Vector Analysis - Palm Beach State College

MAC2313 Calculus III - Chapter 15

1

Vector Analysis15

Copyright © Cengage Learning. All rights reserved.15.1

Vector Fields

Copyright © Cengage Learning. All rights reserved.

15.1

Objectives

Understand the concept of a vector field.

Determine whether a vector field is conservative.

Find the curl of a vector field.

Find the divergence of a vector field.

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Vector Fields

Functions that assign a vector to a point in the plane or a point in space are called vector fields.

Some common physical examples of vector fields are velocity fields, gravitational fields, and electric force fields.

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The gradient is one example of a vector field.

For example, if

f(x, y, z) = x2 + y2 + z2

then gradient of f

f(x, y, z) = fx(x, y, z)i + fy(x, y, z)j + fz(x, y, z)k

= 2xi + 2yj + 2zk Vector field in space

is a vector field in space.

Vector Fields

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1. Velocity fields describe the motions of systems of particles in the plane or in space.

Example: the vector field determined by a wheel rotating on an axle.

Notice that the velocity vectors are determined by the locations of their initial points—the farther a point is from the axle, the greater its velocity.

Vector Fields

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Velocity fields are also determined by the flow of liquids through a container or by the flow of air currents around a moving object.

Vector Fields

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2. Gravitational fields are defined by Newton’s Law of Gravitation, which states that the force of attraction exerted on a particle of mass m1 located at (x, y, z)

by a particle of mass m2 located at (0, 0, 0) is given by

where G is the gravitational constant and u is the unit vector in the direction from the origin to (x, y, z).

Vector Fields

15.1 8

The gravitational field F has the properties that F(x, y, z) always points toward the origin, and that the magnitude of F(x, y, z) is the same at all points equidistant from the origin.

A vector field with these

two properties is called

a central force field.

Vector Fields

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3. Electric force fields are defined by Coulomb’s Law,

which states that the force exerted on a particle with

electric charge q1 located at (x, y, z) by a particle with

electric charge q2 located at (0, 0, 0) is given by

where r = xi + yj + zk, u = r/||r||, and c is a constant that

depends on the choice of units for ||r||, q1, and q2.

Vector Fields

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Example 1 – Sketching a Vector Field

Sketch some vectors in the vector field given by

F(x, y) = –yi + xj.

Solution:

You could plot vectors at several random points in the plane.

However, it is more enlightening to plot vectors of equal magnitude.

This corresponds to finding level curves in scalar fields.

In this case, vectors of equal magnitude lie on circles.

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Example 1 – Solution

To begin making the sketch, choose a value for c and plot several vectors on the resulting circle.

For instance, the following vectors occur on the unit circle.

These and several other vectors

in the vector field are shown in

Figure 15.4. Figure 15.4

cont’d

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Conservative Vector Fields

Some vector fields can be represented as the gradients of differentiable functions and some cannot—those that can are called conservative vector fields.

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Example 4 – Conservative Vector Fields

The vector field given by F(x, y) = 2xi + yj is conservative.

To see this, consider the potential function

Because

f = 2xi + yj = F

it follows that F is conservative.

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Conservative Vector Fields

The following important theorem gives a necessary and sufficient condition for a vector field in the plane to be conservative.

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Example 5 – Testing for Conservative Vector Fields in the Plane

Decide whether the vector field given by F is conservative.

a. F(x, y) = x2yi + xyj b. F(x, y) = 2xi + yj

Solution:

a. The vector field given by F(x, y) = x2yi + xyj is not conservative because

and

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b. The vector field given by F(x, y) = 2xi + yj is conservative

because

and

Example 5 – Solutioncont’d

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Examples

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Examples

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Examples

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Curl of a Vector Field

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Example 7 – Finding the Curl of a Vector Field

Find curl F of the vector field given by

F(x, y, z) = 2xyi + (x2 + z2)j + 2yzk.

Is F irrotational?

Solution:

The curl of F is given by

curl F(x, y, z) = F(x, y, z)

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Because curl F = 0, F is irrotational.

cont’d

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Examples

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Examples

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Divergence of a Vector Field

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Example 9 – Finding the Divergence of a Vector Field

Find the divergence at (2, 1, –1) for the vector field

Solution:

The divergence of F is

At the point (2, 1, –1), the divergence is

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Example

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Line Integrals


15.2

Objectives

Understand and use the concept of a piecewise smooth curve.

Write and evaluate a line integral.

Write and evaluate a line integral of a vector field.

Write and evaluate a line integral in differential form.

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Example 1 – Finding a Piecewise Smooth Parametrization

Find a piecewise smooth parametrization of the graph of Cshown in Figure 15.7.

