷

1
Result 2.1A holds for any function that is continuous on the domain of integration. We shall not discuss formally the
definition of continuity of functions. It suffices to know that most functions we commonly encounter are continuous, and
we shall assume that result 2.1A holds for all functions we consider in this course.
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Using the technique demonstrated above, we can prove the following result (to be discussed in
Tutorial 1) which is a direct consequence of Result 2.1A.

Result 2.1B
Let f be defined on the rectangular domain {(𝑥, 𝑦): 𝑎 ≤ 𝑥 ≤ 𝑏 , 𝑐 ≤ 𝑦 ≤ 𝑑 }.
If 𝑓(𝑥, 𝑦) = 𝑎(𝑥) 𝑏(𝑦) for some continuous functions 𝑎(𝑥) and 𝑏(𝑦), then
b d
b  d 

a c
f ( x, y ) dydx    a ( x ) dx   b( y ) dy 
a  c 

Video Example 2.1.1

Find the volume of the solid bounded by the surface
𝑧 = 9𝑥 2 𝑦 2 + 3
defined over the region { (𝑥, 𝑦): − 1 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2}
𝐴𝑛𝑠𝑤𝑒𝑟: 28

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Remarks (Triple Integrals)

1. For a function 𝑓(𝑥, 𝑦, 𝑧) of three variables defined on a domain of the form
{(𝑥, 𝑦, 𝑧): 𝑎 ≤ 𝑥 ≤ 𝑏 , 𝑐 ≤ 𝑦 ≤ 𝑑, 𝑓 ≤ 𝑧 ≤ 𝑔 },
we can compute an iterated triple integral such as
𝑔 𝑑 𝑏

∫ ∫ ∫ 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧
𝑓 𝑐 𝑎

d b
by first evaluating the inner double integral   f ( x, y, z) dxdy
c a
which is a function of 𝑧,
g

say 𝑘 (𝑧) , and then calculating  k ( z ) dz .

f

2. Result 2.1A can be extended to triple integrals: the order of integration for an iterated
triple integral does not matter. For example,

𝑔 𝑑 𝑏 𝑑 𝑔 𝑏

∫ ∫ ∫ 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧 = ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑧𝑑𝑦

𝑓 𝑐 𝑎 𝑐 𝑓 𝑎

Note that there are a total of 6 different ways given by the order in which we perform
the integration.

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Changing Oder of Integration for Rectangular Domains

Since the two iterated integrals
𝑏 𝑑 𝑑 𝑏

∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦𝑑𝑥 , ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦,

𝑎 𝑐 𝑐 𝑎

are numerically equal, it is natural to ask: which one to use ?

One way to answer this question is to inspect the inner integral of each iterated integral and
look for the one that can be calculated more directly or easily. We illustrate this with an
example.

Video Example 2.1.2

Find the value of
1 𝜋/4

∫ ∫ 𝑦 cos(2𝑥𝑦) 𝑑𝑦𝑑𝑥
0 0

1
Answer : 4

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2.2 Double Integrals over General Domains

We now extend the definition of double integrals to non-rectangular domains.

Type I Domain {(𝒙, 𝒚): 𝒂 ≤ 𝒙 ≤ 𝒃, 𝒈(𝒙) ≤ 𝒚 ≤ 𝒉(𝒙)}

A region 𝐷 is said to be a type I region if it is bounded below two curves, say 𝑦 = 𝑔(𝑥) and
𝑦 = ℎ(𝑥) , over an interval of the form 𝑎 ≤ 𝑥 ≤ 𝑏, where one curve lies entirely above the
other, as shown in the above diagram.

The double integral ∬𝐷 𝑓(𝑥, 𝑦)𝑑𝐴 over the above region is given by the formula

𝑏 ℎ(𝑥)

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

𝐷
𝑎 𝑔(𝑥)

Note that any rectangular region

𝐷 = {(𝑥, 𝑦): 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑},

is a special case of a type I region where the functions 𝑔 and ℎ are constants : 𝑔(𝑥) = 𝑐 and
ℎ(𝑥) = 𝑑.

