Multiple integrals; vectors
4
4.1
Multiple integrals
Let’s review this subject by doing various examples of integrating a function f (x, y)
over a region of 2-space:
Ex. 1:
Z
Z
√
I=
4
y xdydx =
region
√
dx x
Z
√
Z
x
4
y dy =
0
0
dx
0
y
√
1
32
x· x=
2
5
(1)
y=x1/2
2
4 x
Figure 1: Ex. 1
Q: how about changing the order of integration? We could do the x-integral
first obviously. But we need to be careful of the order of the limits.:
Z
Z
2
4
√
dy y
I=
x dx =
x=y 2
y=0
32
.
5
(2)
Q: Why change the order? Sometimes it’s important:
Ex. 2:
Z
Z
ln 16
I=
dx
y=ex/2
x=0
Can’t do
R
4
dy
ln y
(3)
dy/ ln y, so switch:
(4)
1
y
4
y=ex/2
x
Figure 2: Ex. 1
So let’s draw a new figure (always draw a figure if you switch!).
Z 4
Z 2 ln y
Z 4
dy
dx =
dy · 2 = 6
I=
y=1
y=1 ln y 0
4.2
(5)
Change of variables: the Jacobian
First, let’s do a standard example where we don’t get into formalities:
y
y= 1-x2
1
rdθ
dr
dθ
1 x
Figure 3: Ex. 1
Ex. 3:
Z
1
Z
√
1−x2
I=
x=0
e−(x
y=0
2
2
+y 2 )
dydx
(6)
This will certainly be easier in polar coordinates x = r cos θ, y = r sin θ,
Z π/2 Z 1
π
2
I=
e−r rdrdθ = (1 − 1/e).
4
θ=0
r=0
(7)
Note the measure for the integral in polar coords.,
dx dy = r dr dθ,
(8)
is just the size of the little area element in Fig. 3, but can be obtained formally.
In general the change of variables in 2D is defined
¯ µ
¶¯
Z Z
Z Z
¯
¯
u,
v
¯ dr ds,
f (u, v)du dv =
f (u(r, s), v(r, s)) · ¯¯J
(9)
r, s ¯
R
R0
where R is a region in 2D and R0 is the transformed region, which need not have
the same shape. J is the “Jacobian of transformation”
µ
¶ ¯ ∂u ∂u ¯
¯
¯
u, v
1
∂r ∂s ¯ =
³ ´,
(10)
J
= ¯¯ ∂v
∂v ¯
r,s
r, s
∂r ∂s
J u,v
i.e. the determinant of the partial derivatives as shown. If you have forgotten
how to take a determinant, go ahead and look it up in Boas p. 89. Of course the
concept can be generalized to higher-D transformations (x, y, z...) → (u, v, w...),
but the determinental form remains, one just has to do a bit more work. Let’s
specialize now to polar coordinates. Take u = x = r cos θ, v = y = r sin θ, so
¯
µ
¶ ¯
¯ cos θ −r sin θ ¯
x, y
¯ = r(cos2 θ + sin2 θ) = r,
J
= ¯¯
(11)
sin θ r cos θ ¯
r, θ
so dxdy = rdrdθ, what we worked out already in more intuitive fashion.
Here’s a slightly tricky example where the shape of the transformed region is
quite different.
Ex. 5:
Z
Z
1
I=
x
dx
0
0
(x + y)ex+y
dy
x2
(12)
Let’s make a transformation x, y → u, v where u = y/x, v = x + y. The choice
of v is somewhat straightforward; why u is chosen I have no idea. Anyway, 1st task
3
Figure 4: Ex. 5. Original and transformed regions.
is to calculate Jacobian:
¯
¶
µ
¯ −y/x2 1/x
u, v
= ¯¯
J
1
1
x, y
¯ µ
¶¯
¯
¯
¯J x, y ¯ =
⇒
¯
u, v ¯
So
¯
¯
¯ = −x + y
¯
x2
(13)
x2
.
x+y
(14)
Z Z
Z Z
vev x2 (v)
I=
dudv =
ev dudv,
(15)
x2 (v) v
nice and simple, but what are the transformed limits? Well, starting from the
original ones, we learn
y = 0 ⇒ u = 0 if x 6= 0 ; y = x ⇒ u = 1
x = 1 ⇒ v = 1 + u ; y = x = 0 ⇒ v = 0.
So
Z
I=
Z
1
1+u
du
0
Z
1
v
dv e =
0
du(e1+u − 1) = e2 − e − 1.
(16)
(17)
(18)
0
measures for cylindrical and spherical coords.: (Derived in Boas)–check!
Spherical x = r sin θ cos φ ; y = r sin θ sin φ ; z = r cos θ:
¯ µ
¶¯
¯
¯
x,
y,
z
¯J
¯ = r2 sin θ
¯
r, θ, φ ¯
i.e. dxdydz → r2 sin θdrdθdφ
4
(19)
(20)
Cylindrical x = ρ cos θ ; y = ρ sin θ; z = z :
¯ µ
¶¯
¯
¯
x,
y,
z
¯J
¯=ρ
¯
ρ, θ, z ¯
i.e. dxdydz → ρdρdθdz
(21)
(22)
(N.B. θ in spherical coords. is not the same θ as for cylindrical coords!
Might want to use φ for cylindrical coords instead.)
Ex. 6:
Calculate the moment of inertia of a cone of height equal to its base radius
h = R. Take the density of the material to be ρ0 , assumed homogeneous. Moment
of inertia is then
Z Z Z
Z h
Z z
Z 2π
πh5
.
