Chapter 16 – Vector Calculus
16.6 Parametric Surfaces and their Areas
Objectives:
 Understand the various
types of parametric
surfaces.
 Compute the area using
vector functions.
16.6 Parametric Surfaces and their Areas
1
Vector Calculus

So far, we have considered special types of
surfaces:
◦
◦
◦
◦
Cylinders
Quadric surfaces
Graphs of functions of two variables
Level surfaces of functions of three variables
16.6 Parametric Surfaces and their
Areas
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Vector Calculus

Here, we use vector functions to describe
more general surfaces, called parametric
surfaces, and compute their areas.

Then, we take the general surface area
formula and see how it applies to special
surfaces.
16.6 Parametric Surfaces and their
Areas
3
Introduction

We describe a space curve by a vector
function r(t) of a single parameter t.

Similarly, we can describe a surface by
a vector function r(u, v) of two parameters
u and v.
16.6 Parametric Surfaces and their
Areas
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Equation 1

We suppose that
r(u, v) = x(u, v) i + y(u, v) j + z (u, v) k
is a vector-valued function defined
on a region D in the uv-plane.
16.6 Parametric Surfaces and their
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Equation 2

The set of all points (x, y, z) in 3 such that
x = x(u, v)
y = y(u, v)
z = z(u, v)
and (u, v) varies throughout D, is called
a parametric surface S.
◦ Equations 2 are called parametric
equations of S.
16.6 Parametric Surfaces and their
Areas
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Example 1 – pg. 1132 #2

Determine whether the points P and
Q lie on the given surface.
r u , v   u  v, u  v, u  v
2
2
P  3,  1, 5  , Q   1, 3, 4 
16.6 Parametric Surfaces and their
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Parametric Surfaces



Each choice of u and v gives a point on S.
By making all choices, we get all of S.
In other words, the surface S is traced out by the
tip of the position vector r(u, v) as (u, v) moves
throughout the region D.
16.6 Parametric Surfaces and their
Areas
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Example 2 – pg. 1132 # 5

Indentify the surface with the given
vector equation.
r (s, t )  s, t, t  s
2
2
16.6 Parametric Surfaces and their
Areas
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Example 3 – pg. 1132

Match the equations with the graphs labeled I – VI
and give reasons for your answers.
16.6 Parametric Surfaces and their
Areas
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Parametric Representation

In Example 1 we were given a vector
equation and asked to graph the
corresponding parametric surface.
◦ In the following examples, however, we are
given the more challenging problem of
finding a vector function to represent a
given surface.
◦ In the rest of the chapter, we will often
need to do exactly that.
16.6 Parametric Surfaces and their
Areas
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Example 4

Find a parametric representation of
the sphere
x 2 + y 2 + z2 = a 2
16.6 Parametric Surfaces and their
Areas
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Applications – Computer Graphics


One of the uses of parametric surfaces is in
computer graphics.
The figure shows the result of trying
to graph the sphere x2 + y2 + z2 = 1
by:
◦ Solving the equation
for z.
◦ Graphing the top and
bottom hemispheres
separately.
16.6 Parametric Surfaces and their
Areas
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Computer Graphics

Part of the sphere appears to be
missing because of the rectangular
grid system used by the computer.
16.6 Parametric Surfaces and their
Areas
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Computer Graphics

The much better picture here was
produced by a computer using the
parametric equations found in the
example 2.
16.6 Parametric Surfaces and their
Areas
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Parameters

In general, a surface given as the graph of a
function of x and y—an equation of the form
z = f(x, y)—can always be regarded as a parametric
surface by:
◦ Taking x and y as parameters.
◦ Writing the parametric equations as
x = x y = y z = f(x, y)
16.6 Parametric Surfaces and their
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Example 5

Find a parametric representation for
the surface.
20. T he part low er half of the ellipsoid 2 x  4 y  z  1.
2
2
2
16.6 Parametric Surfaces and their
Areas
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Tangent Planes

We now find the tangent plane to a
parametric surface S traced out by a
vector function
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
at a point P0 with position vector r(u0, v0).
16.6 Parametric Surfaces and their
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Tangent Planes

Keeping u constant by putting u = u0,
r(u0, v) becomes a vector function of the
single parameter v and defines a grid curve
C1 lying on S.
16.6 Parametric Surfaces and their
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Tangent Planes – Equation 4

The tangent vector to C1 at P0 is obtained
by taking the partial derivative of r with
respect to v:
rv 
x
v
(u 0 , v0 ) i 
y
v
(u 0 , v0 ) j 
z
v
(u 0 , v0 ) k
16.6 Parametric Surfaces and their
Areas
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Tangent Planes
Similarly, keeping v constant by putting
v = v0, we get a grid curve C2 given by
r(u, v0) that lies on S.
 Its tangent vector at P0 is:

ru 
x
u
(u 0 , v0 ) i 
y
u
(u 0 , v0 ) j 
z
u
(u 0 , v0 ) k
16.6 Parametric Surfaces and their
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Smooth Surface

If ru x rv is not 0, then the surface is called smooth
(it has no “corners”).
◦ For a smooth surface, the tangent plane is the
plane that contains the tangent vectors ru and
rv, and the vector ru x rv is a normal vector to
the tangent plane.
16.6 Parametric Surfaces and their
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Example 6

Find an equation of the tangent plane
to the given parametric surface at
the specified point..
36. r ( u , v )  uv i  u sin v j  v cos u k
u  0, v  
16.6 Parametric Surfaces and their
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Definition 6 – Surface Area

Suppose a smooth parametric surface S is:
◦ Given by the equation :
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
(u, v) D
◦ Covered just once as (u, v) ranges throughout
the parameter domain D.
16.6 Parametric Surfaces and their
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Definition 6 continued

Then, the surface area of S is
A(S ) 
|
r

r
|
dA
u
v

D
where:
ru 
x
u
i
y
u
j
z
u
k
rv 
x
v
i
y
v
j
16.6 Parametric Surfaces and their
Areas
z
v
k
25
Surface Area of the Graph of a
Function

Now, consider the special case of a surface
S with equation z = f(x, y), where (x, y) lies
in D and f has continuous partial
derivatives.
◦ Here, we take x and y as parameters.
◦ The parametric equations are:
x=x
y=y
z = f(x, y)
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Surface Area of the Graph of a
Function

Then, the surface area formula in
Definition 6 becomes:
2
A( S ) 

D

2
 z 
 z 
1 
 dA
 
 x 
 y 
(this is formula 9)
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Example 7

Find the area of the surface.
a . T h e p art o f th e p lan e 2 x  5 y  z  1 0 th at lie s in sid e
th e cylin d er x  y  9 .
2
2
b . T h e p art o f th e p lan e w ith th e vecto r eq u atio n
r (u , v )  1  v , u  2 v , 3  5u  v
th at is g iven b y
0  u  1, 0  v  1 .
16.6 Parametric Surfaces and their
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Example 8 – pg. 1133

Find the area of the surface.
4 4 . T h e p art o f th e su rface z  1  3 x  2 y
th e trian g le w ith vertices
2
th at lies ab o ve
 0, 0  ,  0,1  , an d  2,1  .
4 6. T h e p art o f th e p arab o lo id x  y  z
2
2
th at lies in sid e
th e cylin d er y  z  9 .
2
2
4 8 . T h e h elico id (o r sp iral ram p ) w ith th e ve cto r eq u atio n
r ( u , v )  u co s v , u sin v , v
0  u  1, 0  v   .
th at is g iven b y
16.6 Parametric Surfaces and their
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