Double integration

1. Double integrals
Notice: this material must not
be used as a substitute for attending
the lectures
1

2. 0.1 What is a double integral?
Recall that asingle integralis something of the form
Z
b
a
f(x)dx
Adouble integralis something of the form
Z Z
R
f(x; y)dx dy
whereRis called theregion of integrationand is a region in the (x; y) plane. The
double integral gives us the volume under the surfacez=f(x; y), just as a single
integral gives the area under a curve.
0.2 Evaluation of double integrals
To evaluate a double integral we do it in stages, starting from the inside and working
out, using our knowledge of the methods for single integrals. The easiest kind of
regionRto work with is a rectangle. To evaluate
Z Z
R
f(x; y)dx dy
proceed as follows:
work out the limits of integration if they are not already known
work out the inner integral for a typicaly
work out the outer integral
0.3 Example
Evaluate
Z
2
y=1
Z
3
x=0
(1 + 8xy)dx dy
Solution.In this example the \inner integral" is
R
3
x=0
(1 + 8xy)dxwithytreated as a
constant.
integral =
Z
2
y=1
0
B
B
B
B
@
Z
3
x=0
(1 + 8xy)dx
|
{z}
work out treatingyas constant
1
C
C
C
C
A
dy
=
Z
2
y=1
"
x+
8x
2
y
2
#
3
x=0
dy
=
Z
2
y=1
(3 + 36y)dy
2

3. =
"
3y+
36y
2
2
#
2
y=1
= (6 + 72)(3 + 18)
= 57
0.4 Example
Evaluate
Z
=2
0
Z
1
0
ysinx dy dx
Solution.
integral =
Z
=2
0
Z
1
0
ysinx dy

dx
=
Z
=2
0
"
y
2
2
sinx
#
1
y=0
dx
=
Z
=2
0
1
2
sinx dx
=
h

1
2
cosx
i
=2
x=0
=
1
2
0.5 Example
Find the volume of the solid bounded above by the planez= 4xyand below
by the rectangleR=f(x; y) : 0x1 0y2g.
Solution.The volume under any surfacez=f(x; y) and above a regionRis given by
V=
Z Z
R
f(x; y)dx dy
In our case
V=
Z
2
0
Z
1
0
(4xy)dx dy
=
Z
2
0
h
4x
1
2
x
2
yx
i
1
x=0
dy=
Z
2
0
(4
1
2
y)dy
=
"
7y
2

y
2
2
#
2
y=0
= (72)(0) = 5
The double integrals in the above examples are the easiest types to evaluate because
they are examples in which all four limits of integration are constants. This happens
when the region of integration is rectangular in shape. In non-rectangular regions
of integration the limits are not all constant so we have to get used to dealing with
non-constant limits. We do this in the next few examples.
3

4. 0.6 Example
Evaluate
Z
2
0
Z
x
x
2
y
2
x dy dx
Solution.
integral =
Z
2
0
Z
x
x
2
y
2
x dy dx
=
Z
2
0
"
y
3
x
3
#
y=x
y=x
2
dx
=
Z
2
0

x
4
3

x
7
3
!
dx=
"
x
5
15

x
8
24
#
2
0
=
32
15

256
24
=
128
15
0.7 Example
Evaluate
Z

=2
Z
x
2
0
1
x
cos
y
x
dy dx
Solution.Recall from elementary calculus the integral
R
cosmy dy=
1
m
sinmyform
independent ofy. Using this result,
integral =
Z

=2
2
4
1
x
sin
y
x1x
3
5
y=x
2
y=0
dx
=
Z

=2
sinx dx= [cosx]

x==2= 1
0.8 Example
Evaluate
Z
4
1
Zp
y
0
e
x=
py
dx dy
Solution.
integral =
Z
4
1
"
e
x=
p
y
1=
p
y
#
x=
p
y
x=0
dy
=
Z
4
1
(
p
ye
p
y)dy= (e1)
Z
4
1
y
1=2
dy
= (e1)
"
y
3=2
3=2
#
4
y=1
=
2
3
(e1)(81)
=
14
3
(e1)
4

