Contents

Multiple
integration
1. Introduction to Surface Integrals
2. Multiple integrals over non-rectangular regions
3. Volume integrals
4. Changing coordinates
Learning
outcomes
needs doing.
Time
allocation
You are expected to spend approximately thirteen hours of independent study on the
material presented in this workbook. However, depending upon your ability to concentrate
and on your previous experience with certain mathematical topics this time may vary
considerably.
1
Introduction to Surface
Integrals
27.1
Introduction
Often in Engineering, it is necessary to find the sum of a quantity over an area or surface. This
can be achieved by means of a surface integral or double integral i.e. a function is integrated
twice, once with respect to one variable and subsequently with respect to another variable. This
Section looks at the concept of the double integral and how to evaluate a double integral over
a rectangular area.
#
Prerequisites
Before starting this Section you should . . .
"
① thoroughly understand the various
techniques of integration
② be familiar with the concept of a function
of two variables
Learning Outcomes
✓ understand the concept of a surface integral
After completing this Section you should be
able to . . .
✓ integrate a function over a rectangular
region
!
1. An Example of a Surface Integral
An engineer involved with the construction of a dam to hold back the water in a reservoir needs
to be able to calculate the total force the water exerts on the dam so that the dam is built with
sufficient strength.
In order to calculate this force, two results are required:
(a) The pressure p of the water is proportional to the depth. That is
p = kd
(1)
where k is a constant.
(b) The force on an area subjected to constant pressure is given by
f orce = pressure × area
(2)
The diagram shows the face of the dam. The depth of water is h and δA is a small area in the
face of the dam with coordinates (x, y).
y
δA
h−y
h
y
x
x
Using (1), the pressure at δA ∼ k(h − y). Using (2), the force on an area δA ∼ k(h − y)δA.
Both of these expressions are approximate as y is slightly different at the top of δA to the
bottom.
Now
Total force on dam
=
≈
sum of forces on all areas δA making up the face of the dam
k(h − y)δA
all δA
For a better approximation let δA become smaller, and for the exact result find the limit as
δA → 0. Then
Total force on the dam = lim
k(h − y)δA
δA→0
k(h − y) dA
=
A
where A k(h − y) dA stands for the surface integral of k(h − y) over the area A. Surface
integrals are evaluated using double integrals. The following Section shows a double integral
being developed in the case of the volume under a surface.
3
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
2. Single and Double Integrals.
As has been seen in Section 14.3, the area under the curve y = f (x) between x = a and x = b is
b
given by a f (x)dx (assuming that the curve lies above the axis for all x in the range a ≤ x ≤ b).
This is illustrated by the figure below.
y
area required
y = f (x)
a
x
b
In a similar manner, the volume under a surface (given by a function of two variables) and above
the xy plane can be found by integrating the function twice, once with respect to x and once
with respect to y.
f (x, y)
y
y=d
y=c
x=a
x=b
x
The above figure shows the part of a surface given by f (x, y) which lies above the rectangle
a ≤ x ≤ b, c ≤ y ≤ d. This rectangle is shaded and the volume above this rectangle but below
the surface can be seen.
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
4
f (x, y)
δx
y
y=d
y=c
x=a
x=b
x
Imagine a vertical slice taken through this volume at right angles to the x-axis (figure above).
This slice has thickness δx and lies at position x. Assuming that δx is small enough that the
areas of both sides (left and right) of this slice are virtually the same, the area of each face of
the slice is given by the integral
y=d
f (x, y) dy
(where x measures the position of the slice)
y=c
and the volume of the slice will be given by
y=d
f (x, y) dy
δx
y=c
To find the total volume between the surface and the xy plane, this quantity should be summed
over all possible such slices, each for a different value of x. Thus
y=d
V ≈
f (xi , y) dy δx
i
y=c
When δx becomes infinitesmally small, it can be considered to be dx and the summation will
change into an integral. Hence
x=b y=d
f (x, y) dydx
V =
x=a
y=c
Thus the volume is given by integrating the function twice, once with respect to x and once
with respect to y.
The procedure shown here considers the volume above a rectangular area and below the surface.
The volume beneath the surface over a non-rectangular area can also be found by integrating
twice (see Section 36.2).
5
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Key Point
The volume under the surface z = f (x, y) and above the xy plane (that is the rectangle
a ≤ x ≤ b, c ≤ y ≤ d) is given by the integral:
b
d
V =
f (x, y) dydx
x=a
y=c
3. ‘Inner’ and ‘Outer’ Integrals.
A typical double integral may be expressed as
x=b y=d
I=
f (x, y) dy
dx
x=a
y=c
where the part in the centre i.e.
y=d
f (x, y) dy
y=c
(known as the inner integral) is the integral of a function of x and y with respect to y. As the
integration takes place with respect to y, the variable x may be regarded as a fixed quantity (a
constant) but for every different value of x, the inner integral will take a different value. Thus,
y=d
the inner integral will be a function of x e.g. g (x) = y=c f (x, y) dy.