Figure 15.7

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Because C consists of three line segments C1, C2, and C3, you can construct a smooth parametrization for each segment and piece them together by making the last t-value in Ci correspond to the first t-value in Ci + 1, as follows.


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So, C is given by

Because C1, C2, and C3 are smooth, it follows that C is piecewise smooth.

Parametrization of a curve induces an orientation to the curve.


15.2 32

Examples: Find parametrization

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Line Integrals

You will study a new type of integral called a line integral

for which you integrate over a piecewise smooth curve C.

Consider the mass of a wire of finite length, given by a curve C in space. The density (mass per unit length) of the wire at the point (x, y, z) is given by f(x, y, z).15.2 36

Line Integrals

Note that if f(x, y, z) = 1, the line integral gives the arc length of the curve C. That is,

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Evaluate

where C is the line segment shown in Figure 15.9.

Example 2 – Evaluating a Line Integral

Figure 15.915.2 38

Begin by writing a parametric form of the equation of the line segment:

x = t, y = 2t, and z = t, 0 ≤ t ≤ 1.

Therefore, x'(t) = 1, y'(t) = 2, and z'(t) = 1, which implies that


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So, the line integral takes the following form.


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Evaluate the line integrals

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Evaluate the line integrals

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Example: Mass of coil

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Line Integrals of Vector Fields

One of the most important physical applications of line integrals is that of finding the work done on an object moving in a force field.

For example, Figure 15.12 shows an inverse square force field similar to the gravitational field of the sun.

Figure 15.1215.2 44


Figure 15.13

To see how a line integral can be used to find work done in a force field F, consider an object moving along a path C in the field, as shown in Figure 15.13.

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Find the work done by the force field

on a particle as it moves along the helix given by

from the point (1, 0, 0) to (–1, 0, 3π).

Example 6 – Work Done by a Force

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Because

r(t) = x(t)i + y(t)j + z(t)k

= cos ti + sin tj + tk

it follows that x(t) = cos t, y(t) = sin t, and z(t) = t.

So, the force field can be written as


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To find the work done by the force field in moving a particle along the curve C, use the fact that

r'(t) = –sin ti + cos tj + k

and write the following.


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Example

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Line Integrals in Differential Form

A second commonly used form of line integrals is derived from the vector field notation used in the preceding section.

If F is a vector field of the form F(x, y) = Mi + Nj, and C is given by r(t) = x(t)i + y(t)j, then F • dr is often written as M dx + N dy.

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Line Integrals in Differential Form

This differential form can be extended to three variables. The parentheses are often omitted, as follows.

15.2 52

Example 8 – Evaluating a Line Integral in Differential Form

Let C be the circle of radius 3 given by

r(t) = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π

as shown in Figure 15.17. Evaluate the line integral

Figure 15.1715.2

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Because x = 3 cos t and y = 3 sin t, you have dx = –3 sin t dtand dy = 3 cos t dt. So, the line integral is

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Example

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Conservative Vector Fields and Independence of Path


15.3

Objectives:

Understand and use the Fundamental Theorem of Line Integrals.

Understand the concept of independence of path.

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Fundamental Theorem of Line Integrals

In a gravitational field the work done by gravity on an object moving between two points in the field is independent of the path taken by the object.

In this section, you will study an important generalization of this result—it is called the Fundamental Theorem of Line Integrals.

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Find the work done by the force field

on a particle that moves from (0, 0) to (1, 1) along each path, as shown in Figure 15.19.

a. C1: y = x b. C2: x = y2 c. C3: y = x3

Example 1 – Line Integral of a Conservative Vector Field

Figure 15.1915.3

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Find the work done by the force field from (0, 0) to (1, 1)

a. C1: y = x b. C2: x = y2 c. C3: y = x3


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a. C1: y = x


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c. C3: y = x3


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Example 1(a) – Solution

Let r(t) = ti + tj for 0 ≤ t ≤ 1,

so that

Then, the work done is

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Let r(t) for 0 ≤ t ≤ 2,

so that


So, the work done by a conservative vector field is the same for all paths.

Example 1(c) – Solutioncont’d

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b. C2: x = y2


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Let r(t) = ti + for 0 ≤ t ≤ 1,

so that


Example 1(b) – Solutioncont’d

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Verify that F is conservative.

So the potential function is

So an alternate way of evaluating the line integral is to evaluate the potential function at the end point and the start point. The line integral is the difference.


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In space:

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Example 2 – Using the Fundamental Theorem of Line Integrals

Evaluate where

C is a piecewise smooth

curve from (−1, 4) to (1, 2)

and F(x, y) = 2xyi + (x2 − y)j

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You know that F is the gradient of f

where

Consequently, F is conservative, and by the Fundamental

Theorem of Line Integrals, it follows that


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Independence of Path

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Example 4 – Finding Work in a Conservative Force Field

For the force field given by

F(x, y, z) = ex cos yi − ex sin yj + 2k

show that is independent of path, and calculate the work done by F on an object moving along a curve C from (0, /2, 1) to (1, , 3).