5. 0.9 Evaluating the limits of integration
When evaluating double integrals it is very commonnotto be told the limits of
integration but simply told that the integral is to be taken over a certain specied
regionRin the (x; y) plane. In this case you need to work out the limits of integration
for yourself. Great care has to be taken in carrying out this task. The integration
can in principle be done in two ways: (i) integrating rst with respect toxand then
with respect toy, or (ii) rst with respect toyand then with respect tox. The
limits of integration in the two approaches will in general be quite dierent, but both
approaches must yield the same answer. Sometimes one way round is considerably
harder than the other, and in some integrals one way works ne while the other leads
to an integral that cannot be evaluated using the simple methods you have been
taught. There are no simple rules for deciding which order to do the integration in.
0.10 Example
Evaluate Z Z
D
(3xy)dA [dAmeansdxdyordydx]
whereDis the triangle in the (x; y) plane bounded by thex-axis and the linesy=x
andx= 1.
Solution.A good diagram is essential.
Method 1 : do the integration with respect toxrst. In this approach we select a typical
yvalue which is (for the moment) considered xed, and we draw ahorizontal
line across the regionD; this horizontal line intersects theyaxis at the typical
yvalue. Find out the values ofx(they will depend ony) where the horizontal
lineentersandleavesthe regionD(in this problem it enters atx=yand
leaves atx= 1). These values ofxwill be the limits of integration for the inner
integral. Then you determine what valuesyhas to range between so that the
horizontal line sweeps the entire regionD(in this caseyhas to go from 0 to 1).
This determines the limits of integration for the outer integral, the integral with
respect toy. For this particular problem the integral becomes
Z Z
D
(3xy)dA=
Z
1
0
Z
1
y
(3xy)dx dy
=
Z
1
0
"
3x
x
2
2
yx
#
x=1
x=y
dy
=
Z
1
0


3
1
2
y



3y
y
2
2
y
2
!!
dy
=
Z
1
0

5
2
4y+
3
2
y
2

dy=
"
5y
2
2y
2
+
y
3
2
#
y=1
y=0
=
5
2
2 +
1
2
= 1
5

6. Method 2 : do the integration with respect toyrst and thenx. In this approach we
select a ypicalx" and draw a vertical line across the regionDat that value
ofx.
Vertical line entersDaty= 0 and leaves aty=x. We then need to letx
go from 0 to 1 so that the vertical line sweeps the entire region. The integral
becomes
Z Z
D
(3xy)dA=
Z
1
0
Z
x
0
(3xy)dy dx
=
Z
1
0
"
3yxy
y
2
2
#
y=x
y=0
dx
=
Z
1
0

3xx
2

x
2
2
!
dx=
Z
1
0

3x
3x
2
2
!
dx
=
"
3x
2
2

x
3
2
#
1
x=0
= 1
Note that Methods 1 and 2 give the same answer. If they don't it means something
is wrong.
0.11 Example
Evaluate Z Z
D
(4x+ 2)dA
whereDis the region enclosed by the curvesy=x
2
andy= 2x.
Solution.Again we will carry out the integration both ways,xrst theny, and then
vice versa, to ensure the same answer is obtained by both methods.
Method 1 : We do the integration rst with respect toxand then with respect toy. We
shall need to know where the two curvesy=x
2
andy= 2xintersect. They
intersect whenx
2
= 2x, i.e. whenx= 0;2. So they intersect at the points (0;0)
and (2;4).
For a typicaly, the horizontal line will enterDatx=y=2 and leave atx=
p
y.
Then we need to letygo from 0 to 4 so that the horizontal line sweeps the
entire region. Thus
Z Z
D
(4x+ 2)dA=
Z
4
0
Z
x=
p
y
x=y=2
(4x+ 2)dx dy
=
Z
4
0
h
2x
2
+ 2x
i
x=
p
y
x=y=2
dy=
Z
4
0

