Section 16.2 Line Integrals - Emory Universitympcarr/math211/print/ch16.pdf · Parameterizations...

Section 16.2 Line Integrals

Section 16.2

Line Integrals

Goals:

Compute line integrals of multi variable functions.

Compute line integrals of vector functions.

Interpret the physical implications of a line integral.

Multivariable Calculus 1 / 130


Types of Integrals

We have integrated a function over

The real number line∫ ba f (x)dx

The plane∫∫D f (x , y)dA

Three space∫∫∫R f (x , y , z)dV



Line Integrals

In this chapter we will integrate a function over a curve (in either two orthree dimensions, though more are possible).

A two-variable function f (x , y) over a plane curve r(t).



Parameterizations and the Line Integral

The naive approach to integrating a function f (x , y) over a curver(t) = x(t)i + y(t)j would be to plug in x(t) and y(t). Now we areintegrating a function of t.

But this is dependent on our choice of parameterizations. x(2t)i + y(2t)jdefines the same curve but moves twice as fast. Integrating thiscomposition would give half the ∆t and half the area:



Integrating Independent of Parameterization

Instead, our variable of integration should measure somethinggeometrically inherent to the curve:



Arc Length

A more geometrically relevant measure is to integrate with respect todistance traveled. The rectangles approximate the (signed) area betweenC and the graph z = f (x , y).



Integrating with Respect to Arc Length

Area ≈∑

f (x∗i , y∗i )√

∆x2 + ∆y2

=∑

f (x(t∗i ), y(t∗i ))

√(∆x

∆t

)2

+

(∆y

∆t

)2

∆t

∆t→0−−−−→∫

f (x(t), y(t))

√(dx

dt

)2

+

(dy

dt

)2

dt

Alternately:

∫f (r(t))|r′(t)|dt



Integrating with Respect to x or y

We can also integrate with respect to change in just x or just y .∑f (x∗i , y

∗i )∆x =

∑f (x(t∗i ), y(t∗i ))

∆x

∆t∆t

∆t→0−−−−→∫

f (x(t), y(t))x ′(t)dt



Comparison to Integrating a Vector Function

Remark

A plane curve C is defined by a vector function r(t), but we should notconfuse integrating r(t) itself with integrating a function f (x , y) over C .These two types of integrals are almost completely unrelated.

We visualize

∫ b

ar(t)dt as the total displacement achieved by

traveling with velocity r(t) (a vector).

We visualize

∫Cf (x , y)ds as the area between the graph of f and the

curve C (a number).



Applications

We defined∫C fds as an area. It can also be useful for integrating any

function that is a rate with respect to distance:

Example

Over varied terrain, if p(x , y) gives the price per mile to build railroadtracks at point (x , y), then

∫C p(x , y)ds gives the total cost to construct a

railroad following C .

Example

Over varied terrain, if f (x , y) gives the fuel consumption per mile traveledat the point (x , y), then

∫C f (x , y)ds gives the total fuel consumption to

travel along C .



Example 1

Let C be the line segment from (0, 0) to (3, 4). Letf (x , y) = x2 + cos(πy). Compute the line integral∫

Cf (x , y)ds.



Parameterizations to Know

In Chapter 13 we learned parameterizations for the following plane curves:

A line segment from A to B

A circle of radius a

The graph of an explicit function y = f (x)



Example 2

What does integrating∫C 1ds compute?

Formula∫C

1ds = arc length× height

= arc length



Example 3

Calculate the arc length of r(t) = (t2 − t)i + 23 (2t)3/2j on the interval

0 ≤ t ≤ 4.



Summary Questions

Why do we convert to a different differential when setting up a lineintegral?

What does ds mean? What is its differential in terms of dt?

How do we compute arc length?


Section 16.1 Vector Fields

Section 16.1

Vector Fields

Goals:

Recognize real world phenomena that are modeled by vector fields.

Determine the geometric behavior of a vector field from its equation.

Compute line integrals of a vector field over a curve.



The Definition of a Vector Field

Definition

A vector field in R2 is a function that assigns a two-dimensional vectorF(x , y) to each point in R2.

A vector field in R3 is a function that assigns a three-dimensional vectorF(x , y , z) to each point in R3.



Drawing a Vector Field

We draw a vector field by attaching the vectors F(x , y) to the points(x , y) by the tail. For obvious reasons, we only draw these vectors from afinite set of points.



