Transcript Parametric Surfaces and Their Areas

16
VECTOR CALCULUS
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
VECTOR CALCULUS
Here, we use vector functions to
describe more general surfaces, called
parametric surfaces, and compute their
areas.
VECTOR CALCULUS
Then, we take the general surface
area formula and see how it applies
to special surfaces.
VECTOR CALCULUS
16.6
Parametric Surfaces
and their Areas
In this section, we will learn about:
Various types of parametric surfaces
and computing their areas using vector functions.
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.
INTRODUCTION
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.
INTRODUCTION
So x, y, and z—the component functions
of r—are functions of the two variables
u and v with domain D.
PARAMETRIC SURFACE
Equations 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.
PARAMETRIC SURFACES
Each choice of u and v gives a point
on S.
By making all choices, we get all of S.
PARAMETRIC SURFACES
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.
PARAMETRIC SURFACES
Example 1
Identify and sketch the surface with
vector equation
r(u, v) = 2 cos u i + v j + 2 sin u k
 The parametric equations for this surface
are:
x = 2 cos u
y=v
z = 2 sin u
PARAMETRIC SURFACES
Example 1
So, for any point (x, y, z) on the surface,
we have:
x2 + z2 = 4 cos2u + 4 sin2u
=4
 This means that vertical cross-sections parallel
to the xz-plane (that is, with y constant) are all
circles with radius 2.
PARAMETRIC SURFACES
Example 1
Since y = v and no restriction is placed on v,
the surface is a circular cylinder with radius 2
whose axis is the y-axis.
PARAMETRIC SURFACES
In Example 1, we placed no restrictions
on the parameters u and v.
So, we obtained the entire cylinder.
PARAMETRIC SURFACES
If, for instance, we restrict u and v by writing
the parameter domain as
0 ≤ u ≤ π/2
0≤v≤3
then
x≥0
z≥0
0≤y≤3
PARAMETRIC SURFACES
In that case, we get the quarter-cylinder
with length 3.
PARAMETRIC SURFACES
If a parametric surface S is given by a vector
function r(u, v), then there are two useful
families of curves that lie on S—one with u
constant and the other with v constant.
 These correspond to vertical and
horizontal lines in the uv-plane.
PARAMETRIC SURFACES
Keeping u constant by putting u = u0, r(u0, v)
becomes a vector function of the single
parameter v and defines a curve C1 lying on S.
GRID CURVES
Similarly, keeping v constant by putting v = v0,
we get a curve C2 given by r(u, v0) that lies
on S.
 We call these curves grid curves.
GRID CURVES
In Example 1, for instance, the grid
curves obtained by:
 Letting u be constant are horizontal lines.
 Letting v be constant are circles.
GRID CURVES
In fact, when a computer graphs a parametric
surface, it usually depicts the surface by
plotting these grid curves—as we see in
the following example.
GRID CURVES
Example 2
Use a computer algebra system to graph
the surface
r(u, v) = <(2 + sin v) cos u,
(2 + sin v) sin u, u + cos v>
 Which grid curves have u constant?
 Which have v constant?
GRID CURVES
We graph the portion of
the surface with parameter
domain
0 ≤ u ≤ 4π, 0 ≤ v ≤ 2π
 It has the appearance
of a spiral tube.
Example 2
GRID CURVES
Example 2
To identify the grid curves, we write
the corresponding parametric equations:
x = (2 + sin v) cos u
y = (2 + sin v) sin u
z = u + cos v
GRID CURVES
Example 2
If v is constant, then sin v and cos v
are constant.
 So, the parametric equations resemble those
of the helix in Example 4 in Section 13.1
GRID CURVES
So, the grid curves with v
constant are the spiral
curves.
 We deduce that the grid
curves with u constant
must be the curves that
look like circles.
Example 2
Example 2
GRID CURVES
Further evidence for this assertion is that,
if u is kept constant, u = u0, then the equation
z = u0 + cos v
shows that the z-values vary from u0 – 1
to u0 + 1.
PARAMETRIC REPRESENTATION
In Examples 1 and 2 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.
PARAMETRIC REPRESENTATIONS Example 3
Find a vector function that represents
the plane that:
 Passes through the point P0 with position
vector r0.
 Contains two nonparallel vectors a and b.
PARAMETRIC REPRESENTATIONS Example 3
If P is any point in the plane, we can get
from P0 to P by moving a certain distance
in the direction of a and another distance
in the direction of b.
 So, there are scalars u and v such that:
P0 P = ua + vb
PARAMETRIC REPRESENTATIONS Example 3
The figure illustrates how this works,
by means of the Parallelogram Law, for
the case where u and v are positive.
 See also
Exercise 40
in Section 12.2
PARAMETRIC REPRESENTATIONS Example 3
If r is the position vector of P,
then
r  O P0  P0 P  r0  u a  v b
 So, the vector equation of the plane can be
written as:
r(u, v) = r0 + ua + vb
where u and v are real numbers.
PARAMETRIC REPRESENTATIONS Example 3
If we write
r = <x, y, z>
r0 = <x0, y0, z0>
a = <a1, a2, a3>
b = <b1, b2, b3>
