#10 (09/25/2023)

Curves, Surfaces and Volumes

If the position vector, R, is expressed by one parameter, u, a set of all the points of R varying u forms a curved line. If R is expressed by two parameters, u and v, as R(u,v), a set of all the points of R by varying both u and v forms a curves surface. (obvious statement for R(u,v,w)).

Arc length

linelength.jpg
The length of a curved line from u = u to u = u + du can be computed by carrying out integration of ds (line segment)

a
=
R(u + d u ) − R(u)
=
R(u) + d R
du
 du + …− R(u)
 ∼ 
dR
d u
 du
=
eu d u,
(1)
where
eu ≡ dR
d u
 .
Therefore,

ds
=

√
 
a ·a
 
=

√
 
eu d u ·eu d u
 
=

√
 
eu ·eu
 
 d u.
(2)
and

s = ⌠
⌡ 		u=u1

u=u0 
√
 
eu ·eu
 
 d u.
(3)
Example Compute the length of y = x2 from x = 0 to x = 1.
lengthexample.jpg
Solution As
R = ⎛
⎜
⎜
⎜
⎝
x
y
z
⎞
⎟
⎟
⎟
⎠ 		= ⎛
⎜
⎜
⎜
⎝
u
u2
0
⎞
⎟
⎟
⎟
⎠ 		,
it follows

eu = ⎛
⎜
⎜
⎜
⎝
1
2u
0
⎞
⎟
⎟
⎟
⎠ 		,
and

ds =
√
 
eu ·eu
 
 d u =
√
 
1 + 4 u2
 
 du,
so

s = ⌠
⌡ 		1

0 
√
 
1 + 4 u2
 
 du
Let the computer carry out this integral.
Enter the following 
Sqrt[1 + 4 u^2] u 0 1
exactly (case sensitive, square bracket) in the boxes then press the query button.
Integrate with respect to
from to .

Surface area element

surfacearea.jpg
To obtain the surface area expressed by two parameters, u and v, one can first compute the surface element spanned by the two vectors shown above:

R(u + d u, v)− R(u, v)
=
R(u, v) + ∂R
∂u
 du − R(u, v)
=
eu du,
(4)
R(u, v + dv) − R(u, v)
=
R(u, v) + ∂R
∂v
 dv − R(u, v)
=
ev dv,
(5)
where
eu
≡
∂R
∂u
 ,
(6)
ev
≡
∂R
∂v
 .
(7)
Remembering that the area of a parallelogram spanned by two vectors, a and b, is

ds
=
  ⎛
 ⎥
√
det ⎛
⎜
⎜
⎝
eu ·eu ,
eu ·ev
ev ·eu ,
ev ·ev
⎞
⎟
⎟
⎠
 
 du dv
=

√
 
E F − G2
 
 du dv.
(8)
where
E
=
eu ·eu
F
=
ev ·ev
G
=
eu ·ev.
Alternatively, recall that the magnitude of a ×b is the area of the parallegram spanned by a and b, Equation (8) can be also written as

d s = | eu ×ev | du dv.

Example of surface integral

As an example of surface integrals, we attempt to compute the surface area of a sphere x2 + y2 + z2 = a2. By setting x = u and y = v, a position vector on the sphere is expressed as

R(u, v) = ⎛
⎜
⎜
⎜
⎜
⎝
u
v

√
 
a2 − u2 − v2
 
⎞
⎟
⎟
⎟
⎟
⎠ 		,
(9)
so
eu
=
⎛
⎜
⎜
⎜
⎜
⎜
⎝
1
0
−u

√
 
 [ˉ(a2 − u2 − v2)]
⎞
⎟
⎟
⎟
⎟
⎟
⎠
(10)
ev
=
⎛
⎜
⎜
⎜
⎜
⎜
⎝
0
1
−v

√
 
 [ˉ(a2 − u2 − v2)]
⎞
⎟
⎟
⎟
⎟
⎟
⎠
(11)
and

eu ·eu
=
1+ u2
a2 − u2 − v2
ev ·ev
=
1+ v2
a2 − u2 − v2
eu ·ev
=
uv
a2 − u2 − v2
 .
(12)
Hence,

