Change of variables in multiple integrals

6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
y
dM = σb dx
b
dx
a
x
Figure 6.9 A uniform rectangular lamina of mass M with sides a and b can
be divided into vertical strips.
6.3.5 Mean values of functions
In chapter 2 we discussed average values for functions of a single variable. This
is easily extended to functions of several variables. Let us consider, for example,
a function f(x, y) defined in some region R of the xy-plane. Then the average
value f̄ of the function is given by
f(x, y) dA.
(6.10)
f̄ dA =
R
R
This definition is easily extended to three (and higher) dimensions; if a function
f(x, y, z) is defined in some three-dimensional region of space R then the average
value f̄ of the function is given by
f(x, y, z) dV .
(6.11)
f̄ dV =
R
R
A tetrahedron is bounded by the three coordinate surfaces and the plane x/a+y/b+z/c =
1 and has density ρ(x, y, z) = ρ0 (1 + x/a). Find the average value of the density.
From (6.11), the average value of the density is given by
ρ(x, y, z) dV .
ρ̄ dV =
R
R
Now the integral on the LHS is just the volume of the tetrahedron, which we found in
5
subsection 6.3.1 to be V = 16 abc, and the integral on the RHS is its mass M = 24
abcρ0 ,
calculated in subsection 6.3.2. Therefore ρ̄ = M/V = 54 ρ0 . 6.4 Change of variables in multiple integrals
It often happens that, either because of the form of the integrand involved or
because of the boundary shape of the region of integration, it is desirable to
199
MULTIPLE INTEGRALS
y
u = constant
v = constant
R
M
N
L
K
C
x
Figure 6.10 A region of integration R overlaid with a grid formed by the
family of curves u = constant and v = constant. The parallelogram KLMN
defines the area element dAuv .
express a multiple integral in terms of a new set of variables. We now consider
how to do this.
6.4.1 Change of variables in double integrals
Let us begin by examining the change of variables in a double integral. Suppose
that we require to change an integral
f(x, y) dx dy,
I=
R
in terms of coordinates x and y, into one expressed in new coordinates u and v,
given in terms of x and y by differentiable equations u = u(x, y) and v = v(x, y)
with inverses x = x(u, v) and y = y(u, v). The region R in the xy-plane and the
curve C that bounds it will become a new region R and a new boundary C in
the uv-plane, and so we must change the limits of integration accordingly. Also,
the function f(x, y) becomes a new function g(u, v) of the new coordinates.
Now the part of the integral that requires most consideration is the area element.
In the xy-plane the element is the rectangular area dAxy = dx dy generated by
constructing a grid of straight lines parallel to the x- and y- axes respectively.
Our task is to determine the corresponding area element in the uv-coordinates. In
general the corresponding element dAuv will not be the same shape as dAxy , but
this does not matter since all elements are infinitesimally small and the value of
the integrand is considered constant over them. Since the sides of the area element
are infinitesimal, dAuv will in general have the shape of a parallelogram. We can
find the connection between dAxy and dAuv by considering the grid formed by the
family of curves u = constant and v = constant, as shown in figure 6.10. Since v
200
6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
is constant along the line element KL, the latter has components (∂x/∂u) du and
(∂y/∂u) du in the directions of the x- and y-axes respectively. Similarly, since u
is constant along the line element KN, the latter has corresponding components
(∂x/∂v) dv and (∂y/∂v) dv. Using the result for the area of a parallelogram given
in chapter 7, we find that the area of the parallelogram KLMN is given by
∂x ∂y
∂x ∂y dv
du
dAuv = du dv −
∂u ∂v
∂v ∂u
∂x ∂y ∂x ∂y du dv.
= −
∂u ∂v
∂v ∂u Defining the Jacobian of x, y with respect to u, v as
J=
we have
∂(x, y)
∂x ∂y ∂x ∂y
≡
−
,
∂(u, v)
∂u ∂v
∂v ∂u
∂(x, y) du dv.
dAuv = ∂(u, v) The reader acquainted with determinants
be written as the 2 × 2 determinant
∂(x, y) =
J=
∂(u, v)
will notice that the Jacobian can also
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
.
Such determinants can be evaluated using the methods of chapter 8.
So, in summary, the relationship between the size of the area element generated
by dx, dy and the size of the corresponding area element generated by du, dv is
∂(x, y) du dv.
dx dy = ∂(u, v) This equality should be taken as meaning that when transforming from coordinates x, y to coordinates u, v, the area element dx dy should be replaced by the
expression on the RHS of the above equality. Of course, the Jacobian can, and
in general will, vary over the region of integration. We may express the double
integral in either coordinate system as
∂(x, y) du dv.
f(x, y) dx dy =
g(u, v) (6.12)
I=
∂(u, v) R
R
When evaluating the integral in the new coordinate system, it is usually advisable
to sketch the region of integration R in the uv-plane.