we can write the parametric equations
of the plane through the point (x0, y0, z0) as:
x = x0 + ua1 + vb1
y = y0 + ua2 + vb2
z = z0 + ua3 + vb3
PARAMETRIC REPRESENTATIONS Example 4
Find a parametric representation of
the sphere
x2 + y2 + z2 = a2
 The sphere has a simple representation ρ = a
in spherical coordinates.
 So, let’s choose the angles Φ and θ in spherical
coordinates as the parameters (Section 15.8).
PARAMETRIC REPRESENTATIONS Example 4
Then, putting ρ = a in the equations for
conversion from spherical to rectangular
coordinates (Equations 1 in Section 15.8),
we obtain:
x = a sin Φ cos θ
y = a sin Φ sin θ
z = a cos Φ
as the parametric equations of the sphere.
PARAMETRIC REPRESENTATIONS Example 4
The corresponding vector equation is:
r(Φ, θ)
= a sin Φ cos θ i + a sin Φ sin θ j + a cos Φ k
 We have 0 ≤ Φ ≤ π and 0 ≤ θ ≤ 2π.
 So, the parameter domain is the rectangle
D = [0, π] x [0, 2π]
PARAMETRIC REPRESENTATIONS Example 4
The grid curves with:
 Φ constant are the circles of constant latitude
(including the equator).
 θ constant are the meridians (semicircles),
which connect the north and south poles.
APPLICATIONS—COMPUTER GRAPHICS
One of the uses of
parametric surfaces is
in computer graphics.
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.
COMPUTER GRAPHICS
Part of the sphere appears to be missing
because of the rectangular grid system
used by the computer.
COMPUTER GRAPHICS
The much better picture here was produced
by a computer using the parametric equations
found in Example 4.
PARAMETRIC REPRESENTATIONS Example 5
Find a parametric representation for
the cylinder
x2 + y2 = 4
0≤z≤1
 The cylinder has a simple representation r = 2
in cylindrical coordinates.
 So, we choose as parameters θ and z
in cylindrical coordinates.
PARAMETRIC REPRESENTATIONS Example 5
Then the parametric equations of
the cylinder are
x = 2 cos θ
where:
 0 ≤ θ ≤ 2π
0≤z≤1
y = 2 sin θ
z=z
PARAMETRIC REPRESENTATIONS Example 6
Find a vector function that represents
the elliptic paraboloid z = x2 + 2y2
 If we regard x and y as parameters,
then the parametric equations are simply
x=x
y=y
z = x2 + 2y2
and the vector equation is
r(x, y) = x i + y j + (x2 + 2y2) k
PARAMETRIC REPRESENTATIONS
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)
PARAMETRIZATIONS
Parametric representations (also
called parametrizations) of surfaces
are not unique.
 The next example shows two ways
to parametrize a cone.
Example 7
PARAMETRIZATIONS
Find a parametric representation for
the surface
z2 x  y
2
that is,
the top half of the cone
z2 = 4x2 + 4y2
2
PARAMETRIZATIONS
E. g. 7—Solution 1
One possible representation is obtained
by choosing x and y as parameters:
x=x
y=y
z2 x  y
2
 So, the vector equation is:
r ( x, y )  x i  y j  2 x  y k
2
2
2
PARAMETRIZATIONS
E. g. 7—Solution 2
Another representation results
from choosing as parameters the polar
coordinates r and θ.
 A point (x, y, z) on the cone satisfies:
x = r cos θ
y = r sin θ
z  2 x  y  2r
2
2
PARAMETRIZATIONS
E. g. 7—Solution 2
So, a vector equation for the cone is
r(r, θ) = r cos θ i + r sin θ j + 2r k
where:
r≥0
 0 ≤ θ ≤ 2π
PARAMETRIZATIONS
For some purposes, the parametric
representations in Solutions 1 and 2 are
equally good.
In certain situations, though, Solution 2
might be preferable.
PARAMETRIZATIONS
For instance, if we are interested only in
the part of the cone that lies below the plane
z = 1, all we have to do in Solution 2 is
change the parameter domain to:
0≤r≤½
0 ≤ θ ≤ 2π
SURFACES OF REVOLUTION
Surfaces of revolution can be
represented parametrically and thus
graphed using a computer.
SURFACES OF REVOLUTION
For instance, let’s consider the surface S
obtained by rotating the curve
y = f(x)
a≤x≤b
about the x-axis, where f(x) ≥ 0.
SURFACES OF REVOLUTION
Let θ be the angle of rotation
as shown.
SURFACES OF REVOLUTION
If (x, y, z) is a point on S,
then
x=x
y = f(x) cos θ
z = f(x) sin θ
Equations 3
SURFACES OF REVOLUTION
Thus, we take x and θ as parameters
and regard Equations 3 as parametric
equations of S.
 The parameter domain is given by:
a≤x≤b
0 ≤ θ ≤ 2π
SURFACES OF REVOLUTION
Example 8
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.
Example 8
SURFACES OF REVOLUTION
From Equations 3,
 The parametric equations are:
x=x
y = sin x cos θ
z = sin x sin θ
 The parameter domain is:
0 ≤ x ≤ 2π
0 ≤ θ ≤ 2π
SURFACES OF REVOLUTION
Example 8
Using a computer to plot these equations
and rotate the image, we obtain this graph.
SURFACES OF REVOLUTION
We can adapt Equations 3 to represent
a surface obtained through revolution
about the y- or z-axis.
 See Exercise 30.
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).
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.
Equation 4
TANGENT PLANES
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
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.
Equation 5
TANGENT PLANES
Its tangent vector at P0 is:
ru 
x
u
(u 0 , v0 ) i 

