#08 (09/15/2025)

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

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

R(u + d u ) − R(u)

R(u) +

d R

du + …− R(u)

∼

d u

e_u d u,

(1)

where

e_u ≡

d u

Therefore,

√

dR ·dR

√

e_u d u ·e_u d u

√

e_u ·e_u

d u.

(2)

and

s =

⌠
⌡

u=u₁

u=u₀

√

e_u ·e_u

d u.

(3)

Example Compute the length of y = x² from x = 0 to x = 1.

Solution As

R =

⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠

⎛
⎜
⎜
⎜
⎝

u²

⎞
⎟
⎟
⎟
⎠

it follows

e_u =

⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠

and

ds =

√

e_u ·e_u

d u =

√

1 + 4 u²

du,

s =

⌠
⌡

1

0

√

1 + 4 u²

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.

Surface area element

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)

e_u du,

(4)

R(u, v + dv) − R(u, v)

R(u, v) +

∂R

∂v

dv − R(u, v)

e_v dv,

(5)

where

e_u

≡

∂R

∂u

(6)

e_v

≡

∂R

∂v

(7)

Remembering that the area of a parallelogram spanned by two vectors, a and b, is

⎛
⎥
√

det

⎛
⎜
⎜
⎝

e_u ·e_u ,

e_u ·e_v

e_v ·e_u ,

e_v ·e_v

⎞
⎟
⎟
⎠

du dv

√

E F − G²

du dv.

(8)

where

e_u ·e_u

e_v ·e_v

e_u ·e_v.

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 = | e_u ×e_v | 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) =

⎛
⎜
⎜
⎜
⎜
⎝

√

a² − u² − v²

⎞
⎟
⎟
⎟
⎟
⎠

(9)

e_u

⎛
⎜
⎜
⎜
⎜
⎜
⎝

−u

√

[ˉ(a² − u² − v²)]

⎞
⎟
⎟
⎟
⎟
⎟
⎠

(10)

e_v

⎛
⎜
⎜
⎜
⎜
⎜
⎝

−v

√

[ˉ(a² − u² − v²)]

⎞
⎟
⎟
⎟
⎟
⎟
⎠

(11)

and

e_u ·e_u

u²

a² − u² − v²

e_v ·e_v

v²

a² − u² − v²

e_u ·e_v

a² − u² − v²

(12)

Hence,

√

(e_u ·e_u) (e_v ·e_v) − (e_u ·e_v)²

du dv

√

a² − u² − v²

du dv.

(13)

The surface area of a semi-sphere (radius = a) is

⌠
⌡

a du dv

√

a² − u² − v²

⌠
⌡

r dr dθ

√

a² − r²

⌠
⌡

a

0

r dr

√

a² − r²

⌠
⌡

2π

0

1 dθ

2πa

⌠
⌡

a

0

r dr

√

a² − r²

2πa².

(14)

Alternative approach

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

a sin ϕcos θ,

a sin ϕsin θ,

a cos ϕ.

(15)

Hence we can set

R =

⎛
⎜
⎜
⎜
⎝

a sin ϕcos θ

a sin ϕsin θ

a cos ϕ,

⎞
⎟
⎟
⎟
⎠

(16)

It follows

e_ϕ

⎛
⎜
⎜
⎜
⎝

a cos ϕcos θ

a cos ϕsin θ

−a sin ϕ

⎞
⎟
⎟
⎟
⎠

(17)

e_θ

⎛
⎜
⎜
⎜
⎝

−a sin ϕsin θ

a sin ϕcos θ

⎞
⎟
⎟
⎟
⎠

(18)

and

a²

(19)

a² sin ² ϕ

(20)

(21)

√

EF−G²

dθdϕ

(22)

a² sin ϕdθdϕ,

(23)

thus

⌠
⌡

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 + f_x² + f_y²

dx dy,

(26)

where

f_x

≡

∂f

∂x

f_y

≡

∂f

∂y

(27)

Proof

e_x

∂R

∂x

⎛
⎜
⎜
⎜
⎝

f_x

⎞
⎟
⎟
⎟
⎠

e_y

∂R

∂y

⎛
⎜
⎜
⎜
⎝

f_y

⎞
⎟
⎟
⎟
⎠

e_x ·e_x

1 + f_x²,

e_y ·e_y

1 + f_y²,

e_x ·e_y

f_x f_y,

EF−G²

(1 + f_x²)(1 + f_y²) − (f_x f_y)²

1 + f_x² + f_y².

(28)

Volume

The volume element of the parallelepiped spanned by the three vectors above can be expressed as³

⎛
⎥
⎥
√

⎢
⎢
⎢
⎢
⎢

e_u ·e_u, e_u ·e_v, e_u ·e_w

e_v ·e_u, e_v ·e_v, e_v ·e_w

e_w ·e_u, e_w ·e_v, e_w ·e_w

⎢
⎢
⎢
⎢
⎢

du dv dw

|e_u·(e_v×e_w)| du dv dw.