Solution:

Writing the force field in the form F(x, y, z) = Mi + Nj + Pk,

you have

M = ex cos y, N = –ex sin y, and P = 2,

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and it follows that

So, F is conservative.

If f is a potential function of F, then


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By integrating with respect to x, y, and z separately, you obtain

By comparing these three versions of f(x, y, z), you can conclude that

f(x, y, z) = ex cos y + 2z + K.


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Therefore, the work done by F along any curve C from

(0, /2, 1) to (1, , 3) is

cont’d

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Evaluate , where

F(x, y) = (y3 + 1)i + (3xy2 + 1)j

and C1 is the semicircular path

from (0, 0) to (2, 0)

Options for solving:

1. Evaluate a line integral using a parametrization for C1

2. If F is conservative, evaluate the potential function at the end and start points.

3. If F is conservative, take a different path.15.3 76


Evaluate , where

F(x, y) = (y3 + 1)i + (3xy2 + 1)j


from (0, 0) to (2, 0)


1. Evaluate a line integral using a parametrization for C1

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a. Use the method presented in the preceding section to evaluate the line integral along the given curve.

Use r(t) = (1 − cos t)i + sin tj, where 0 ≤ t ≤ dr = r′(t) dt = (sin ti + cos tj) dt,

This integral should dampen your enthusiasm for this option.

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Evaluate , where

F(x, y) = (y3 + 1)i + (3xy2 + 1)j


from (0, 0) to (2, 0)


2. If F is conservative, evaluate the potential function at the end and start points.

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b. Verify that F is conservative. Find a potential functionand evaluate the line integral by the Fundamental Theorem of Line Integrals.

The potential function is f(x, y) = xy3 + x + y + K,

and, by the Fundamental Theorem,

cont’dExample 5 – Solution

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Evaluate , where

F(x, y) = (y3 + 1)i + (3xy2 + 1)j


from (0, 0) to (2, 0)


3. If F is conservative, take a different path.

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c. Knowing that F is conservative, you have a third option. Because the value of the line integral is independent of path, you can replace the semicircular path with a simpler path.

Suppose you choose the straight-line path C2 from (0, 0) to (2, 0).

Then, r(t) = ti, where 0 ≤ t ≤ 2.


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So, dr = i dt and F(x, y) = (y3 + 1)i + (3xy2 + 1)j = i + j,

so that

Of the three options, obviously the third one is the easiest.

cont’dExample 5 – Solution

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Conservation of Energy

The English physicist Michael Faraday wrote, “Nowhere is there a pure creation or production of power without a corresponding exhaustion of something to supply it.”

This statement represents the first formulation of one of the most important laws of physics—the Law of Conservation of Energy.

In modern terminology, the law is stated as follows: In a conservative force field, the sum of the potential and kinetic energies of an object remains constant from point to point.

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Green’s Theoremnamed after the English mathematician George Green (1793–1841).


15.4

Objectives:

Use Green’s Theorem to evaluate a line integral.

Use alternative forms of Green’s Theorem.


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Example 1

Evaluate the line integral

where C is the path from

(0, 0) to (1, 1) along the

graph of y = x3 and from

(1, 1) to (0, 0) along the

graph of y = x.

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Green’s Theorem

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Green’s Theorem

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Example 1 – Using Green’s Theorem

Use Green’s Theorem to

evaluate the line integral

where C is the path from

(0, 0) to (1, 1) along the

graph of y = x3 and from

(1, 1) to (0, 0) along the

graph of y = x.

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Because M = y3 and N = x3 + 3xy2, it follows that

Applying Green’s Theorem, you then have

90



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Example

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Parametric Surfaces


15.5Objectives:

Understand the definition of a parametric surface, and sketch the surface.

Find a set of parametric equations to represent a surface.

Find a normal vector and a tangent plane to a parametric surface.

Find the area of a parametric surface.15.5

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Parametric Surfaces

Recall the vector valued functions that define a path:r(t) = x(t)i + y(t)j Plane curve

r(t) = x(t)i + y(t)j + z(t)k Space curve

For curves, the vector-valued function r is a function of a single parameter t. For surfaces, the vector-valued function is a function of two parameters u and v.

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If S is a parametric surface given by the vector-valued function r, then S is traced out by the position vector r(u, v) as the point (u, v) moves throughout the domain D.

Parametric Surfaces

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Identify and sketch the parametric surface S given by

r(u, v) = 3 cos ui + 3 sin uj + vk

where 0 u 2 and 0 v 4.