y
u
z
u
(u 0 , v0 ) j
(u 0 , v0 ) k
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.
Example 9
TANGENT PLANES
Find the tangent plane to the surface with
parametric equations
x = u2
y = v2
at the point (1, 1, 3).
z = u + 2v
Example 9
TANGENT PLANES
We first compute the tangent vectors:
ru 
rv 
x
u
x
v
i
i
y
u
y
v
j
j
z
u
z
v
k  2u i  k
k  2v j  2 k
Example 9
TANGENT PLANES
Thus, a normal vector to the tangent plane
is:
i
ru  rv  2 u
0
j
k
0
1
2v
2
  2 v i  4 u j  4 uv k
TANGENT PLANES
Example 9
Notice that the point (1, 1, 3) corresponds
to the parameter values u = 1 and v = 1.
 So, the normal vector there is:
–2 i + 4 j + 4 k
TANGENT PLANES
Example 9
Therefore, an equation of the tangent plane
at (1, 1, 3) is:
–2(x – 1) – 4(y – 1) + 4(z – 3) = 0
or
x + 2y – 2z + 3 = 0
TANGENT PLANES
The figure shows the self-intersecting
surface in Example 9 and its tangent plane
at (1, 1, 3).
SURFACE AREA
Now, we define the surface area of
a general parametric surface given by
Equation 1.
SURFACE AREAS
For simplicity, we start by considering
a surface whose parameter domain D
is a rectangle, and we divide it into
subrectangles Rij.
SURFACE AREAS
Let’s choose (ui*, vj*) to be the lower left
corner of Rij.
PATCH
The part Sij of the surface S that corresponds
to Rij is called a patch and has the point Pij
with position vector r(ui*, vj*) as one of its
corners.
SURFACE AREAS
Let
ru* = ru(ui*, vj*)
and
rv* = rv(ui*, vj*)
be the tangent vectors at Pij as given by
Equations 5 and 4.
SURFACE AREAS
The figure shows how the two edges
of the patch that meet at Pij can be
approximated by vectors.
SURFACE AREAS
These vectors, in turn, can be approximated
by the vectors Δu ru* and Δv rv* because
partial derivatives can be approximated by
difference quotients.
 So, we approximate Sij by the parallelogram
determined by the vectors Δu ru* and Δv rv*.
SURFACE AREAS
This parallelogram is shown here.
 It lies in the
tangent plane
to S at Pij.
SURFACE AREAS
The area of this parallelogram is:
| (u r )  (v r ) |  | r  r | u v
*
u
*
v
*
u
*
v
So, an approximation to the area of S is:
m
n
 |r
*
u
i 1
j 1
 r | u v
*
v
SURFACE AREAS
Our intuition tells us that this approximation
gets better as we increase the number of
subrectangles.
Also, we recognize the double sum as
a Riemann sum for the double integral
|
r