(2y+ 2
py)

y
2
2
+y
!!
dy
=
Z
4
0

y+ 2y
1=2

y
2
2
!
dy=
"
y
2
2
+
2y
3=2
3=2

y
3
6
#
4
0
= 8
6

7. Method 2 : Integrate rst with respect toyand thenx, i.e. draw a vertical line acrossD
at a typicalxvalue. Such a line entersDaty=x
2
and leaves aty= 2x. The
integral becomes
Z Z
D
(4x+ 2)dA=
Z
2
0
Z
2x
x
2
(4x+ 2)dy dx
=
Z
2
0
[4xy+ 2y]
y=2x
y=x
2dx
=
Z
2
0

8x
2
+ 4x



4x
3
+ 2x
2

dx
=
Z
2
0
(6x
2
4x
3
+ 4x)dx=
h
2x
3
x
4
+ 2x
2
i
2
0
= 8
The example we have just done shows that it is sometimes easier to do it one way
than the other. The next example shows that sometimes the dierence in eort is
more considerable. There is no general rule saying that one way is always easier than
the other; it depends on the individual integral.
0.12 Example
Evaluate Z Z
D
(xyy
3
)dA
whereDis the region consisting of the squaref(x; y) :1x0;0y1g
together with the trianglef(x; y) :xy1;0x1g.
Method 1 : (easy). integrate with respect toxrst. A diagram will show thatxgoes
from1 toy, and thenygoes from 0 to 1. The integral becomes
Z Z
D
(xyy
3
)dA=
Z
1
0
Z
y
1
(xyy
3
)dx dy
=
Z
1
0
"
x
2
2
yxy
3
#
x=y
x=1
dy
=
Z
1
0

y
3
2
y
4
!
(
1
2
y+y
3
)
!
dy
=
Z
1
0


y
3
2
y
4

1
2
y
!
dy=
"

y
4
8

y
5
5

y
2
4
#
1
y=0
=
23
40
Method 2 : (harder). It is necessary to break the region of integrationDinto two sub-
regionsD1(the square part) andD2(triangular part). The integral overDis
given by
Z Z
D
(xyy
3
)dA=
Z Z
D1
(xyy
3
)dA+
Z Z
D2
(xyy
3
)dA
7

8. which is the analogy of the formula
R
c
a
f(x)dx=
R
b
a
f(x)dx+
R
c
b
f(x)dxfor
single integrals. Thus
Z Z
D
(xyy
3
)dA=
Z
0
1
Z
1
0
(xyy
3
)dy dx+
Z
1
0
Z
1
x
(xyy
3
)dy dx
=
Z
0
1
"
xy
2
2

y
4
4
#
1
y=0
dx+
Z
1
0
"
xy
2
2

y
4
4
#
1
y=x
dx
=
Z
0
1

1
2
x
1
4

dx+
Z
1
0


x
2

1
4



x
3
2

x
4
4
!!
dx
=
"
x
2
4

x
4
#
0
1
+
"
x
2
4

x
4

x
4
8
+
x
5
20
#
1
0
=
1
2

3
40
=
23
40
In the next example the integration can only be done one way round.
0.13 Example
Evaluate
Z Z
D
sinx
x
dA
whereDis the trianglef(x; y) : 0yx;0xg.
Solution.Let's try doing the integration rst with respect toxand theny. This gives
Z Z
D
sinx
x
dA=
Z

0
Z

y
sinx
x
dx dy
but we cannot proceed because we cannot nd an indenite integral for sinx=x. So,
let's try doing it the other way. We then have
Z Z
D
sinx
x
dA=
Z

0
Z
x
0
sinx
x
dy dx
=
Z

0

sinx
x
y
x
y=0
dx=
Z

0
sinx dx
= [cosx]

0
= 1(1) = 2
0.14 Example
Find the volume of the tetrahedron that lies in the rst octant and is bounded by the
three coordinate planes and the planez= 52xy.
Solution.The given plane intersects the coordinate axes at the points (
5
2
;0;0), (0;5;0)
and (0;0;5). Thus, we need to work out the double integral
Z Z
D
(52xy)dA
8