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Type II Domain {(𝒙, 𝒚): 𝒄 ≤ 𝒚 ≤ 𝒅, 𝒈(𝒚) ≤ 𝒙 ≤ 𝒉(𝒚)}

A region 𝐷 is said to be a type II region if it is bounded between two curves, say 𝑥 = 𝑔(𝑦)
and = ℎ(𝑦) , over an interval of the form 𝑎 ≤ 𝑦 ≤ 𝑏, where one curve lies entirely to the
right of other, as shown in the above diagram.

The double integral ∬𝐷 𝑓(𝑥, 𝑦)𝑑𝐴 over the above region is given by the formula

𝑑 ℎ(𝑦)

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦

𝐷
𝑐 𝑔(𝑦)

Note that any rectangular region

𝐷 = {(𝑥, 𝑦): 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑},

is a special case of a type II region where the functions 𝑔 and ℎ are constants : 𝑔(𝑦) = 𝑎 and
ℎ(𝑦) = 𝑏.

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We illustrate the above formulae in the following example.

Video Example 2.2.1

  xy dA , where 𝐷 is the region in the first quadrant bounded by the curves 𝑦 = 𝑥 3

2
Find
D

and 𝑦 = √𝑥.
5
Answer:
77

Remarks
If the domain of integration 𝐷 happens to be a union of two or more regions, e.g. 𝐷 = 𝐷1 ∪ 𝐷2
then, the additive nature of integrals gives

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 + ∬ 𝑓(𝑥, 𝑦)𝑑𝐴

𝐷 𝐷1 𝐷2

An example of such a region is shown below. This shaded region is made up of two type I regions,
namely
𝐷1 = {(𝑥, 𝑦): 0 ≤ 𝑥 ≤ 1, 3 − 3𝑥 ≤ 𝑦 ≤ (𝑥 + 1)(3 − 𝑥)}
𝐷2 = {(𝑥, 𝑦): 1 ≤ 𝑥 ≤ 2, 3𝑥 − 3 ≤ 𝑦 ≤ (𝑥 + 1)(3 − 𝑥)}

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Changing Oder of Integration Between Type I and II Domains

Domains which are of both types I and II are not uncommon. An example of such a domain is
the shown below.

When a given iterated integral is difficult or even impossible to compute and the domain of integration
is both of type I and type II, we change the order of integration.

To change the order of integration

Step 1 Sketch the domain of integration
Step 2 If the given iterated integral is of type I, rewrite it as a Type II integral.
If the given iterated integral is of type II, rewrite it as a Type I integral.

Using the above region as an example, the double integral

∬ 𝑓 (𝑥, 𝑦)𝑑𝐴
𝐷

can be found by computing one of the following iterated integrals, whichever is possible or
more efficient to calculate.
2 𝑥3 8 2

(Type I) ∫ ∫ 𝑓 (𝑥, 𝑦) 𝑑𝑦𝑑𝑥 OR (Type II) ∫ ∫ 𝑓 (𝑥, 𝑦) 𝑑𝑥𝑑𝑦

0 0 0 𝑦 1/3

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We illustrate the technique of changing the order of integration in the following video
example.
Video Example 2.2.2
Determine and sketch the region of integration of

 

 
0 y
sin x 2 dx dy

Hence, evaluate this integral.

Answer: 1

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2.3 Double Integrals in Polar Coordinates

Polar Coordinates

Take any point 𝑃(𝑥, 𝑦) (other than the origin O) in the coordinate plane. Let 𝑟 be the length of the
line segment 𝑂𝑃 , and let 𝛼 (0 ≤ α ≤ π) be the angle made between OP and the positive 𝑥 –
axis. Then, 𝑃(𝑥, 𝑦) can be represented by the ordered pair (𝑟, 𝜃) where

α if 𝑦 ≥ 0
𝜃={
−α if 𝑦 < 0

The origin O, also known as the pole, has polar coordinates (0, 𝜃) for any 𝜃.