This ”inner integral”,
x=b being a function of x, once evaluated, can take its place within the outer
integral i.e. I = x=a g (x) dx which can then be integrated with respect to x to give the value
of the double integral.
The limits on the outer integral will be constants; the limits on the inner integral may be
constants (in which case the integration takes place over a rectangular area) or may be functions
of the variable used for the outer integral (in this case x). In this latter case, the integration
takes place over a non-rectangular area (see Section 36.2). In the examples quoted in this Section
or in the early parts of the next Section, the limits include the name of the relevant variable;
this can be omitted once more familiarity has been gained with the concept. It will be assumed
that the limits on the inner integral apply to the variable used to integrate the inner integral
and the limits on the outer integral apply to the variable used to integrate this outer integral.
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
6
4. Integration over rectangular areas
Consider the double integral
5 1
I=
(2x + y) dydx
x=0
y=−1
This represents an integral over the rectangle shown below.
y
1
5
x
−1
Here, the inner integral is
1
g (x) =
(2x + y) dy
−1
and the outer integral is
5
g (x) dx
I=
x=0
Looking in more detail at the inner integral
1
g (x) =
(2x + y) dy
−1
the function (2x + y) can be integrated with respect to y (keeping x constant) to give 2xy +
1 2
y + C (where C is a constant and can be omitted as the integral is a definite integral) i.e.
2
1
1
1
1
1
1 2
= 2x +
− −2x +
= 2x + + 2x − = 4x.
g (x) = 2xy + y
2
2
2
2
2
−1
This is a function of x as expected. This inner integral can be placed into the outer integral to
get
5
I=
4x dx
x=0
which becomes
5
I = 2x2 0 = 2 × 52 − 2 × 02 = 2 × 25 − 0 = 50
Hence the double integral
7
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
5
1
I=
(2x + y) dydx = 50
x=0
y=−1
Key Point
Evaluate the inner integral first and substitute the result into the outer integral.
Example
Evaluate the double integral
2 3
I=
x2 y dydx
x=−1
y=−2
This integral is evaluated over the area shown below.
y
3
−1
2
x
−2
Solution
Here, the inner integral is
2 3
9
4
5
2y
g (x) =
x y dy = x
= x2 − x2 = x2
2 −2 2
2
2
y=−2
3
2
and hence the outer integral is
2
2
5 2
51 3
5
5
15
I=
= × 8 − (−1) =
x dx =
x
23
6
6
2
x=−1 2
−1
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
8
Example Use the above approach to evaluate the double integral
5
1
x2 cos
I=
x=0
y=−1
πy
dydx
2
Note that the limits are the same as in a previous case but that the function
itself has changed.
Solution
The inner integral is
1
1
πy
πy
2
2
4
2 2
2
x cos dy =
= x2 1 − x2 (−1) = x2
x sin
2
π
2 −1 π
π
π
y=−1
so the outer integral becomes
5
5
4 2
4 3
4
4
500
I=
=
x dx =
x
125 −
0=
≈ 53.1
3π
3π
3π
3π
x=0 π
0
Clearly, variables other than x and y may be used.
Example Evaluate the double integral
4
π
I=
s sin t dtds
s=1
t=0
Solution
This integral becomes (dispensing with the step of formally writing the inner integral),
4
4
4
π
[−s cos t]0 ds =
[−s cos π + s cos 0] ds =
[−s (−1) + s (1)] ds
I =
s=1
1
1
4
4
2s ds = s2 1 = 16 − 1 = 15
=
1
Clearly, evaluating the integrals can involve further tools of integration, e.g. integration by parts
or by substitution.
Example Evaluate the double integral
2
3
I=
−1
9
−2
xye−x
dydx
y2 + 1
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Here, the limits have not formally been linked with a variable name but the limits on the outer
integral apply to x and the limits on the inner integral apply to y. As the integrations are more
complicated, the inner integral will be evaluated explicitly.
Solution
3
Inner Integral =
−2
xye−x
dy
y2 + 1
which can be evaluated by means of the substitution U = y 2 + 1.
1
If U = y 2 + 1 then dU = 2ydy so ydy = dU .
2
Also if y = −2 then U = 5 and if y = 3 then U = 10.
So the inner integral becomes (remembering that x may be treated as a constant)
10
xe−x
xe−x 10 dU
xe−x
1 xe−x
ln 2
=
dU =
=
[ln U ]10
(ln 10 − ln 5) = xe−x
5
2 U
2
U
2
2
2
5
5
and so the double integral becomes
2
ln 2 2 −x
−x ln 2
I=
xe
xe dx
dx =
2
2 −1
−1
which can be evaluated by integration by parts.