(29)

Example: Volume of sphere

R =

⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠

⎛
⎜
⎜
⎜
⎝

r sin ϕcos θ

r sin ϕsin θ

r cos ϕ

⎞
⎟
⎟
⎟
⎠

(30)

e_r =

⎛
⎜
⎜
⎜
⎝

sinϕ cos θ

sinϕ sin θ

cos ϕ

⎞
⎟
⎟
⎟
⎠

, e_ϕ =

⎛
⎜
⎜
⎜
⎝

r cosϕ cos θ

r cosϕ sin θ

− r sin ϕ

⎞
⎟
⎟
⎟
⎠

, e_θ =

⎛
⎜
⎜
⎜
⎝

− r sin ϕ sin θ

r sin ϕ cos θ

⎞
⎟
⎟
⎟
⎠

(31)

Noting that

e_ϕ ×e_θ =

⎛
⎜
⎜
⎜
⎝

r² sin² ϕ cos θ

r² sin² ϕ sin θ

r² sin ϕ cos ϕ

⎞
⎟
⎟
⎟
⎠

(32)

it follows

e_r ·e_ϕ ×e_θ = r² sin ϕ.

(33)

⌠
⌡

2π

0

⌠
⌡

π

0

⌠
⌡

a

0

r² sin ϕdr dϕdθ

⌠
⌡

2π

0

dθ

⌠
⌡

π

0

sin ϕdϕ

⌠
⌡

a

0

r² dr

4πa³

(34)

Summary

Length of Arc

dl

=

√

e_u ·e_u

du

=

|e_u | du.
Area

d s

=

⎛
⎥
√

|

e_u ·e_u,
e_u ·e_v

e_u ·e_v,
e_v ·e_v
|

du dv

=

|e_u ×e_v| du dv.

Volume

d v

⎛
⎥
⎥
√

⎢
⎢
⎢
⎢
⎢

e_u ·e_u,

e_u ·e_v,

e_u ·e_w

e_v ·e_u,

e_v ·e_v,

e_v ·e_w

e_w ·e_u,

e_w ·e_v,

e_w ·e_w

⎢
⎢
⎢
⎢
⎢

du dv dw

|e_u ·(e_v ×e_w) | du dv dw.

Divergence

The divergence operator can be defined on a vector, u, by

div u ≡

lim
V→ 0

⌠
⌡

n·u dS

(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.

1-D

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)

(36)

lim
h→ 0

u(x + h) − u(x)

(37)

u′(x).

(38)

This is of course the ordinary differentiation in 1-D.

2-D

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 = − u_y(x, y). Thus

⌠
⌡

x + h

x

− u_y(x, y) dx → −u_y(x, y) h.

(39)

Path B

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

⌠
⌡

y + k

y

u_x(x + h, y) dy → u_x(x + h, y) k.

(40)

Path C

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

⌠
⌡

x

x + h

u_y(x, y + k) (−d x) → u_y(x, y + k) h.

(41)

Path D

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

⌠
⌡

y

y + k

− u_x(x, y) (−dy) → − u_x(x, y) k.

(42)

So the contour integral is

⌠
⌡

A + B + C + D

= k ( u_x(x + h, y)−u_x(x, y)) +h ( u_y(x, y + k)−u_y(x, y)).

(43)

Thus

⌠
⌡

n·u dS

u_x(x + h, y) − u_x(x, y)

u_y(x, y + k) − u_y(x, y)

(44)

⇒

∂u_x

∂x

∂u_y

∂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 =

∂u_x

∂x

∂u_y

∂y

∂u_z

∂z

(46)

The divergence operator, div, is often written as

div u = ∇·u,

(47)

where ∇ ⁴ 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

⌠
⌡

n u dS

(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

⌠
⌡

n ×u dS

(53)

Using the nabla operator, the above can be written as

curl u

∇×u

(54)

⎢
⎢
⎢
⎢
⎢
⎢
⎢

∂

∂x

∂

∂y

∂

∂z

u_x

u_y

u_z

⎢
⎢
⎢
⎢
⎢
⎢
⎢

(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.

⇒

⎧
⎪
⎨
⎪
⎩

∇·u

Divergence

∇u

Gradient

∇×u

Curl

Footnotes:

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

√t

(− 2 r)

⌠
⌡

a²

0

√t

= a.

(58)

³ Let a=(a₁, a₂, a₃), b=(b₁, b₂, b₃), c=(c₁, c₂, c₃).

a·(b×c) = (−a₃ b₂ + a₂ b₃) c₁ + (a₃ b₁ − a₁ b₃) c₂ + (−a₂ b₁ + a₁ b₂) c₃ =

⎢
⎢
⎢
⎢
⎢

a₁

a₂

a₃

b₁

b₂

b₃

c ₁

c₂

c₃

⎢
⎢
⎢
⎢
⎢

On the other hand,

⎢
⎢
⎢
⎢
⎢

(a, a)

(a, b)

(a, c)

(b, a)

(b, b)

(b, a)

(b, b)

⎢
⎢
⎢
⎢
⎢

=(−a₃ b₂ c₁ + a₂ b₃ c₁ + a₃ b₁ c₂ − a₁ b₃ c₂ − a₂ b₁ c₃ + a₁ b₂ c₃)² =

⎢
⎢
⎢
⎢
⎢

a₁

a₂

a₃

b₁

b₂

b₃

c ₁

c₂

c₃

⎢
⎢
⎢
⎢
⎢

⁴ ναβλα (nabla) = harp in Phoenicia (Greek). Also called atled or simply del.

File translated from T_EX by T_TH, version 4.03.
On 14 Sep 2025, 19:34.