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MAC2313 Calculus III - Chapter 15
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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.
15.1
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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.
15.1 4
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
15.1 10
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
15.1 12
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
15.1 16
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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Example 7 – Solution
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
15.1 28
Line Integrals
Copyright © Cengage Learning. All rights reserved.
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
15.2 30
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.
Example 1 – Solution
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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.
Example 1 – Solutioncont’d
15.2 32
Examples: Find parametrization
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Examples: Find parametrization
15.2 34
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,
15.2
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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
Example 2 – Solution
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So, the line integral takes the following form.
Example 2 – Solutioncont’d
15.2 40
Evaluate the line integrals
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Evaluate the line integrals
15.2 42
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
Line Integrals of Vector Fields
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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Line Integrals of Vector Fields
15.2 46
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
15.2
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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
Example 6 – Solution
15.2 48
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.
Example 6 – Solutioncont’d
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Example
15.2 50
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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Example 8 – Solution
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
15.2 54
Example 8 – Solutioncont’d
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Example
15.2 56
Conservative Vector Fields and Independence of Path
Copyright © Cengage Learning. All rights reserved.
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.
15.3 58
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
Example 1 – Line Integral of a Conservative Vector Field
15.3 60
Find the work done by the force field from (0, 0) to (1, 1)
a. C1: y = x
Example 1 – Line Integral of a Conservative Vector Field
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Find the work done by the force field from (0, 0) to (1, 1)
c. C3: y = x3
Example 1 – Line Integral of a Conservative Vector Field
15.3 62
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
Then, the work done is
So, the work done by a conservative vector field is the same for all paths.
Example 1(c) – Solutioncont’d
15.3 64
Find the work done by the force field from (0, 0) to (1, 1)
b. C2: x = y2
Example 1 – Line Integral of a Conservative Vector Field
15.3
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Let r(t) = ti + for 0 ≤ t ≤ 1,
so that
Then, the work done is
Example 1(b) – Solutioncont’d
15.3 66
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.
Fundamental Theorem of Line Integrals
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Fundamental Theorem of Line Integrals
In space:
15.3 68
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
Example 2 – Solution
15.3 70
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,
15.3 72
and it follows that
So, F is conservative.
If f is a potential function of F, then
Example 4 – Solutioncont’d
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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.
Example 4 – Solutioncont’d
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Example 4 – Solution
Therefore, the work done by F along any curve C from
(0, /2, 1) to (1, , 3) is
cont’d
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Example 5 – Evaluating a Line Integral
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
Example 5 – Evaluating a Line Integral
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
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Example 5 – Solution
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.
15.3 78
Example 5 – Evaluating a Line Integral
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:
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
15.3 80
Example 5 – Evaluating a Line Integral
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:
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.
Example 5 – Solutioncont’d
15.3 82
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.
15.3 84
Green’s Theoremnamed after the English mathematician George Green (1793–1841).
Copyright © Cengage Learning. All rights reserved.
15.4
Objectives:
Use Green’s Theorem to evaluate a line integral.
Use alternative forms of Green’s Theorem.
MAC2313 Calculus III - Chapter 15
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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
87
Green’s Theorem
88
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.
89
Example 1 – Solution
Because M = y3 and N = x3 + 3xy2, it follows that
Applying Green’s Theorem, you then have
90
Example 1 – Solutioncont’d
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Example
92
Parametric Surfaces
Copyright © Cengage Learning. All rights reserved.
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.
15.5 94
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
Example 1 – Solutioncont’d
Figure 15.3615.5
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Examples:
15.5 98
Examples:
15.5
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Examples:
15.5 100
Examples:
15.5
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Write a set of parametric equations for the cone given by
Example 3 – Representing a Surface Parametrically
15.5 102
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.
Example 3 – Solution
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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
15.5 104
Normal Vectors and Tangent Planes
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.
15.5
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Normal Vectors and Tangent Planes
15.5 106
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
15.5
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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.
Example 5 – Solutioncont’d
15.5 108
The tangent plane is shown in Figure 15.41.
Example 5 – Solutioncont’d
Figure 15.4115.5
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Example:
15.5 110
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
15.5 112
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
15.5
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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
Example 6 – Solutioncont’d
15.5 114
Finally, the surface area of the sphere is
Example 6 – Solutioncont’d
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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
Copyright © Cengage Learning. All rights reserved.
15.6Objectives:
Evaluate a surface integral as a double integral.
Evaluate a surface integral for a parametric surface.
117
Surface Integrals
118
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
119
Using the partial derivatives and
you can write
Figure 15.45
Example 1 – Solutioncont’d
120
Example 1 – Solution
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
Example 1 – Solutioncont’d
122
So, the surface integral is
Example 1 – Solutioncont’d
123
Example
124
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
126
Example:
MAC2313 Calculus III - Chapter 15
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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.
128
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
Example 4 – Solution
129
So, the surface integral can be evaluated as follows.
Example 4 – Solutioncont’d