VECTOR CALCULUS 17. 2 VECTOR CALCULUS So far, we have considered special types of surfaces: Cylinders Quadric surfaces Graphs of functions of two.

VECTOR CALCULUSVECTOR CALCULUS

17

2

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

3

Here, we use vector functions to

describe more general surfaces, called

parametric surfaces, and compute their

areas.

VECTOR CALCULUS

4

Then, we take the general surface

area formula and see how it applies

to special surfaces.

VECTOR CALCULUS

5

17.6Parametric Surfaces

and their Areas

VECTOR CALCULUS

In this section, we will learn about:

Various types of parametric surfaces

and computing their areas using vector functions.

6

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.

7

INTRODUCTION

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.

Equation 1

8

INTRODUCTION

So x, y, and z—the component functions

of r—are functions of the two variables

u and v with domain D.

9

PARAMETRIC SURFACE

The set of all points (x, y, z) in 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.

° 3

Equations 2

10

PARAMETRIC SURFACES

Each choice of u and v gives a point

on S.

By making all choices, we get all of S.

11

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.

Fig. 17.6.1, p. 1106

12

PARAMETRIC SURFACES

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

Example 1

13

PARAMETRIC SURFACES

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.

Example 1

14

PARAMETRIC SURFACES

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.

Example 1

Fig. 17.6.2, p. 1106

15

PARAMETRIC SURFACES

In Example 1, we placed no restrictions

on the parameters u and v.

So, we obtained the entire cylinder.

16

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

17

PARAMETRIC SURFACES

In that case, we get the quarter-cylinder

with length 3.

Fig. 17.6.3, p. 1107

18

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.

19

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.

Fig. 17.6.4, p. 1107

20

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.

Fig. 17.6.4, p. 1107

21

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.

22

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.

23

GRID CURVES

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?

Example 2

24

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

Fig. 17.6.5, p. 1107

25

GRID CURVES

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

Example 2

26

GRID CURVES

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 14.1

Example 2

27

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

Fig. 17.6.5, p. 1107

28

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.

Example 2

29

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.

30

PARAMETRIC REPRESENTATIONS

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.

Example 3

31


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:

= ua + vb

Example 3

0P P��

32


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 13.2

Example 3

Fig. 17.6.6, p. 1108

33


If r is the position vector of P,

then

So, the vector equation of the plane can be written as: r(u, v) = r0 + ua + vb

where u and v are real numbers.

Example 3

0 0 0OP P P u v r r a b��

34


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

Example 3

35


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).

Example 4

36


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.

Example 4

37


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]

Example 4

38


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.

Example 4

39

APPLICATIONS—COMPUTER GRAPHICS

One of the uses of

parametric surfaces is

in computer graphics.

40

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.

Fig. 17.6.7, p. 1108

41

COMPUTER GRAPHICS

Part of the sphere appears to be missing

because of the rectangular grid system

used by the computer.

Fig. 17.6.7, p. 1108

42

COMPUTER GRAPHICS

The much better picture here was produced

by a computer using the parametric equations

found in Example 4.

Fig. 17.6.8, p. 1108

43


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.

Example 5

44


Then the parametric equations of

the cylinder are

x = 2 cos θ y = 2 sin θ z = z

where: 0 ≤ θ ≤ 2π 0 ≤ z ≤ 1

Example 5

45


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

Example 6

46


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)

47

PARAMETRIZATIONS

Parametric representations (also

called parametrizations) of surfaces

are not unique.

The next example shows two ways to parametrize a cone.

48

PARAMETRIZATIONS

Find a parametric representation for

the surface

that is,

the top half of the cone

z2 = 4x2 + 4y2

Example 7

2 22z x y

49

PARAMETRIZATIONS

One possible representation is obtained

by choosing x and y as parameters:

x = x y = y

So, the vector equation is:

E. g. 7—Solution 1

2 22z x y

2 2( , ) 2x y x y x y r i j k

50

PARAMETRIZATIONS

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


2 22 2z x y r

51

PARAMETRIZATIONS

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


52

PARAMETRIZATIONS

For some purposes, the parametric

representations in Solutions 1 and 2 are

equally good.

In certain situations, though, Solution 2

might be preferable.

53

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

54

SURFACES OF REVOLUTION

Surfaces of revolution can be

represented parametrically and thus

graphed using a computer.

55


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.

56


Let θ be the angle of rotation

as shown.

Fig. 17.6.9, p. 1110

57


If (x, y, z) is a point on S,

then

x = x

y = f(x) cos θ

z = f(x) sin θ

Equations 3

Fig. 17.6.9, p. 1110

58


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

59


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

60


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

Example 8

61


Using a computer to plot these equations

and rotate the image, we obtain this graph.

Example 8

Fig. 17.6.10, p. 1110

62


We can adapt Equations 3 to represent

a surface obtained through revolution

about the y- or z-axis.

See Exercise 30.

63

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).

64

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.

Fig. 17.6.11, p. 1110

65

TANGENT PLANES

The tangent vector to C1 at P0 is obtained

by taking the partial derivative of r with

respect to v:

0 0 0 0 0 0( , ) ( , ) ( , )v

x y zu v u v u v

v v v

r i j k

Equation 4

66

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.

Fig. 17.6.11, p. 1110

67

TANGENT PLANES

Its tangent vector at P0 is:

Equation 5

0 0 0 0

0 0

( , ) ( , )

( , )

u

x yu v u v

u uzu v

u

r i j

k

68

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.

69

TANGENT PLANES

Find the tangent plane to the surface with

parametric equations

x = u2 y = v2 z = u + 2v

at the point (1, 1, 3).