Solution:

Because x = 3 cos u and y = 3 sin u, you know that for each point (x, y, z) on the surface, x and y are related by the equation x2 + y2 = 32.

Example 1 – Sketching a Parametric Surface

15.5 96

In other words, each cross section of S taken parallel to the xy-plane is a circle of radius 3, centered on the z-axis.

Because z = v, where 0 v 4, you can see that the surface is a right circular cylinder of height 4.

The radius of the cylinder is 3,

and the z-axis forms the

axis of the cylinder, as shown

in Figure 15.36


Figure 15.3615.5


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Examples:

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Examples:

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Examples:

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Examples:

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Write a set of parametric equations for the cone given by

Example 3 – Representing a Surface Parametrically

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Because this surface is given in the form z = f(x, y), you can let x and y be the parameters.

Then the cone is represented by the vector-valued function

where (x, y) varies over the entire xy-plane.


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103

Let S be a parametric surface given byr(u, v) = x(u, v)i + y(u, v)j + z(u, v)k

The partial derivatives of r with respect to u and v are defined as

and

Each can be interpreted geometrically in terms of tangent vectors.

Normal Vectors and Tangent Planes

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If the normal vector ru rv is not 0 for any (u, v) in D, the surface S is called smooth and will have a tangent plane.

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Find an equation of the tangent plane to the paraboloid given by

r(u, v) = ui + vj + (u2 + v2)k

at the point (1, 2, 5).

Solution:

The point in the uv-plane that is mapped to the point

(x, y, z) = (1, 2, 5) is (u, v) = (1, 2).

The partial derivatives of r are

ru = i + 2uk and rv = j + 2vk.

Example 5 – Finding a Tangent Plane to a Parametric Surface

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The normal vector is given by

ru rv = = –2ui – 2vj + k

which implies that the normal vector at (1, 2, 5) is ru rv = –2i – 4j + k.

So, an equation of the tangent plane at (1, 2, 5) is

–2(x – 1) – 4(y – 2) + (z – 5) = 0

–2x – 4y + z = –5.


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The tangent plane is shown in Figure 15.41.


Figure 15.4115.5


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Example:

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Area of a Parametric Surface

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Area of a Parametric Surface

This implies that the surface area of S is

Surface area

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Find the surface area of the unit sphere given by

r(u, v) = sin u cos vi + sin u sin vj + cos uk

where the domain D is given by 0 u and 0 v 2.

Solution:

Begin by calculating ru and rv.

ru = cos u cos vi + cos u sin vj – sin uk

rv = –sin u sin vi + sin u cos vj

Example 6 – Finding Surface Area

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The cross product of these two vectors is

ru rv =

= sin2 u cos vi + sin2 u sin vj + sin u cos uk

which implies that

ru rv

sin u > 0 for 0 u


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Finally, the surface area of the sphere is


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ExampleFind the surface area of

r(u, v) = 4u cos v i + 4u sin vj + u2 k

where the domain D is given by

0 u 2 and 0 v 2.

15.5 116

Surface Integrals


15.6Objectives:

Evaluate a surface integral as a double integral.

Evaluate a surface integral for a parametric surface.

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Surface Integrals

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Example 1 – Evaluating a Surface Integral

Evaluate the surface integral

where S is the first-octant portion of the plane

2x + y + 2z = 6.

Solution:

Begin by writing S as

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Using the partial derivatives and

you can write

Figure 15.45


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Using Figure 15.45 and Theorem 15.10, you obtain

cont’d


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An alternative solution to Example 1 would be to project Sonto the yz-plane, as shown in Figure 15.46.

Then,

and

Figure 15.46


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So, the surface integral is


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Example

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If the function f defined on the surface S is simply

f(x, y, z) = 1, the surface integral yields the surface area of S.

Area of surface =

If S is a lamina of variable density and ρ(x, y, z) is the density at the point (x, y, z), then the mass of the lamina is given by

Mass of lamina =

Surface Integrals

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Parametric Surfaces and Surface Integrals

For a surface S given by the vector-valued function

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k Parametric surface

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Example:


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Example 4 – Evaluating Surface Integral

Evaluate the surface integral

in parametric form where S is the first-octant portion of the cylinder y2 + z2 = 9 between x = 0 and x = 4.

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In parametric form, the surface is given by

r(x, θ) = xi + 3 cos θj + 3 sin θk

where 0 ≤ x ≤ 4 and 0 ≤ θ ≤ /2.

To evaluate the surface integral in parametric form, begin by calculating the following.

rx = i

rθ = – 3 sin θj + 3 cos θk


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So, the surface integral can be evaluated as follows.


Vector Analysis - Palm Beach State College

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