The positive 𝑥-axis is called the polar axis and (𝑟, 𝜃) is the polar coordinates of the point 𝑃.

We have described the so-called polar coordinate system of representing points on the 𝑥𝑦-plane.
Some examples of points in both rectangular (Cartesian) coordinates and polar coordinates are given
in the table below.

Cartesian coordinates Polar Coordinates

A(1, 1) 𝜋
A (√2, )
4
B(-1, 1) 3𝜋
B (√2, )
4
C(-1, -1) 3𝜋
C(√2, − 4 )
D(1, -1) 𝜋
D(√2, − 4 )

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Conversion between Polar and Rectangular Coordinates

Let a point 𝑃(𝑥, 𝑦) have polar coordinates (𝑟, 𝜃)
It can be shown easily that
𝑥 = 𝑟cos𝜃 , 𝑦 = 𝑟sin𝜃 and 𝑟 2 = 𝑥 2 + 𝑦 2
𝜋 𝟓√𝟑 5
For example, the point in polar coordinates (5, − ) has Cartesian coordinates ( ,− ) using
6 𝟐 2
𝜋
the above formulae for 𝑥 and 𝑦 with 𝑟 = 5 and 𝜃 = − 6.

Graphs of Polar Equations (Polar Curves)

The graph of a polar equation 𝑟 = 𝑓(𝜃) , or implicity 𝑓(𝑟, 𝜃) = 0, consists of points (𝑟 , 𝜃 )
satisfying the given equation.

Examples of commonly encountered polar curves are

 𝑟 = 𝑎 (𝑎 > 0) : a circle centered at the origin O(0, 0) and having a radius of 𝑎 , since every
point in this graph has the same distance 𝑎 from the origin.
 𝜃 = 𝛼 : a ray from the pole (but excluding the pole) which makes an angle |𝛼| with the
positive x-axis (note that 𝛼 < 0 if the ray lies in the third or forth quadrant), e.g.

Video Example 2.3.1

Find the Cartesian equation of the polar curve 𝑟 = 𝑎 cos 𝜃 (𝑎 > 0).
𝑎 𝑎2
Answer:(𝑥 − 2)2 + 𝑦 2 = 4

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Polar Rectangles
A polar rectangle is a region defined in polar coordinates by

{(𝑟, 𝜃): 𝑎 ≤ 𝑟 ≤ 𝑏 and 𝛼 ≤ 𝜃 ≤ 𝛽} , where 0 ≤ 𝛽 − 𝛼 ≤ 2𝜋

A special case when 𝑎 = 0 gives a sector of a circle centered at O and radius 𝑏 :

An application of polar coordinates is the calculation of a double integral whose domain is a polar
rectangle,

∬ 𝑓 (𝑥, 𝑦)𝑑𝐴
𝐷

There are at least two reasons why polar coordinates is preferred to Cartesian coordinates ( in terms of
x and 𝑦) :
(1) In general. a polar rectangle is neither a single type I nor a single type II domain (defined in
2.2), as illustrated in the diagram below.
(2) In polar form, the equation of a circle of radius 𝑎 > 0 is simply 𝑟 = 𝑎, whereas the
Cartesian equation of the same circle takes a more complicated form of 𝑥 2 + 𝑦 2 = 𝑅 2 .

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To illustrate the advantage of using polar coordinates in double integral, we consider the shaded
region below.

Clearly, the shaded region is a union of two type I regions, namely

1
𝐷1 = {(𝑥, 𝑦): − ≤ 𝑥 ≤ 0, −𝑥 ≤ 𝑦 ≤ √1 − 𝑥 2 }
√2
1
𝐷2 = {(𝑥, 𝑦): 0 ≤ 𝑥 ≤ , 𝑥 ≤ 𝑦 ≤ √1 − 𝑥 2 }
√2
In Cartesian coordinates, we have
1
0 √1−𝑥 2 √2 √1−𝑥 2

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 + ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

𝐷 1 −𝑥 0 𝑥
−
√2

for any given 𝑓(𝑥, 𝑦). Such integrals are typically tedious to calculate.