dS
=

√
 
(eu ·eu) (ev ·ev) − (eu ·ev)2
 
 du dv
=
a

√
a2 − u2 − v2
 du dv.
(13)
The surface area of a semi-sphere (radius = a) is

S
=
⌠
⌡ 		⌠
⌡ 		a du dv

√
a2 − u2 − v2
=
a ⌠
⌡ 		⌠
⌡ 		r dr dθ

√
a2 − r2
 1
=
a ⌠
⌡ 		a

0 
 r dr

√
a2 − r2
 ⌠
⌡ 		2π

0 
 1 dθ
=
2πa ⌠
⌡ 		a

0 
 r dr

√
a2 − r2
 2
=
2πa2.
(14)
Alternative approach
spherical.jpg
A bit faster approach than the one above is to use the spherical coordinate system. Noting that the equation, x2 + y2 + z2 = a2 is automatically satisfied if

x
=
a sin ϕcos θ,
y
=
a sin ϕsin θ,
z
=
a cos ϕ.
(15)
Hence we can set
R = ⎛
⎜
⎜
⎜
⎝
a sin ϕcos θ
a sin ϕsin θ
a cos ϕ,
⎞
⎟
⎟
⎟
⎠ 		.
(16)
It follows
eϕ
=
⎛
⎜
⎜
⎜
⎝
a cos ϕcos θ
a cos ϕsin θ
−a sin ϕ
⎞
⎟
⎟
⎟
⎠
(17)
eθ
=
⎛
⎜
⎜
⎜
⎝
−a sin ϕsin θ
a sin ϕcos θ
0
⎞
⎟
⎟
⎟
⎠ 		,
(18)
and
E
=
a2
(19)
F
=
a2 sin 2 ϕ
(20)
G
=
0
(21)
dS
=

√
 
EF−G2
 
 dθdϕ
(22)
=
a2 sin ϕdθdϕ,
(23)
thus
S
=
⌠
⌡ 		π/2

0 
 ⌠
⌡ 		2π

0 
 a2 sin ϕdθdϕ
=
2πa2 ⌠
⌡ 		π/2

0 
 sin ϕdϕ
=
2πa2.
(24)

z = f(x, y)

If the equation of a curved surface is given as
z = f(x, y),
(25)
the area element can be expressed as
d S =
√
 
1 + fx2 + fy2
 
 dx dy,
(26)
where
fx
≡
∂f
∂x
 ,
fy
≡
∂f
∂y
 .
(27)
Proof

ex
=
∂R
∂x
=
⎛
⎜
⎜
⎜
⎝
1
0
fx
⎞
⎟
⎟
⎟
⎠
ey
=
∂R
∂y
=
⎛
⎜
⎜
⎜
⎝
0
1
fy
⎞
⎟
⎟
⎟
⎠ 		,
E
=
ex ·ex
=
1 + fx2,
F
=
ey ·ey
=
1 + fy2,
G
=
ex ·ey
=
fx fy,
EF−G2
=
(1 + fx2)(1 + fy2) − (fx fy)2
=
1 + fx2 + fy2.
(28)

Volume

volumeelement.jpg
The volume element of the parallelepiped spanned by the three vectors above can be expressed as

dv
=
  ⎛
 ⎥
 ⎥
√
⎢
⎢
⎢
⎢
⎢
eu ·eu, eu ·ev, eu ·ew
ev ·eu, ev ·ev, ev ·ew
ew ·eu, ew ·ev, ew ·ew
⎢
⎢
⎢
⎢
⎢
 
 du dv dw
=
|eu·(ev×ew)| du dv dw.
(29)
Example: Volume of sphere
spherical2.jpg

R = ⎛
⎜
⎜
⎜
⎝
x
y
z
⎞
⎟
⎟
⎟
⎠ 		= ⎛
⎜
⎜
⎜
⎝
r sin ϕcos θ
r sin ϕsin θ
r cos ϕ
⎞
⎟
⎟
⎟
⎠ 		,
(30)
so