(23)
I=
ρ0 dV (x2 + y 2 ) = ρ0
dz
rdr
dθ r2 = ρ0
10
V
z=0
r=0
0
Note dimensions are correct, since [ρ0 ] = M/L3 , so [ρ0 h5 ] = M L2 .
4.3
4.3.1
Vectors
Properties of vectors
Vector: “set of components which transforms under rotation of a coordinate system
in the same way as the coordinates of a point ~r. ” (“A vector is something that
transforms like a vector”.) Huh? What does that mean? Take the coordinates
x1 , x2 of a point ~r in a Cartesian coordinate system.
Figure 5: Transformation of coordinates x1 , x2 → x01 , x02 .
If we rotate the coordinate axes to a new set x01 , x02 by an angle φ, geometry tells
5
us
x01 = cos φ x1 + sin φ x2
| {z }
|{z}
a11
a12
x02 = − sin φ x1 + cos φ x2
| {z }
| {z }
a21
a22 ,
(24)
(25)
(26)
(27)
so we can write this as a general linear transformation
x0i
= ai1 x1 + ai2 x2 =
2
X
aij xj .
(28)
j=1
What happens to the vector ~r under this transformation?
~r = x1 ê1 + x2 ê2
= x01 ê01 + x02 ê02
2
2
02
x02
1 + x2 = x1 + x2
⇒
a211 + a221 = 1 ; a212 + a222 = 1
a11 a12 + a21 a22 = 0.
(29)
(30)
(31)
(32)
(33)
(34)
(35)
The length of the vector is preserved in the new coordinate system; the transformation is said to be orthogonal. We can write the same thing in a nice compact
notation using the ”Kronecker delta” symbol
½
1 i=j
δij =
(36)
0 i 6= j
as
X
X
aij aik = δjk =
aji aki .
(37)
i
i
If you find this confusing, write this out in terms of indices and see what it means.
You may also wish to ask yourself how it’s related to matrix multiplication.
Inverse transformation: (φ → −φ):
∂xj
∂x0i
=
aij =
∂xj
∂x0i
(Why is this true?)
~ = (A1 , A2 . . . ):
Some properties of general N-D vectors A
6
(38)
~=B
~ ⇒ A i = Bi
• A
~+B
~ =C
~ ⇒ Ai + Bi = Ci
• A
~ = (aA1 , aA2 , . . . aAN )
• aA
~ = (−A1 , −A2 , · · · − aAN )
• −A
~+B
~ =B
~ +A
~
• A
~ + B)
~ = aA
~ + bB
~
• a(A
etc.
Magnitude of a vector:
2
A
=
N
X
A2i
=
N
X
Ai02
check!
i=1
i=1
√
~ =
|A|
A2 = A
4.3.2
(39)
(40)
Products of vectors
1) Dot or scalar product:
~·B
~ =
A
X
A i Bi
(41)
²ijk Aj Bk ,
(42)
i
2) Cross product
~ × B)
~ i=
(A
X
jk
where ²ijk is the so-called Levi-Civita symbol, sometimes called the completely
antisymmetric tensor:

if any two indices are the same
 0
²ijk =
+1 if indices correspond to an even permutation of 123
(43)

−1 if indices correspond to an odd permutation of 123
Very useful identity – worth memorizing!
X
²ijk ²lmk = δi` δjm − δim δj`
k
7
(44)
Ex 1:
~ × B)
~ · (C
~ × D)
~ = ?
(A
(45)
X
XX
X
~
~
~
~
²ijk Aj Bk ·
²i`m C` Dm
(A × B)i (C × D)i =
i
i
=
X
jk
`m
(²ijk ²i`m )Aj Bk C` Dm
ijk`m
=
X
(δj` δkm − δjm δk` )Aj Bk C` Dm
jk`m
=
X
(A` Bm C` Dm − Am B` C` Dm )
`m
= (
X
A` C` )(
X
X
X
Bm Cm ) − (
B` C` )(
Am Dm )
m
`
m
`
~ · C)(
~ B
~ · D)
~ − (B
~ · C)(
~ A
~ · D)
~
= (A
(46)
Some more identities to check using these techniques:
~ · (B
~ × C)
~ =B
~ · (C
~ × A)
~ = (A
~ × B)
~ ·C
~
• A
~ × (B
~ × C)
~ = B(
~ A
~ · C)
~ − C(
~ A
~ · B)
~
• A
(“ BAC -CAB rule”)
Remarks:
~·B
~ = AB cos θAB . If A
~·B
~ = 0, two vectors are “orthogonal”,
• 2 or 3d A
θAB = π/2.
• unit vectors ê1 , ê2 , . . . . We can choose mutually orthogonal, êi · êj = δij . Also
note êi × êj = ²ijk êk .
~ in terms of D orthogonal unit vectors in D dimensions,
• Expand
any vector A
P
~=
A
i Ai êi .
• The scalar or dot product of two vectors is invariant under coordinate transformations:
X
XX
X
X
~0 · B
~0 =
A
A0i Bi0 =
(
aij Ai )(
aik Bk ) =
aij aik Aj Bk
i
=
X
i
δjk Aj Bk =
jk
j
X
k
~·B
~
Ak Bk = A
k
~ × B|
~ = AB sin θAB
• |A
8
ijk
(47)
~ · (B
~ × C)|:
~
~ B,
~
• meaning of |A
volume of parallelogram spanned by vectors A,
~
and C.
9