9. whereDis the triangle in the (x; y) plane with vertices (x; y) = (0;0), (
5
2
;0) and
(0;5). It is a good idea to draw another diagram at this stage showing just the region
Din the (x; y) plane. Note that the equation of the line joining the points (
5
2
;0) and
(0;5) isy=2x+ 5. Then:
volume =
Z Z
D
(52xy)dA=
Z
5
0
Z
(5y)=2
0
(52xy)dx dy
=
Z
5
0
h
5xx
2
yx
i
x=(5y)=2
x=0
dy
=
Z
5
0
"
5

5y
2



5y
2
2
y

5y
2

#
dy
=
Z
5
0

25
4

5y
2
+
y
2
4
!
dy
=
"
25y
4

5y
2
4
+
y
3
12
#
5
0
=
125
12
0.15 Changing variables in a double integral
We know how to change variables in asingleintegral:
Z
b
a
f(x)dx=
Z
B
A
f(x(u))
dx
du
du
whereAandBare the new limits of integration.
Fordouble integralsthe rule is more complicated. Suppose we have
Z Z
D
f(x; y)dx dy
and want to change the variables touandvgiven byx=x(u; v),y=y(u; v). The
change of variables formula is
Z Z
D
f(x; y)dx dy=
Z Z
D

f(x(u; v); y(u; v))jJjdu dv (0.1)
whereJis the Jacobian, given by
J=
@x
@u
@y
@v

@x
@v
@y
@u
andD

is the new region of integration, in the (u; v) plane.
0.16 Transforming a double integral into polars
A very commonly used substitution is conversion into polars. This substitution is
particularly suitable when the region of integrationDis a circle or an annulus (i.e.
region between two concentric circles). Polar coordinatesrandare dened by
x=rcos; y=rsin
9

10. The variablesuandvin the general description above arerandin the polar
coordinates context and the Jacobian for polar coordinates is
J=
@x
@r
@y
@

@x
@
@y
@r
= (cos)(rcos)(rsin)(sin)
=r(cos
2
+ sin
2
) =r
SojJj=rand the change of variables rule (0.1) becomes
Z Z
D
f(x; y)dx dy=
Z Z
D

f(rcos; rsin)r dr d
0.17 Example
Use polar coordinates to evaluate
Z Z
D
xy dx dy
whereDis the portion of the circle centre 0, radius 1, that lies in the rst quadrant.
Solution.For the portion in the rst quadrant we need 0r1 and 0=2.
These inequalities give us the limits of integration in therandvariables, and these
limits will all be constants.
Withx=rcos,y=rsinthe integral becomes
Z Z
D
xy dx dy=
Z
=2
0
Z
1
0
r
2
cossin r dr d
=
Z
=2
0
"
r
4
4
cossin
#
1
r=0
d
=
Z
=2
0
1
4
sincos d=
Z
=2
0
1
8
sin 2 d
=
1
8
"

cos 2
2
#
=2
0
=
1
8
0.18 Example
Evaluate Z Z
D
e
(x
2
+y
2
)
dx dy
whereDis the region between the two circlesx
2
+y
2
= 1 andx
2
+y
2
= 4.
Solution.It is not feasible to attempt this integral by any method other than trans-
forming into polars.
Letx=rcos,y=rsin. In terms ofrandthe regionDbetween the two circles
is described by 1r2, 02, and so the integral becomes
Z Z
D
e
(x
2
+y
2
)
dx dy=
Z
2
0
Z
2
1
e
r
2
r dr d
10

11. =
Z
2
0
h

1
2
e
r
2
i
2
r=1
d
=
Z
2
0


1
2
e
4
+
1
2
e
1

d
=(e
1
e
4
)
0.19 Example: integratinge
x
2
The functione
x
2
has no elementary antiderivative. But we can evaluate
R
1
1
e
x
2
dx
by using the theory of double integrals.
Z
1
1
e
x
2
dx