2
2
−x ln 2
ln 2 −x 2
−2
1
−x
1 × −e
e dx
dx =
−xe −1 −
−2e + (−1) e +
I =
2
2
−1
−1
2 ln 2 ln 2 ln 2 −2e−2 − e1 + −e−x −1 =
−2e−2 − e−1 − e−2 + e1 =
−3e−2 ≈ −0.14
=
2
2
2
Evaluate the following double integrals over rectangular areas.
1 2
• I = x=0 y=0 xy dydx
• I=
• I=
• I=
• I=
• I=
• I=
1 2
−1
0
−2
0
3 4
π1
0
−1
23
0
−1
31
0
0
(x2 + y 2 ) dxdy
y sin2 x dydx
st3 dsdt
4
5z 2 w (w2 − 1) dwdz
2π 1
0
(x2 y + 3y 2 ) dydx
0
ty sin t dydt
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
10
Your solution
1, 52/3, 460/3, 0, 16, 9/2, −π
Note that the last two examples require integration by substitution and integration by parts
respectively.
5. Special cases
If the integrand can be written as
f (x, y) = g (x) h (y)
then the double integral
b d
g (x) h (y) dydx
I=
a
c
can be written as
b
I=
g (x) dx ×
a
d
h (y) dy
c
i.e. the product of the two individual integrals. For example, the integral
2 3
I=
x2 y dydx
x=−1
y=−2
which was evaluated earlier can be written as
3 2 2 3
2
3
x
8 (−1) 9 4
y
2
I=
−
−
x dx ×
ydy =
=
3 −1 2 −2
3
3
2 2
−1
y=−2
15
5
=
2
2
the same result as before.
=3×
11
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Key Point
The integral
b
d
g(x)h(y) dydx can be written as
a
c
b
g(x) dx ×
d
h(y) dy
a
c
Imagine the integral
1 1
2
I=
xe−y dydx
−1
0
Approached directly, this would involve evaluating the integral
1
xe−y dy which cannot be
2
0
done by algebraic means (i.e. it can only be determined numerically). However, the integral can
be re-written as
1
1
1
1
1 21
2
−y 2
−y 2
xdx ×
e dy = [ x ]−1 ×
e dy = 0 ×
e−y dy = 0
I=
2
−1
0
0
0
and the result can be found without the need to evaluate the difficult integral.
If the integrand is independent of one of the variables and is simply a function of the other
variable, then onlyone
integration need be carried out.
d
bd
The integral I1 = a c h (y) dydx may be written as I1 = (b − a) c h (y) dy and the integral
bd
b
I2 = a c g (x) dydx may be written as I2 = (d − c) a g (x) dx i.e. the integral in the variable
upon which the integrand depends multiplied by the length of the range of integration for the
other variable.
Example Evaluate the double integral
2
2
y 2 dydx
I=
0
−1
Solution
As the integral in y can be multiplied by the range of integration in x, the double integral will
equal
3 2
2
3
3
y
2
(−1)
y 2 dy = 2
=2
−
=6
I = (2 − 0)
3 −1
3
3
−1
Note that the two integrations can be carried out in either order as long as the limits are
associated with the correct variable. For example
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
12
1
2
I=
x y dydx =
x=0
1
=
0
and
1
4
y=−1
x=0
2
1
2
x y dxdy =
I=
y=−1
2
=
−1
x4 y 2
2
2
1 4
4
dx =
2x − x dx
2
x=0
−1
1
1
3
3
3 4
3 5
3
x dx =
x
×1−
×0=
=
2
10
10
10
10
0
4
x=0
y=−1
x5 y
5
1
2
y
−1
5
dy =
0
− 0 dy
2 2
4
y
1
3
y
=
dx =
−
=
5
10 −1 10 10
10
Evaluate the following integrals:
11
(a) I = 0 −1 z (w + 1) dwdz
π/2 1
(b) I = 0
(y cos x) dydx
0
3 1 2
(c) I = −8 −1 y dydx
15
(d) I = 0 0 (s + 1)4 dtds
Your solution
1, 1/2, 22/3, 31
3 2
23
Evaluate the integrals −1 0 x3 y dydx and 0 −1 x3 y dxdy and show that
they are equal. As explained above, the order in which the integrations are
carried out does not matter for integrations over rectangular areas.
13
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Your solution
40
6. Applications of surface integration over rectangular areas
Force on a dam
At the beginning of this Section, the total force on a dam was given by the surface integral
k(h − y) dA
A
Imagine that the dam is rectangular in profile with a width of 100m and a height h of 40m.