er = ⎛
⎜
⎜
⎜
⎝
sinϕ cos θ
sinϕ sin θ
cos ϕ
⎞
⎟
⎟
⎟
⎠ 		,   eϕ = ⎛
⎜
⎜
⎜
⎝
r cosϕ cos θ
r cosϕ sin θ
− r sin ϕ
⎞
⎟
⎟
⎟
⎠ 		,   eθ = ⎛
⎜
⎜
⎜
⎝
− r sin ϕ sin θ
r sin ϕ cos θ
0
⎞
⎟
⎟
⎟
⎠
(31)
Noting that
eϕ ×eθ = ⎛
⎜
⎜
⎜
⎝
r2  sin2 ϕ cos θ
r2  sin2 ϕ sin θ
r2  sin ϕ cos ϕ
⎞
⎟
⎟
⎟
⎠
(32)
it follows
er ·eϕ ×eθ = r2 sin ϕ.
(33)
so
V
=
⌠
⌡ 		⌠
⌡ 		⌠
⌡ 		dV
=
⌠
⌡ 		2π

0 
 ⌠
⌡ 		π

0 
 ⌠
⌡ 		a

0 
 r2 sin ϕdr dϕdθ
=
⌠
⌡ 		2π

0 
 dθ⌠
⌡ 		π

0 
 sin ϕdϕ⌠
⌡ 		a

0 
 r2 dr
=
4πa3
3
 .
(34)

Summary

  1. Length of Arc

    dl
    =

    √
     
    eu ·eu
     
    		 du
    =
    |eu | du.
  2. Area

    d s
    =
      ⎛
     ⎥
    √
    |
    eu ·eu,
    eu ·ev
    eu ·ev,
    ev ·ev
    		|
     
    		 du dv
    =
    |eu ×ev| du dv.
  3. Volume

    d v
    =
      ⎛
     ⎥
     ⎥
    √
    ⎢
    ⎢
    ⎢
    ⎢
    ⎢
    eu ·eu,
    eu ·ev,
    eu ·ew
    ev ·eu,
    ev ·ev,
    ev ·ew
    ew ·eu,
    ew ·ev,
    ew ·ew
    		⎢
    ⎢
    ⎢
    ⎢
    ⎢
     
    		 du dv dw
    =
    |eu ·(ev ×ew) | du dv dw.

Divergence

The divergence operator can be defined on a vector, u, by
div u ≡
lim
V→ 0 
⌠
⌡

S 
 n·u dS
V
 ,
(35)
where n is the normal to the boundary and V is the volume of an object and the integral range is over the surface (boundary) of the object. This definition has an advantage over other definitions as it is independent of the coordinate system.
divergence.jpg

1-D

div_1d.jpg
In 1-D, an interval (x, x + h) is V in eq.(35) and its boundary is the two end points x and x + h. The normal, n, at x + h is +1 and is −1 at x = x. Thereby, eq.(35) becomes

div u
=

lim
h→ 0 
 (+1) ×u(x+h) +(−1) ×u(x)
h
(36)
=

lim
h→ 0 
 u(x + h) − u(x)
h
(37)
=
u′(x).
(38)
This is of course the ordinary differentiation in 1-D.