2
=
Z
1
1
e
x
2
dx
Z
1
1
e
x
2
dx

=
Z
1
1
e
x
2
dx
Z
1
1
e
y
2
dy

=
Z
1
1
e
y
2
Z
1
1
e
x
2
dx dy
=
Z
1
1
Z
1
1
e
(x
2
+y
2
)
dx dy
Now transform to polar coordinatesx=rcos,y=rsin. The region of integration
is the whole (x; y) plane. In polar variables this is given by 0r <1, 02.
Thus
Z
1
1
e
x
2
dx

2
=
Z
1
1
Z
1
1
e
(x
2
+y
2
)
dx dy
=
Z
2
0
Z
1
0
e
r
2
r dr d
=
Z
2
0
h

1
2
e
r
2
i
r=1
r=0
d
=
Z
2
0
1
2
d=
We have shown that
Z
1
1
e
x
2
dx

2
=
Hence Z
1
1
e
x
2
dx=
p
:
The above integral is very important in numerous applications.
0.20 Other substitutions
So far we have only illustrated how to convert a double integral into polars. We
will now illustrate some examples of double integrals that can be evaluated by other
substitutions. Unlike single integrals, for a double integral the choice of substitution
is often dictated not only by what we have in the integrand but also by the shape of
the region of integration.
11

12. 0.21 Example
Evaluate Z Z
D
(x+y)
2
dx dy
whereDis the parallelogram bounded by the linesx+y= 0,x+y= 1, 2xy= 0
and 2xy= 3.
Solution.(A diagram to show the regionDwill be useful).
In an example like this the boundary curves ofDcan suggest what substitution to
use. So let us try
u=x+y; v= 2xy:
In these new variables the regionDis described by
0u1;0v3:
We need to work out the Jacobian
J=





@x
@u
@x
@v
@y
@u
@y
@v





To work this out we needxandyin terms ofuandv. From the equationsu=x+y,
v= 2xywe get
x=
1
3
(u+v); y=
2
3
u
1
3
v
Therefore
J=





1
3
1
3
2
3

1
3





=
1
9

2
9
=
1
3
and sojJj=
1
3
(recall it isjJjand notJthat we put into the integral). Therefore the
substitution formula gives
Z Z
D
(x+y)
2
dx dy=
Z
3
0
Z
1
0
u
2
1
3
|{z}
=jJj
du dv=
Z
3
0
"
u
3
9
#
1
0
dv=
Z
3
0
1
9
dv=
1
3
:
0.22 Example
LetDbe the region in the rst quadrant bounded by the hyperbolasxy= 1,xy= 9
and the linesy=x,y= 4x. Evaluate
Z Z
D
r
y
x
+
p
xy

dx dy
Solution.A diagram showingDis useful. We make the substitution
xy=u
2
;
y
x
=v
2
:
12

13. We will needxandyin terms ofuandv. By multiplying the above equations we
gety
2
=u
2
v
2
. Hencey=uvandx=u=v. In the (u; v) variables the regionDis
described by
1u3;1v2:
The Jacobian is
J=





@x
@u
@x
@v
@y
@u
@y
@v





=





1
v

u
v
2
v u





=
u
v
+
u
v
=
2u
v
Therefore
Z Z
D
r
y
x
+
p
xy

dx dy
=
Z Z
(v+u)jJjdu dv=
Z
2
1
Z
3
1
(v+u)