The expression dA is replaced by dxdy and the limits on the variables x and y are 0 to 100m
and 0 to 40m respectively. The constant k may be assumed to be 104 kg m−2 s−2 . The surface
integral becomes the double integral
40 100
k(h − y) dxdy
0
0
that is
40 0
100
104 (40 − y) dxdy
0
As the integral in this double integral does not contain x, the integral may be written
40
40 100
4
10 (40 − y) dxdy = (100 − 0)
104 (40 − y)dy
0
0
0
y 2 40
]
2 0
= 106 [(40 × 40 − 402 /2) − 0]
= 106 × 800 = 8 × 108 N
= 100 × 104 [40y −
that is the total force is 800 Mega Newtons.
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
14
Centre of Pressure
We wish to find the centre of pressure (xp , yp ) of a plane area immersed vertically in a fluid.
Take the x axis to be in the surface of the fluid and the y axis to be vertically down, so that
the plane Oxy contains the area.
O
surface
x
y
x
δA
y
We require the following results:
(a) The pressure p is proportional to the depth h, so that p = ωh where ω is a constant.
(b) The force F on an area δA subjected to constant pressure p is given by F = pδA
Consider a small element of area δA at the position shown. The pressure at δA is ωy. Then
the
force acting on δA is ωyδA. Hence the total force acting on the area A is A ωy dA = ω A y dA.
Moment of force on δA aboutOy = ωxyδA
Total moment of force on δA aboutOy = ω xy dA
A
Moment of force on δA aboutOx = ωy 2 δA
Total moment of force on δA aboutOx = ω y 2 dA
A
Taking moments about Oy:
total force × xp = total moment
ω y dA xp = ω xy dA
A
A
xp
y dA =
xy dA
A
A
Taking moments about Ox:
total force × yp = total moment
ω y dA yp = ω y 2 dA
A
A
y dA =
y 2 dA
yp
A
A
Hence
2
xy dA
y dA
A
xp = and yp = A
.
y dA
y dA
A
A
15
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Example A rectangle of sides a and b is immersed vertically in a fluid with one of its
edges in the surface as shown. Where is the centre of pressure?
O
surface
x
a
b
y
Solution
To express the surface integral as double integrals we will use cartesian coordinates and vertical
slices. We need the following integrals.
b
a
1 2
1 2
1
1 2
y dA =
y dydx =
dx =
y
a dx = a x = a2 b
2
2
2
2
A
0
0
o
0
0
a
b
b a
b
1 2
1
1
1 2
xy dA =
xy dydx =
dx =
xy
xa dx = x2 a2 = a2 b2
2
2
4
4
0
0
o
A
0
0
b
a
b a
b
1 3
1
1
1 3
y
a dx = a3 x = a3 b
y 2 dA =
y 2 dydx =
dx =
3
3
3
3
0
0
o
A
0
0
b
a
b
Hence
2
1 3
1 2 2
y
dA
xy dA
a
b
ab
2
1
xp = A
= 31 2 = a and yp = A
= 41 2 = b
3
2
y dA
y dA
ab
ab
A
A
2
2
The centre of pressure is ( 12 b, 23 a), so is at a depth of 23 a.
Areas and Moments
The surface integral A f (x, y) dA can represent a number of physical quantities, depending on
the function f (x, y) that is used.
(a) If f (x, y) = 1 then the integral represents the area of A.
(b) If f (x, y) = x then the integral represents the first moment of A about the y axis.
(c) If f (x, y) = y then the integral represents the first moment of A about the x axis.
(d) If f (x, y) = x2 then the integral represents the second moment of A about the y axis.
(e) If f (x, y) = y 2 then the integral represents the second moment of A about the x axis.
(f) If f (x, y) = x2 + y 2 then the integral represents the second moment of A about the
z axis.
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
16
As an example, imagine a rectangular lamina of length , width b, thickness t (small) and density
ρ (see figure below). Find the second moment of area of this lamina (moment of inertia) about
the y-axis.
y
b
density: ρ
thickness: t (small)
O
x
By (e) above, the moment of inertia is given by
b
y ρt dxdy = ρt( − 0)
3 b
y
= ρt
3 0
3
b
= ρt
3
2
0
0
b
y 2 dy
0
As the mass of the lamina is M = btρ, the moment of inertia simplifies to 13 M b2 . The t and ρ
are included in the integral to make it a moment of inertia rather than simply a second moment.
By a similar method, find the moment of inertia of the same lamina about the
x-axis.
Your solution
1
1
ρt3 b = M 2
3
3
17
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
Exercises
1. By making use of the form of the integrand, evaluate the following double integrals;
π1
(a) I = 0 0 y cos2 x dydx
3 1
(b) I = −8 −1 y 2 dydx
15
(c) I = 0 0 (s + 1)4 dtds
1.(a)
π
,
4
(b)
22
3
(c) 31
HELM (VERSION 1: March 18, 2004): Workbook Level 2
27.1: Introduction to Surface Integrals
18