2-D

div_2d.jpg
In 2-D, a rectangle is V in eq.(35) and its boundary is the contour integral that surrounds the rectangle. The normals are the normal vectors perpendicular to each side of the rectangle. Therefore, it is necessary to compute eq.(35) in four different segments as

Path A

Along the path A, n = (0,−1) so n·u = − uy(x, y). Thus
⌠
⌡

A 
 = ⌠
⌡ 		x + h

x 
 − uy(x, y) dx → −uy(x, y) h.
(39)

Path B

Along the path B, n = (1, 0) so n·u = ux(x + h, y). Thus
⌠
⌡

B 
 = ⌠
⌡ 		y + k

y 
 ux(x + h, y) dy → ux(x + h, y) k.
(40)

Path C

Along the path C, n=(0,1) so n·u = uy(x,y+k). Thus
⌠
⌡

C 
 = ⌠
⌡ 		x

x + h 
 uy(x, y + k) (−dx) → uy(x, y + k) h.
(41)

Path D

Along the path D, n = (−1, 0) so n·u = −ux(x, y). Thus
⌠
⌡

D 
 = ⌠
⌡ 		y

y + k 
 − ux(x, y) (−dy) → − ux(x, y) k.
(42)
So the contour integral is
⌠
⌡

A + B + C + D 
 = k ( ux(x + h, y)−ux(x, y)) +h ( uy(x, y + k)−uy(x, y)).
(43)
Thus
⌠
⌡

S 
 n·u dS
V
=
ux(x + h, y) − ux(x, y)
h
 + uy(x, y + k) − uy(x, y)
k
(44)
⇒
∂ux
∂x
 + ∂uy
∂y
    as     h→ 0    and     k→ 0.
(45)

3-D

From the result in 2-D, it is obvious that the divergence operator in 3-D is expressed as
div u = ∂ux
∂x
 + ∂uy
∂y
 + ∂uz
∂z
 .
(46)
The divergence operator, div, is often written as

div u = ∇·u,
(47)
where ∇ 3 is a differential operator defined as
∇ ≡ ⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
∂
∂x
∂
∂y
∂
∂z
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠ 		.
(48)
As will be shown in class, the physical interpretation of the divergence operator is the net flow in the rectangle, i.e. the difference between the flows that comes in and the flow that flows out of the rectangle.

Gradient

Similar to the divergence operator, the gradient operator is defined as

grad u ≡
lim
V→ 0 
⌠
⌡

S 
 n u dS
V
 .
(49)
Note that u is a scalar function and the result of gradient is a vector.
By applying this definition to a rectangular element in the Cartesian coordinate system, one can obtain

grad u = ⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
∂u
∂x
∂u
∂y
∂u
∂z
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠ 		.
(50)
Note that the gradient operator operates on a scalar and the result is a vector. Using a symbol , ∇ (nabla), the gradient on a scalar function, u, can be expressed as
grad u = ∇u,
(51)
where
∇ ≡ ⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
∂
∂x
∂
∂y
∂
∂z
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠ 		.
(52)

Curl

The curl operator is defined on a 3-D vector as

curl u ≡
lim
V→ 0 
⌠
⌡

S 
 n ×u dS
V
 .
(53)
Using the nabla operator, the above can be written as

curl u
=
∇×u
(54)
=
⎢
⎢
⎢
⎢
⎢
⎢
⎢
i
j
k
∂
∂x
∂
∂y
∂
∂z
ux
uy
uz
⎢
⎢
⎢
⎢
⎢
⎢
⎢ 		.
(55)
Note that the curl is defined on a 3-D vector and the result is also a 3-D vector.
It should be noted that the three differential operators are reduced to an ordinary differentiation in 1-D.

du
dx
 ⇒ ⎧
⎪
⎨
⎪
⎩
∇·u
Divergence
∇u
Gradient
∇×u
Curl

Footnotes:

1
u = r cos θ,     v = r sin θ,     du dv = r dr dθ.
(56)
2 Let
a2 − r2 ≡ t,
(57)
then − 2 r dr = dt and as r → 0, t → a2 and as r → a, t → 0 so that
⌠
⌡ 		a

0 
 r dr

√
a2 − r2
 = ⌠
⌡ 		0

a2 
 r
√t
 dt
(− 2 r)
 = 1
2
 ⌠
⌡ 		a2

0 
 dt
√t
 = a.
(58)
3 ναβλα (nabla) = harp in Phoenicia (Greek). Also called atled or simply del.

File translated from TEX by TTH, version 4.03.
On 25 Sep 2023, 23:12.