201
MULTIPLE INTEGRALS
Evaluate the double integral
I=
a+
x2 + y 2
dx dy,
R
where R is the region bounded by the circle x2 + y 2 = a2 .
In Cartesian coordinates, the integral may be written
a √a2 −x2
I=
dx √
dy a + x2 + y 2 ,
−a
− a2 −x2
and can be calculated directly. However, because of the circular boundary of the integration
region, a change of variables to plane polar coordinates ρ, φ is indicated. The relationship
between Cartesian and plane polar coordinates is given by x = ρ cos φ and y = ρ sin φ.
Using (6.12) we can therefore write
∂(x, y) dρ dφ,
I=
(a + ρ) ∂(ρ, φ) R
where R is the rectangular region in the ρφ-plane whose sides are ρ = 0, ρ = a, φ = 0
and φ = 2π. The Jacobian is easily calculated, and we obtain
cos φ
∂(x, y)
sin φ J=
= = ρ(cos2 φ + sin2 φ) = ρ.
−ρ sin φ ρ cos φ ∂(ρ, φ)
So the relationship between the area elements in Cartesian and in plane polar coordinates is
dx dy = ρ dρ dφ.
Therefore, when expressed in plane polar coordinates, the integral is given by
I=
(a + ρ)ρ dρ dφ
R
a
2
a
2π
5πa3
aρ
ρ3
dφ
dρ (a + ρ)ρ = 2π
=
=
+
.
2
3 0
3
0
0
6.4.2 Evaluation of the integral I =
∞
−∞
e−x dx
2
By making a judicious change of variables, it is sometimes possible to evaluate
an integral that would be intractable otherwise. An important example of this
method is provided by the evaluation of the integral
∞
2
e−x dx.
I=
−∞
Its value may be found by first constructing I 2 , as follows:
∞
∞
∞
∞
2
2
2
2
I2 =
e−x dx
e−y dy =
dx
dy e−(x +y )
−∞
−∞
−∞
−∞
2
2
=
e−(x +y ) dx dy,
R
202
6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
y
a
−a
a
x
−a
Figure 6.11 The
used to illustrate the convergence properties of the
a regions
2
integral I(a) = −a e−x dx as a → ∞.
where the region R is the whole xy-plane. Then, transforming to plane polar
coordinates, we find
2π
∞
2
2
2 ∞
e−ρ ρ dρ dφ =
dφ
dρ ρe−ρ = 2π − 12 e−ρ
= π.
I2 =
R
0
0
0
√
Therefore the original integral is given by I = π. Because the integrand is an
even function of x, it follows that the value of the integral from 0 to ∞ is simply
√
π/2.
We note, however, that unlike in all the previous examples, the regions of
integration R and R are both infinite in extent (i.e. unbounded). It is therefore
prudent to derive this result more rigorously; this we do by considering the
integral
a
2
e−x dx.
I(a) =
−a
We then have
e−(x +y ) dx dy,
2
I 2 (a) =
2
R
where R is the square of side 2a centred on the origin. Referring to figure 6.11,
since the integrand is always positive the value of the integral taken over the
square lies between the value of the integral taken over the region bounded by
the inner circle √
of radius a and the value of the integral taken over the outer
circle of radius 2a. Transforming to plane polar coordinates as above, we may
203
MULTIPLE INTEGRALS
z
R
T
v = c2
u = c1
S
P
Q
w = c3
C
y
x
Figure 6.12 A three-dimensional region of integration R, showing an element of volume in u, v, w coordinates formed by the coordinate surfaces
u = constant, v = constant, w = constant.
evaluate the integrals over the inner and outer circles respectively, and we find
2
2
π 1 − e−a < I 2 (a) < π 1 − e−2a .
√
Taking the limit a → ∞, we find I 2 (a) → π. Therefore I = π, as we found previ√
ously. Substituting
x = αy shows that the corresponding integral of exp(−αx2 )
has the value π/α. We use this result in the discussion of the normal distribution
in chapter 30.
6.4.3 Change of variables in triple integrals
A change of variable in a triple integral follows the same general lines as that for
a double integral. Suppose we wish to change variables from x, y, z to u, v, w.
In the x, y, z coordinates the element of volume is a cuboid of sides dx, dy, dz
and volume dVxyz = dx dy dz. If, however, we divide up the total volume into
infinitesimal elements by constructing a grid formed from the coordinate surfaces
u = constant, v = constant and w = constant, then the element of volume dVuvw
in the new coordinates will have the shape of a parallelepiped whose faces are the
coordinate surfaces and whose edges are the curves formed by the intersections of
these surfaces (see figure 6.12). Along the line element P Q the coordinates v and
204
6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
w are constant, and so P Q has components (∂x/∂u) du, (∂y/∂u) du and (∂z/∂u) du
in the directions of the x-, y- and z- axes respectively. The components of the
line elements P S and ST are found by replacing u by v and w respectively.