Notation for a Vector Field

A vector field is defined by component functions P and Q (and R inthree dimensions)

F(x , y) = P(x , y)i + Q(x , y)j

or F(x , y) = 〈P(x , y),Q(x , y)〉 .

F(x , y , z) = (0.2x + 0.04y)i + (0.03z − 0.1)j + 0.2 sin(xz)k



Examples of Vector Fields

Wind speed at each point on the ground.

The force exerted by gravity (or magnetism or charge) at each pointin space.

The gradient of a differentiable function.



Example 1

Sketch the two dimensional vector field F(x , y) = y2 i− 1

2 j.



Example 2

How can we visualize the vector field

F(x , y) =x i + y j√x2 + y2

?



Example 3

How can we visualize the vector field

F(x , y) = −y i + x j?



Exercise

Consider the vector field

F =−x i− y j

(x2 + y2)3/2.

1 What direction do the vectors of this field point?

2 How do the lengths depend on the location of (x , y)?

3 Can you sketch the field?



Dot Products and Work

The dot product measures the angle of two vectors, as well as theirmagnitude.

F · s = |F||s| cos θ

This models the work done by a force F on a displacement s, since onlyFproj contributes to work.

W = Fproj · s = F · s



Line Integrals of Vector Fields

The formula W = F · s assumes that F is constant, and the displacement sis along a straight line. A vector field introduces the possibility that F isdifferent at different points. To compute the work done by a vector field,we use an integral.

Definition

The line integral of the vector field F(x , y) over the vector function r(t) isdefined: ∫

CF · dr =

∫F(r(t)) · r′(t)dt

This is sometimes called a work integral.

We’ll see later that line integral of a vector field has applications beyondphysics.



Alternate Notation

If F(x , y) = P(x , y)i + Q(x , y)j we can rewrite this integral without avector operation:

∫C

F(x , y) · dr =

∫C〈P,Q〉 ·

⟨x ′(t), y ′(t)

⟩dt

=

∫CPx ′(t)dt + Qy ′(t)dt

=

∫CPdx + Qdy

Notation

The line integral of a vector field can also be written:∫CP(x , y)dx +

∫CQ(x , y)dy .



Example 4

Suppose an object travels once anticlockwise around the unit circle and isacted on by a force field F(x , y) = y

2 i− 12 j. Does F do positive or negative

work on the object? Calculate the total work done.



Independence of Parameterization

Theorem

If C1 and C2 are two parameterizations of the same curve then∫C1

F(x , y)dr =

∫C2

F(x , y)dr

To prove this, let C1 be given by r1(t). Then there is some function u(t)such that C2 is given by r2(t) = r1(u(t)). The proof is a u-substitution.

This makes physical sense, because work does not care about speed, onlydisplacement. Thus traveling more quickly or slowly along a curve will notchange the total work done.



Summary of Line Integrals

NotationGeometricInterpretation

Calculation∫Cf (x , y) ds Area below the graph

∫ b

a

f (x(t), y(t))√

(x ′(t))2 + (y ′(t))2dt∫Cf (x , y) dx

Area when projectedonto xz plane

∫ b

af (x(t), y(t))x ′(t)dt∫

CF(x , y) · dr Work done by F

∫ b

aF(x(t), y(t)) ·

⟨x ′(t), y ′(t)

⟩dt

or

∫ b

aP(x(t), y(t))x′(t) + Q(x(t), y(t))y′(t) dt∫ b

ar(t) dt

Not a line integral,computes change inposition from velocity

∫ b

ax(t)i + y(t)j dt



Summary Questions

How do we represent a vector field, graphically?

What does a line integral of a vector field measure?

How can we see whether a vector field is doing positive or negativework on a path?

What does dr mean? What is its differential in terms of dt?


Section 16.3 The Fundamental Theorem for Line Integrals

Section 16.3

The Fundamental Theorem for Line Integrals

Goals:

Use the fundamental theorem to evaluate line integrals ofconservative vector fields.

Determine when a vector field is conservative.



Choosing a Path

We asserted previously that two parameterizations of the same curve orvector function yield equal line integrals. However, changing the course ofthe curve will usually change the value of the integral, even if the startingand ending points are left the same.



