Surfaces and Surfaces Integrals.
Surfaces
In previous sections we define a surface as all points in (x, y, z) ∈ R3 that
verifies f (x, y, z) = c, for some smooth scalar field f and constant c.
For example the sphere of radius a in R3 , S 2 (a):
S 2 (a) = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = a2 }.
In this case f (x, y, z) = x2 + y 2 + z 2 and c = a2 .
Recall that to describe curves in the xy-plane or R3 we use a parameterization,
for example the ellipse
x2 y 2
C = {(x, y) : 2 + 2 = 1}
a
b
could be parameterizided by
~r(t) = (a cos t, b sin t),
for t ∈ [0, 2π].
Now, we will try to use a similar idea to describe surfaces using two parameters, u, v say, for u ∈ I1 and v ∈ I2 , where I1 , I2 are intervals in R. A surface
described using two parameters is a parametric representation of it.
Example 1 Consider the circular cylinder:
S = {(x, y, z) : x2 + y 2 = a2 , 0 ≤ z ≤ 3}
If we cut the cylinder with the plane z = b (for any b ∈ R), we see that x and
y vary around the circumference of a circle of radius a, and z does not depend
on x and y. z represents the height in which we are in the cylinder.
A possible parametric representation could be:
~r(u, v) = (a cos u, a sin u, v),
1
u ∈ [0, 2π], v ∈ [0, 3]
Example 2 The sphere of radius a is:
S 2 (a) = {((x, y, z) : x2 + y 2 + z 2 = a2 )}
Let P = (x0 , y0 , z0 ) ∈ S 2 (a), and let P̂ be its projection in the xy-plane. Define
π
v ∈ [ −π
2 , 2 ] as the angle that P makes with the z-axis and u ∈ [0, 2π] as the
angle that P̂ makes with the x-axis in the xy-plane. Then, each point of the
sphere of radius a cab be written as
~r(u, v) = (a cos u cos v, a sin u cos u, a sin v),
π
for some (u, v) ∈ [0, 2π] × [ −π
2 , 2 ].
This way of writing the points of S 2 (a) is called spherical coordinates. (See
next page).
2
Example 3 Let S be the cone in R3 , defined as
(x, y, z) : x2 + y 2 = z 2 , x, y, z ∈ R
(See next page) Then a parametric representation could be:
~r(u, v) = (u cos v, u sin v, u),
u ∈ R, v ∈ [0, 2π]
Notice that if we consider (u cos v, u sin v, u) = (x, y, z) then
x2 + y 2 = (u cos v)2 + (u sin v)2 = u2 = z 2
as desired.
2
y2 z2
x
Example 4 The ellipsoid S = (x, y, z) : 2 + 2 + 2 = 1 . To find a paraa
b
c
metric representation of S we recall that the parametric representation of the
3
unit sphere is
(cos u cos v, sin u cos v, sin v), (u, v) ∈ [0, 2π] × [
−π π
, ]
2 2
Notice that the unit sphere is defined as all the points in the space (x, y, z) such
that x2 + y 2 + z 2 = 1, so we must have that
(cos u cos v)2 + (sin u cos v)2 + sin2 v = 1
(1)
2
2
2
y
A point in the ellipsoid verifies x2 + 2 + z2 = 1, so if we consider the following
a
b
c
parametric representation:
~r(u, v) = (a cos u cos v, b sin u cos v, c sin v), (u, v) ∈ [0, 2π] × [
then
a2 (cos u cos v)2 b2 (sin u cos u)2 c2 sin2 v
+
+
= 1,
a2
b2
c2
4
−π π
, ]
2 2
using (??), which means that all points of the form (a cos u cos v, b sin u cos v, c sin v)
are indeed in the ellipsoid.
Recall that if we had a curve in the space parameterized by
~r(t) = (x(t), y(t), z(t)),
then a tangent vector to the curve at the point P = (x0 , y0 , z0 ) = (x(t0 ), y(t0 ), z(t0 ))
was:
d~r
|t=t = (x0 (t0 ), y 0 (t0 ), z 0 (t0 )).
dt 0
Associated to a point of a smooth surface S with parametric representation
~r(u, v) = (x(u, v), y(u, v), z(u, v)), P = ~r(u0 , v0 ) there are infinitely many tangent vectors which are contained in a plane, called the tangent plane: It can be
proven that
~ru (u0 , v0 ) =
∂~r
|(u,v)=(u0 ,v0 )
∂u
5
and
~rv (u0 , v0 ) =
∂~r
|(u,v)=(u0 ,v0 )
∂v
are two linearly independent tangent vectors to the surface at the point P =
(x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 )) that span the tangent plane to the surface at that
point. This means that if ~a is a tangent vector to the surface at the point P ,
then it can be written as a linear combination of ~ru (u0 , v0 ) and ~rv (u0 , v0 ). Thus
there exists λ and β real numbers such that
~a = λ~ru (u0 , v0 ) + β~rv (u0 , v0 ).
(For further information check in the recommended book.)
Furthermore since ~ru (u0 , v0 ) and ~rv (u0 , v0 ) are tangent to the surface, their
cross product must be perpendicular to the surface. Thus
~ |(u ,v ) = ~ru (u0 , v0 ) × ~rv (u0 , v0 )
N
0 0
6
is a normal vector to the surface at P = ~r(u0 , v0 ) = (x(u0 , v0 ), y(u0 , v0 ), z(u0 , v0 )).
And a unit normal to the surface at the point P would be:
~n =
~
N
~ru (u0 , v0 ) × ~rv (u0 , v0 )
|(u0 ,v0 ) =
~|
|~ru (u0 , v0 ) × ~rv (u0 , v0 )|
|N
Remark: Recall that if the surface is represented as
S = {(x, y, z) : f (x, y, z) = c}
~ (x0 , y0 , z0 ). And
then a normal vector to the surface at the point (x0 , y0 , z0 ) is ∇f
therefore a unit normal to the surface at a point P would be:
m
~ =
~ (x0 , y0 , z0 )
∇f
.
~ (x0 , y0 , z0 )|
|∇f
This means that if we have a parametric representation of the same surface S
with unit normal at P
~n =
~
N
~ru (u0 , v0 ) × ~rv (u0 , v0 )
|(u0 ,v0 ) =
~|
|~ru (u0 , v0 ) × ~rv (u0 , v0 )|
|N
then we must have:
m
~ = ~n or m
~ = −~n.
Example 5 Compute a unit normal to the sphere of radius a
S = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = a2 }.
In this case:
m
~ =
~ (x, y, z)
(2x, 2y, 2z)
(x, y, z)
x y z
∇f
=p
=p
=( , , )
~ (x, y, z)|
a a a
|∇f
4x2 + 4y 2 + 4z 2
x2 + y 2 + z 2
Example 6 The elliptic cylinder has a parametric representation:
~r(u, v) = (a cos u, b sin u, v),
7
u ∈ [0, 2π], v ∈ R
then:
~ru = (−a sin u, b cos u, 0)
and
~rv = (0, 0, 1)
And a normal to the elliptic cylinder would be:
~k ~i
~j
~ru × ~rv = −a sin u b cos u 0 = b cos u~i + a sin u~j = (b cos u, a sin u, 0)
0
0
1 In particular if we have a = b, we get the circular cylinder of radius a and a
normal to it would be (a cos u, a sin u, 0)
Surface Integrals
Suppose S is a surface with parametric representation ~r(u, v), (u, v) ∈ R and
~
~ = ~ru × ~rv and unit normal ~n = N
normal vector N
. Then given a vector field
~|
|N
F~ (x, y, z) we define the surface integral of F~ along S as follows:
ZZ
ZZ
~ v)) · N
~
~ dudv.
F · ~n dA =
F~ (r(u,
S
The integral
RR
S
R
F~ ·~n dA is the flux across S when F~ = ρ~v , where ρ is the density
of particles passing through S and ~v is their mean velocity. This is the reason
why sometimes surfaces integrals are called flux integrals.
Example 7 Let S be the parabolic cylinder defined by
S = {(x, y, z) : y = x2 , 0 ≤ x ≤ 2, 0 ≤ z ≤ 3}
and the vector field F~ (x, y, z) = (3z 2 , 6, 6xz) A parametric representation of the
parabolic cylinder could be:
~r(u, v) = (u, u2 , v),
(u, v) ∈ [0, 2] × [0, 3]
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Then:
~ru = (1, 2u, 0) and
~rv = (0, 0, 1)
and the unit normal:
~ = ~ru × ~rv = N
~k 1 2u 0 = (2u, −1, 0)
0 0 1 ~i
~j
Furthermore:
F~ (~r(u, v)) = F~ (u, u2 , v) = (3v 2 , 6, 6uv).
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Therefore:

RR
2u





2
(3v
,
6,
6uv)
 −1  dvdu
0 0


0
R2R3
R2
v=3
= 0 0 (6uv 2 − 6)dvdu = 0 [2uv 3 − 6v]v=0 du
R2
2
= 0 (54u − 18)du = [27u2 − 18u]0 = 72
F~ · ~n dA =
S
R2R3
Example 8 Let F~ (x, y, z) = (x2 , 0, 3y 2 ) and
S = {(x, y, z) : x + y + z = 1, x, y, z > 0}
This surface is an equilateral triangle with vertices at (1, 0, 0), (0, 1, 0) and
(0, 0, 1).
A possible parametric representation of S would be:
~r(u, v) = (u, v, 1 − u − v),
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(u, v) ∈ [0, 1] × [0, 1]
and
~ru = (1, 0, −1) and ~rv = (0, 1, −1)
~ ~ ~ i j k ~
N = ~ru × ~rv = 1 0 −1 = (1, 1, 1)
0 1 −1 Then F~ (~r(u, v)) = F~ (u, v, 1 − u − v) = (u2 , 0, 3v 2 ) and:

RR
1

 
R 1 R 1−v 2

1
2 
~
√
(u , 0, 3v )  1  dudv
F · ~n dA =
S
0 0
 
3
1
iu=1−v
R 1 R 1−v 2
R 1 h u3
2
2
= 0 0 (u + 3v )dudv = 0 3 + 3v u
dv
u=0
3
R 1 (1 − v)
= 0(
+ 3v 2 (1 − v))dv = 31
3
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