r
|
d
u
d
v
u
v

D
 This motivates the following definition.
SURFACE AREAS
Definition 6
Suppose a smooth parametric
surface S is:
 Given by
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.
Definition 6
SURFACE AREAS
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
z
v
k
SURFACE AREAS
Example 10
Find the surface area of a sphere of
radius a.
 In Example 4, we found
x = a sin Φ cos θ, y = a sin Φ sin θ, z = a cos Φ
where the parameter domain is:
D = {(Φ, θ) | 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π)
Example 10
SURFACE AREAS
We first compute the cross product of
the tangent vectors:
r  r

i
j
k
x
y
z



x
y
z



Example 10
SURFACE AREAS

i
j
k
a cos  cos 
a cos  sin 
 a sin 
 a sin  sin 
a sin  cos 
0
 a sin  cos  i  a sin  sin  j
2
2
2
2
 a sin  cos  k
2
Example 10
SURFACE AREAS
Thus,
| r  r |

a sin  cos   a sin  sin   a sin  cos 

a sin   a sin  cos 
a
2
4
4
4
4
2
4
4
2
4
2
2
sin   a sin 
2
2
since sin Φ ≥ 0 for 0 ≤ Φ ≤ π.
4
2
2
Example 10
SURFACE AREAS
Hence, by Definition 6, the area of the sphere
is:
A

dA
|
r

r
|



2
 
0

a sin  d  d 
2
0
D
a
2

2
0
d


sin  d 
0
 a (2  )2  4  a
2
2
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)
Equation 7
GRAPH OF A FUNCTION
Thus,
 f 
ry  j  
k
 y 
 f 
rx  i  
k
 x 
and
i
rx  r y  1
0
j
0
1
k
f
x
f
y

f
x
i
f
y
j k
Equation 8
GRAPH OF A FUNCTION
Thus, we have:
2
2
| rx  r y | 
 f 
 f 
 1

 
 x 
 y 
2

 z 
 z 
1 

 
 x 
 y 
2
Formula 9
GRAPH OF A FUNCTION
Then, the surface area formula in
Definition 6 becomes:
2
A( S ) 

D
2
 z 
 z 
1 
 dA
 
 x 
 y 
GRAPH OF A FUNCTION
Example 11
Find the area of the part of
the paraboloid z = x2 + y2 that lies
under the plane z = 9.
 The plane intersects the paraboloid
in the circle
x2 + y2 = 9, z = 9
GRAPH OF A FUNCTION
Example 11
Therefore, the given surface lies
above the disk D
with center the origin
and radius 3.
Example 11
GRAPH OF A FUNCTION
Using Formula 9, we have:
2
2
A

D


 z 
 z 
1 
 dA
 
 x 
 y 
1   2 x    2 y  dA
2
2
D


D
1  4( x  y ) dA
2
2
Example 11
GRAPH OF A FUNCTION
Converting to polar coordinates,
we obtain:
A
2
 
0


2

6
1  4r r dr d
2
0
d
0
 2

3
 
1
8

3
r 1  4r dr
2
0
2
3
(1  4 r )
(3 7 3 7  1)
2
3/2
3

0
SURFACE AREA
The question remains:
 Is our definition of surface area (Definition 6)
consistent with the surface area formula
from single-variable calculus (Formula 4
in Section 8.2)?
SURFACE AREA
We consider the surface S obtained by
rotating the curve
y = f(x), a ≤ x ≤ b
about the x-axis,
where:
 f(x) ≥ 0.
 f’ is continuous.
SURFACE AREA
From Equations 3, we know that parametric
equations of S are:
x=x
y = f(x) cos θ
a≤x≤b
z = f(x) sin θ
0 ≤ θ ≤ 2π
SURFACE AREA
To compute the surface area of S,
we need the tangent vectors
r x  i  f '( x ) co s  j  f '( x ) sin  k
r   f ( x ) sin  j  f ( x ) co s  k
SURFACE AREA
Thus,
r x  r
i
j
k
 1
f '( x ) cos 
f '( x ) sin 
0
 f ( x ) sin 
f ( x ) cos 
 f ( x ) f '( x ) i  f ( x ) cos  j  f ( x ) sin  k
SURFACE AREA
Hence,
| r x  r |
 [ f ( x )] [ f '( x )]  [ f ( x )] cos   [ f ( x )] sin 
2
2
2
 [ f ( x )] [1  [ f '( x )] ]
2
 f ( x ) 1  [ f ( x )]
2
2
because f(x) ≥ 0.
2
2
2
SURFACE AREA
Thus, the area of S is:
A
 | r
 r | dA
x
D

2
 
0
 2

b
f ( x ) 1  [ f '( x )] dx d 
2
a
b
a
f ( x ) 1  [ f '( x )] dx
2
SURFACE AREA
This is precisely the formula that was used
to define the area of a surface of revolution
in single-variable calculus (Formula 4 in
Section 8.2).