Lesson 22: Parametric Surfaces
July 27th, 2015
Section 16.6
In this lesson we discuss the idea of parametrizing a surface.
We will need this notion in the next few lesson when we talk
about surface integrals.
We have already computed some surface integrals, namely in
the special case where our surface is imbedded in the plane.
These are just the double integrals that we have looked at.
We want to look at surfaces that live in space.
The move from surfaces in the plane to surfaces in space is
exactly analogous to the move from curves that are imbedded
in the line, i.e. intervals, to curves that live in the plane and
space.
Section 16.6
A parametric surface is given by a vector function r(u, v ) of
two parameters, u and v . Since we will be considering surfaces
living in 3-space, these vector functions have the form
r(u, v ) = x (u, v ) i + y (u, v ) j + z(u, v ) k
Here we are identifying the position vector
hx (u, v ), y (u, v ), z(u, v )i
with the point (x (u, v ), y (u, v ), z(u, v )).
Notice the similarity between this and the vector equation of a
curve C given by r(t) = x (t) i + y (t) j + z(t) k.
Section 16.6
The parametric surface S given by the vector function r(u, v )
is the set of all points (x , y , z) in R3 such that
x = x (u, v )
y = y (u, v )
z = z(u, v )
as u and v vary through some domain D in the uv -plane.
Just as with a curve, we can think of r(u, v ) as tracing out the
surface S.
Section 16.6
Example
Determine whether or not the points P(7, 10, 4) and
Q(5, 22, 5) lie on the surface
r(u, v ) = h2u + 3v , 1 + 5u − v , 2 + u + v i
Section 16.6
Example
Identify and sketch the surface with vector equation
r(u, v ) = 2 cos u i + v j + 2 sin u k
Section 16.6
The domain in the previous problem was all of the uv -plane, or
rather all of R2 . If we want just a portion of the surface, we
can restrict the domain of u and v
Example
Sketch the surface r(u, v ) = 2 cos u i + v j + 2 sin u k for
0 ≤ u ≤ π/2 and 0 ≤ v ≤ 3.
Section 16.6
The domain in the previous problem was all of the uv -plane, or
rather all of R2 . If we want just a portion of the surface, we
can restrict the domain of u and v
Example
Sketch the surface r(u, v ) = 2 cos u i + v j + 2 sin u k for
0 ≤ u ≤ π/2 and 0 ≤ v ≤ 3.
Notice to get the full surface we can restrict from all of the
uv -plane to 0 ≤ u ≤ 2π and −∞ ≤ v ≤ ∞.
Section 16.6
If we fix one of the variables u or v in r(u, v ), the result is a
curve that lives on S. We call these grid curves, and they
can be useful in graphing parametric surfaces.
For the surface in the previous examples, if we fix u = u0 , we
get the curve
r(u0 , v ) = 2 cos u0 i + v j + 2 sin u0 k.
This has constant x and z components and y component
y = v . Thus it’s a line parallel to the y -axis.
For v = v0 we get the curve
r(u, v0 ) = 2 cos u i + v0 j + 2 sin u k.
This is a circle of radius two centered around the y -axis and
parallel to the xz-plane.
Section 16.6
So far we have discussed how to find the surface corresponding
to a given vector equation.
Often a more difficult question is to find a vector equation
corresponding to a given surface.
Section 16.6
Example
Find a vector function that represents the plane that passes
through the point P0 with position vector r0 and that contains
two nonparallel vectors a and b.
Section 16.6
Example
Identify the surface with the vector equation
r(u, v ) = (u + v ) i + (2 + u) j + (1 + 4u + 5v ) k
Section 16.6
Example
Find a parametric representation of the sphere
x 2 + y 2 + z 2 = a2
and determine its grid curves.
Section 16.6
Example
Find a parametric representation for the cylinder
x2 + y2 = 4
0≤z ≤1
Section 16.6
Example
Find a vector function that represents the elliptic paraboloid
z = x 2 + 2y 2
Section 16.6
The previous example illustrates how we may parametrize
surfaces that are given as graphs of functions of two variables.
Namely, if S is the graph of the function z = f (x , y ), then S
is given by the parametric equations
x =x
y =y
z = f (x , y )
or equivalently by the vector equation
r(x , y ) = x i + y j + f (x , y ) k
This is completely analogous to the way we parametrize a
curve that is the graph of a function y = f (x ).
It is also true (like with curves) that parametric
representations (or parametrizations) of surfaces are not
unique, as the next example illustrates.
Section 16.6
Example
Find two different parametrizations for the cone
q
z = 2 x2 + y2
Section 16.6
Recall that a surface of revolution is a surface that is obtained
by rotating the graph of a function about an axis. It is fairly
easy to parametrize a surface of revolution S obtained by
rotating y = f (x ) around the x -axis where f (x ) ≥ 0.
A point (x , y , z) on S is given by
x =x
y = f (x ) cos θ
z = f (x ) sin θ
This gives a parametrization of S with parameters x and θ
with a ≤ x ≤ b and 0 ≤ θ ≤ 2π.
Section 16.6
Example
Find parametric equations for the surface generated by
rotating the curve y = sin x , 0 ≤ x ≤ 2π, about the x -axis.
Use these equations to graph the surface of revolution.
Section 16.6