2u
v

du dv
=
Z
2
1
Z
3
1

2u+
2u
2
v
!
du dv=
Z
2
1
"
u
2
+
2u
3
3v
#
u=3
u=1
dv
=
Z
2
1

9 +
18
v



1 +
2
3v

dv=

8v+
52
3
lnv
2
1
= 8 +
52
3
ln 2:
0.23 Application of double integrals: centres of gravity
We will show how double integrals may be used to nd the location of the centre
of gravity of a two-dimensional object. Mathematically speaking, aplateis a thin
2-dimensional distribution of matter considered as a subset of the (x; y) plane. Let
= mass per unit area
This is the denition ofdensityfor two-dimensional objects. If the plate is all made
of the same material (a sheet of metal, perhaps) thenwould be a constant, the value
of which would depend on the material of which the plate is made. However, if the
plate is not all made of the same material thencould vary from point to point on
the plate and therefore be a function ofxandy,(x; y). For some objects, part of the
object may be made of one material and part of it another (some currencies have coins
that are like this). But(x; y) could quite easily vary in a much more complicated
way (a pizza is a simple example of an object with an uneven distribution of matter).
The intersection of the two thin strips denes a small rectangle of lengthxand width
y. Thus
mass of little rectangle = (mass per unit area)(area)
=(x; y)dx dy
Therefore the total mass of the plateDis
M=
Z Z
D
(x; y)dx dy:
Suppose you try to balance the plateDon a pin. Thecentre of massof the plate is
the point where you would need to put the pin. It can be shown that the coordinates
13

14. (x;y) of the centre of mass are given by
x=
Z Z
D
x (x; y)dA
Z Z
D
(x; y)dA
;y=
Z Z
D
y (x; y)dA
Z Z
D
(x; y)dA
(0.2)
0.24 Example
A homogeneous triangle with vertices (0;0), (1;0) and (1;3). Find the coordinates of
its centre of mass.
[`Homogeneous' means the plate is all made of the same material which is uniformly
distributed across it, so that(x; y) =, a constant.]
Solution.A diagram of the triangle would be useful. Withconstant, we have
x=
Z Z
D
x dA
Z Z
D
dA
=

Z
1
0
Z
3x
0
x dy dx

Z
1
0
Z
3x
0
dy dx
=
Z
1
0
[xy]
y=3x
y=0
dx
Z
1
0
[y]
y=3x
y=0
dx
=
Z
1
0
3x
2
dx
Z
1
0
3x dx
=
1
3=2
=
2
3
and
y=
Z Z
D
y dA
Z Z
D
dA
=

Z
1
0
Z
3x
0
y dy dx

Z
1
0
Z
3x
0
dy dx
=
Z
1
0
"
y
2
2
#
y=3x
y=0
dxZ
1
0
[y]
y=3x
y=0
dx
=
Z
1
0
9x
2
2
dxZ
1
0
3x dx
=
3=2
3=2
= 1:
So the centre of mass is at (x;y) = (
2
3
;1).
0.25 Example
Find the centre of mass of a circle, centre the origin, radius 1, if the right half is made
of material twice as heavy as the left half.
Solution.By symmetry, it is clear that the centre of mass will be somewhere on the
x-axis, and so y= 0. In order to model the fact that the right half is twice as heavy,
we can take
(x; y) =
(
2 x >0
x <0
with thein the right hand side of the above expression being any positive constant.
14

15. From the general formula,
x=
Z Z
D
x (x; y)dA
Z Z
D
(x; y)dA
: (0.3)
Let us work out the integral in the numerator rst. We shall need to break it up as
follows
Z Z
D
x(x; y)dA=
Z Z
right half
+
Z Z
left half
=
Z Z
right
2x dA+
Z Z
left
x dA
The circular geometry suggests we convert to plane polars,x=rcos,y=rsin.
Recall that, in this coordinate system,dA=r dr d. The right half of the circle
is described by=2=2, 0r1, and the left half similarly but with
=23=2. Thus
Z Z
D
x(x; y)dA=
Z
=2
=2
Z
1
0
2(rcos)r dr d+
Z
3=2
=2
Z
1
0
(rcos)r dr d
= 2
Z
=2
=2
"
r
3
3
cos
#
r=1
r=0
d+
Z
3=2
=2
"
r
3
3
cos
#
r=1
r=0
d
=
2
3
Z
=2
=2
cos d+

3
Z
3=2
=2
cos d
=
4
3

2
3
=
2
3
:
Finally, we work out the denominator in (0.3):
Z Z
D
(x; y)dA=
Z Z
left half
dA+
Z Z
right half
2 dA
=
Z Z
left half
dA+ 2
Z Z
right half
dA
=(area of left half) + 2(area of right half)
=(=2) + 2(=2)
=
3
2
Therefore thexcoordinate of the centre of mass of the object is
x=
2=3
3=2
=
4
9
:
15