The expression for the volume of a parallelepiped in terms of the components
of its edges with respect to the x-, y- and z-axes is given in chapter 7. Using this,
we find that the element of volume in u, v, w coordinates is given by
∂(x, y, z) du dv dw,
dVuvw = ∂(u, v, w) where the Jacobian of x, y, z with respect to u, v, w is a short-hand for a 3 × 3
determinant:
∂x ∂y ∂z ∂u ∂u ∂u ∂(x, y, z) ∂x ∂y ∂z .
≡
∂(u, v, w) ∂v
∂v
∂v ∂x ∂y ∂z ∂w ∂w ∂w
So, in summary, the relationship between the elemental volumes in multiple
integrals formulated in the two coordinate systems is given in Jacobian form by
∂(x, y, z) du dv dw,
dx dy dz = ∂(u, v, w) and we can write a triple integral in either set of coordinates as
∂(x, y, z) du dv dw.
I=
f(x, y, z) dx dy dz =
g(u, v, w) ∂(u, v, w) R
R
Find an expression for a volume element in spherical polar coordinates, and hence calculate the moment of inertia about a diameter of a uniform sphere of radius a and mass M.
Spherical polar coordinates r, θ, φ are defined by
x = r sin θ cos φ,
y = r sin θ sin φ,
z = r cos θ
(and are discussed fully in chapter 10). The required Jacobian is therefore
sin θ sin φ
cos θ ∂(x, y, z) sin θ cos φ
J=
= r cos θ cos φ r cos θ sin φ −r sin θ .
∂(r, θ, φ) −r sin θ sin φ r sin θ cos φ
0
The determinant is most easily evaluated by expanding it with respect to the last column
(see chapter 8), which gives
J = cos θ(r2 sin θ cos θ) + r sin θ(r sin2 θ)
= r2 sin θ(cos2 θ + sin2 θ) = r 2 sin θ.
Therefore the volume element in spherical polar coordinates is given by
dV =
∂(x, y, z)
dr dθ dφ = r2 sin θ dr dθ dφ,
∂(r, θ, φ)
205
MULTIPLE INTEGRALS
which agrees with the result given in chapter 10.
If we place the sphere with its centre at the origin of an x, y, z coordinate system then
its moment of inertia about the z-axis (which is, of course, a diameter of the sphere) is
2
2
I=
x + y 2 dM = ρ
x + y 2 dV ,
where the integral is taken over the sphere, and ρ is the density. Using spherical polar
coordinates, we can write this as
2 2 2
I=ρ
r sin θ r sin θ dr dθ dφ
V
2π
π
0
0
= ρ × 2π ×
a
dθ sin3 θ
dφ
=ρ
4
3
dr r4
0
× 15 a5 =
8
πa5 ρ.
15
Since the mass of the sphere is M = 43 πa3 ρ, the moment of inertia can also be written as
I = 25 Ma2 . 6.4.4 General properties of Jacobians
Although we will not prove it, the general result for a change of coordinates in
an n-dimensional integral from a set xi to a set yj (where i and j both run from
1 to n) is
∂(x1 , x2 , . . . , xn ) dy1 dy2 · · · dyn ,
dx1 dx2 · · · dxn = ∂(y1 , y2 , . . . , yn ) where the n-dimensional Jacobian can be written as an n × n determinant (see
chapter 8) in an analogous way to the two- and three-dimensional cases.
For readers who already have sufficient familiarity with matrices (see chapter 8)
and their properties, a fairly compact proof of some useful general properties
of Jacobians can be given as follows. Other readers should turn straight to the
results (6.16) and (6.17) and return to the proof at some later time.
Consider three sets of variables xi , yi and zi , with i running from 1 to n for
each set. From the chain rule in partial differentiation (see (5.17)), we know that
∂xi ∂yk
∂xi
=
.
∂zj
∂yk ∂zj
n
(6.13)
k=1
Now let A, B and C be the matrices whose ijth elements are ∂xi /∂yj , ∂yi /∂zj and
∂xi /∂zj respectively. We can then write (6.13) as the matrix product
cij =
n
aik bkj
or
C = AB.
(6.14)
k=1
We may now use the general result for the determinant of the product of two
matrices, namely |AB| = |A||B|, and recall that the Jacobian
Jxy =
∂(x1 , . . . , xn )
= |A|,
∂(y1 , . . . , yn )
206
(6.15)