On the other hand, in the polar form, the domain 𝐷 = 𝐷1 ∪ 𝐷2 is simply a single polar rectangle
𝜋 3𝜋
𝐷 = {(𝑟, 𝜃): 0 ≤ 𝑟 ≤ 1, ≤𝜃≤ }
4 4
So we need only calculate one double integral in which the integration limits are numbers. Further, we
do not have to deal with the term √1 − 𝑥 2 since the polar equation of the circle is simply 𝑟 = 1.

We now present the integral formula in polar form for the double integral

∬ 𝑓(𝑥, 𝑦)𝑑𝐴.
𝐷

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Double Integrals in Polar Coordinates

The integral,   f ( x, y) dA
D
over a domain bounded by two polar curves can be computed in polar

coordinates using the following result.

Result 2.3A
Let 𝑓 be a continuous function defined on the polar rectangle
𝐷 = {(𝑟, 𝜃): 𝑔(𝜃) ≤ 𝑟 ≤ ℎ(𝜃) and 𝛼 ≤ 𝜃 ≤ 𝛽 }, where 0 ≤ 𝛽 − 𝛼 ≤ 2𝜋.
Then,
𝛽 ℎ(𝜃)

∬ 𝑓(𝑥, 𝑦) = ∫ ∫ 𝑓(𝑟cos𝜃, 𝑟sin 𝜃) 𝑟𝑑𝑟𝑑𝜃

𝐷
𝛼 𝑔(𝜃)

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The special case when 𝑔(𝜃) = 0 is presented in result 2.3B

Result 2.3B
Let 𝑓 be a continuous function defined on the polar domain
𝐷 = {(𝑟, 𝜃): 0 ≤ 𝑟 ≤ ℎ(𝜃) and 𝛼 ≤ 𝜃 ≤ 𝛽 }, where 0 ≤ 𝛽 − 𝛼 ≤ 2𝜋.
Then,
𝛽 ℎ(𝜃)

∬ 𝑓(𝑥, 𝑦) = ∫ ∫ 𝑓(𝑟cos𝜃, 𝑟sin 𝜃) 𝑟𝑑𝑟𝑑𝜃

𝐷
𝛼 0

Some commonly encountered polar regions and their definitions are given below.

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The mathematical proof of result 2.3A is beyond the scope of this course. Nonetheless, we can
understand the formula
𝛽 ℎ(𝜃)

∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫ ∫ 𝑓(𝑟cos𝜃, 𝑟sin 𝜃) 𝑟𝑑𝑟𝑑𝜃

𝐷
𝛼 𝑔(𝜃)

by considering the effect of the following pair of substitutions

𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃
which convert Cartesian coordinates to polar coordinates.
Under these substitutions, the integrand 𝑓(𝑥, 𝑦) will be converted to 𝑓(𝑟 cos 𝜃, 𝑟 sin 𝜃) .
As for 𝑑𝐴, we can think of it as being “replaced with” 𝑟𝑑𝑟𝑑𝜃 .

Finally, the limits of integration are obtained from the following pair of inequalities
𝑔(𝜃) ≤ 𝑟 ≤ ℎ(𝜃) and 𝛼 ≤ 𝜃 ≤ 𝛽
that defined the polar rectangle.

We illustrate the use of polar coordinates in double integral with the following video example.

Video Example 2.3.2

Calculate the volume of the solid bounded between the surface 𝑧 = 1 − 𝑥 2 − 𝑦 2 and the 𝑥 − 𝑦
plane, 𝑧 = 0
𝜋
Answer : 2

End of Chapter 2

“Mathematics consists in proving the most obvious thing in the least obvious way.”

— George Polya

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