Math 251 Section 14.6 Parametric Surfaces and Their Areas
A surface in
R 3 has a 3-component vector equation which depends on two independent parameters.




r (u , v )  x (u , v ) i  y (u , v ) j  z (u , v ) k
z  x 2  y 2 0  z  9 , has vector equation
Example1: The surface of the paraboloid



(u , v ) in D .

r ( x, y )  x i  y j  ( x 2  y 2 ) k
D  {( x , y ) :
x 2  y 2  9}
The parameter domain is an important part of the definition of the surface.
Example 2: Parameterize the surface of the cylinder,
x2  y2  4
between the planes
z  0 and
x  z  3.
A point on this surface lies on the circle of radius 2 centered at
Let the parameters be  and


z.
Then

x  2 cos  ,

( 0 ,0 , z ) .
y  2 sin  ,
r ( , z )  2 cos  i  2 sin  j  z k for 0    2 ,
Example 3: Parameterize the surface of the plane
zz
0  z  3  2 cos 
x  z  3 over the disk x 2  y 2  4,
z  0.
S,
Tangent Planes to Surfaces: Let the surface,



be given by

r (u , v )  x (u , v ) i  y (u , v ) j  z (u , v ) k

Let


(u , v ) in D
and
P ( a , b , c ) a point on S .

r ( u 0 , v0 )  a i  b j  c k .
r
(u0 , v0 ),
u
r
(u0 , v0 )
v
are both tangent to
S at P ( a , b , c ) .
So their cross product is a normal vector to the tangent plane to
S at P ( a , b , c ) .
The vectors
Example2: Find the tangent plane to the surface
z  4  x2  y2 ,
z  0 at P (1, 1,2).
Example 3: Find the tangent plane to the surface parameterized by




r (u , v )  u i  ue j  ( v  u ) e k
2
v
v
for

If the surface is given by
12, the normal is
| u | 2,
| v | 2

at the point


P (1,  1,1) .
z  f ( x , y ) then r ( x , y )  x i  y j  f ( x , y ) k and as in Chapter
f  f    

i
j  k  rx  ry
x
y
evaluated at the given point.
Surface Areas:

Definition: A surface given by



r (u , v )  x (u , v ) i  y (u , v ) j  z (u , v ) k
(u , v ) in D

is called smooth if ru  rv is never zero in
surface area 


D . For such a surface, the surface area is

 | ru  rv | dA ,
(assuming the surface is covered exactly once on the parameter
D
domain. )

In the case that the surface is given by
2



r ( x, y )  x i  y j  f ( x, y ) k
then
2
 f   f 
surface area         1 dA
 y 
D  x 
Example4: Find the surface area of the part of the paraboloid
cylinder
y2  z2  9 .
x  y2  z2
that lies inside the