Example 9

70

TANGENT PLANES

We first compute the tangent vectors:

Example 9

2

2 2

u

v

x y zu

u u u

x y zv

v v v

r i j k i k

r i j k j k

71

TANGENT PLANES

Thus, a normal vector to the tangent plane

is:

Example 9

2 0 1

0 2 2

2 4 4

u v u

v

v u uv

i j k

r r

i j k

72

TANGENT PLANES

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

Example 9

73

TANGENT PLANES

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

Example 9

74

TANGENT PLANES

The figure shows the self-intersecting

surface in Example 9 and its tangent plane

at (1, 1, 3).

Fig. 17.6.12, p. 1111

75

SURFACE AREA

Now, we define the surface area of

a general parametric surface given by

Equation 1.

76

SURFACE AREAS

For simplicity, we start by considering

a surface whose parameter domain D

is a rectangle, and we divide it into

subrectangles Rij.

77

SURFACE AREAS

Let’s choose (ui*, vj*) to be the lower left

corner of Rij.

Fig. 17.6.13, p. 1111

78

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.

Fig. 17.6.13, p. 1111

79

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.

80

SURFACE AREAS

The figure shows how the two edges

of the patch that meet at Pij can be

approximated by vectors.

Fig. 17.6.14a, p. 1112

81

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*.

82

SURFACE AREAS

This parallelogram is shown here.

It lies in the tangent plane to S at Pij.

Fig. 17.6.14b, p. 1112

83

SURFACE AREAS

The area of this parallelogram is:

So, an approximation to the area of S is:

* * * *| ( ) ( ) | | |u v u vu v u v r r r r

* *

1 1

| |m n

u vi j

u v

r r

84

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

This motivates the following definition.

| |u v

D

r r du dv

85

SURFACE AREAS

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

86

SURFACE AREAS

Then, the surface area of S is

where:

Definition 6

( ) | |u v

D

A S dA r r

u v

x y z x y z

u u u v v v

r i j k r i j k

87

SURFACE AREAS

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

88

SURFACE AREAS

We first compute the cross product of

the tangent vectors:

Example 10

x y z

x y z

r r

i j k

89

SURFACE AREAS

Example 10

2 2 2 2

2

cos cos cos sin sin

sin sin sin cos 0

sin cos sin sin

sin cos

a a a

a a

a a

a

i j k

i j

k

90

SURFACE AREAS

Thus,

since sin Φ ≥ 0 for 0 ≤ Φ ≤ π.

Example 10

4 4 2 4 4 2 4 2 2

4 4 4 2 2

2 2 2

| |

sin cos sin sin sin cos

sin sin cos

sin sin

r r

a a a

a a

a a

91

SURFACE AREAS

Hence, by Definition 6, the area of the sphere

is:

Example 10

2 2

0 0

22

0 0

2 2

| | sin

sin

(2 )2 4

D

A r r dA a d d

a d d

a a

92

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)

93

GRAPH OF A FUNCTION

Thus,

and

Equation 7

1 0

0 1

x y

f f f

x x yf

y

i j k

r r i j k

x y

f f

x y

r i k r j k

94

GRAPH OF A FUNCTION

Thus, we have:

Equation 8

22

22

| | 1

1

x y

f f

x y

z z

x y

r r

95

GRAPH OF A FUNCTION

Then, the surface area formula in

Definition 6 becomes:

Formula 9

22

( ) 1D

z zA S dA

x y

96

GRAPH OF A FUNCTION

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

Example 11

97

GRAPH OF A FUNCTION

Therefore, the given surface lies

above the disk D

with center the origin

and radius 3.

Example 11

Fig. 17.6.15, p. 1113

98

GRAPH OF A FUNCTION

Using Formula 9, we have:

Example 11

22

2 2

2 2

1

1 2 2

1 4( )

D

D

D

z zA dA

x y

x y dA

x y dA

99

GRAPH OF A FUNCTION

Converting to polar coordinates,

we obtain:

Example 11

2 3 2

0 0

2 3 2

0 0

32 3/ 21 28 3 0

1 4

1 4

2 (1 4 )

(37 37 1)6

A r r dr d

d r r dr

r

100

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 9.2)?

101

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.

102

SURFACE AREA

From Equations 3, we know that parametric

equations of S are:

x = x y = f(x) cos θ z = f(x) sin θ

a ≤ x ≤ b 0 ≤ θ ≤ 2π

103

SURFACE AREA

To compute the surface area of S,

we need the tangent vectors

'( ) cos '( )sin

( )sin ( ) cos

x f x f x

f x f x

r i j k

r j k

104

SURFACE AREA

Thus,

1 '( ) cos '( )sin

0 ( )sin ( )cos

( ) '( ) ( ) cos ( )sin

x

f x f x

f x f x

f x f x f x f x

r r

i j k

i j k

105

SURFACE AREA

Hence,

because f(x) ≥ 0.

2 2 2 2 2 2

2 2

2

| |

[ ( )] [ '( )] [ ( )] cos [ ( )] sin

[ ( )] [1 [ '( )] ]

( ) 1 [ ( )]

x

f x f x f x f x

f x f x

f x f x

r r

106

SURFACE AREA

Thus, the area of S is:

2 2

0

2

| |

( ) 1 [ '( )]

2 ( ) 1 [ '( )]

x

D

b

a

b

a

A dA

f x f x dx d

f x f x dx

r r

107

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 9.2).

VECTOR CALCULUS 17. 2 VECTOR CALCULUS So far, we have considered special types of surfaces: Cylinders Quadric surfaces Graphs of functions of two.

Documents