Example 1/Exercise

Consider the vector field F(x , y) = ∇f (x , y), where f (x , y) =x2 + y2

4. If

A = (−4, 0) and B = (4, 0), what is the work done by F traveling from Ato B along:

1 A line segment?

2 A semicircle of radius 4?



Line Integrals of Gradient Fields

Gradient fields have the following property.

Theorem (The Fundamental Theorem of Line Integrals)

If F = ∇f and C travels from point A to point B, then∫C

F · dr = f (B)− f (A)



Proof of the Fundamental Theorem

F = ∇f =∂f

∂xi +

∂f

∂yj

dr = r′(t)dt =

(dx

dti +

dy

dtj

)dt

∫C

F · dr =

∫ b

a

∂f

∂x

dx

dt+∂f

∂y

dy

dtdt

=

∫ b

a

d

dtf (r(t))dt

= f (r(b))− f (r(a))

= f (B)− f (A)



Visualizing the Fundamental Theorem

By our previous calculation, F · r′(t) computes the rate of change off (r(t)) with respect to t. This can be realized as the change in height ofthe graph z = f (x , y).



Conservative Vector Fields

Definition

A vector field is conservative if line integrals depend only the endpointsof the curve.

Theorem

For a vector field F = P i + Qj on a simply connected (no holes) domain,the following are equivalent (if one is true, the others are true).

1 F is conservative.

2 F = ∇f for some function f .

3∫C F · dr = 0 for all closed curves (start and end at same pt).

4 Py = Qx

The function f is sometimes called a potential function for F.



Proving the Equivalence

Here’s an outline of how we’d prove the theorem.

F conservative F = ∇f

Closed curves integrate to 0 Py = Qx

define f (x , y) =

∫ (x,y)

(0,0)

F · dr

FTLI

compare toconstant curve

combine C1 anda backward C2

fxy = fyxhard



Examples of Conservative Vector Fields

There are a couple cases of conservative fields that are easy to recognize:

A constant field F(x , y) = ai + bj.

A sum of the form F(x , y) = P(x)i + Q(y)j.

What is the potential function for each of these?



Example 2

1 Is F = (3x2 − 2xy)i + (cos y − x2)j conservative?

2 What is its potential function?

3 If C is a path from (1, 0) to (3, 0), what is∫C F · dr?



Exercise

For each vector field, determine whether it is conservative. If it is, find apotential function.

1 F1 = (√xy − y)i + (

√xy − x)j (for x , y ≥ 0)

2 F2 = (ey + 2x)i + (xey − 4y3)j



Summary Questions

What does is mean for a vector field to be conservative?

What is the relationship between the gradient and a conservativevector field?

How do we test that a vector field is conservative?

What does the Fundamental Theorem of Line Integrals say?


Section 16.4 Green’s Theorem

Section 16.4

Green’s Theorem

Goals:

Use Green’s Theorem to replace a line integral with a double integralor vice versa.



Conclusions from Line Integrals of Closed Curves



Comparison to the Fundamental Theorem of Calculus

The fundamental theorem of calculus says that integrating (adding up insmall pieces) a rate of change on the interval [a, b] gives the total changebetween the boundary points a and b.

∫ b

af ′(x)dx = f (b)− f (a)

We will attempt to find a similar correspondence for two-dimensionaldomains.



Green’s Theorem

Theorem

Suppose D is a simply connected (no holes), bounded region and r(t)defines C , a piecewise smooth curve that traces the boundary of Dcounterclockwise. If F = P i + Qj is a vector field, then∫

CF · dr =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA



Alternate Line Integral Notation

To avoid mentioning vectors, your textbook uses the notation∫CPdx + Qdy

Thus we can also write Green’s Theorem

Theorem ∫CPdx + Qdy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA



Significance of ∂Q∂x −

∂P∂y

If ∂Q∂x > 0, then the upward work on

the right side of C outweighs theupward work on the left side of C .

If ∂P∂y < 0, then the rightward work on

the bottom of C outweighs therightward work on the top of C .



Proof of Green’s Theorem

We approximate the line integral about C by summing line integralsaround ∆x by ∆y rectangles. Notice that the interior edges cancel eachother out. The remaining outer edges approximate C .

∫C

F·dr = lim∆x ,∆y→0

n∑i=1

∫Ci

F·dri




In order to approximate the line integral around a ∆x by ∆y rectangle, welinearize F. We’ll use differential notation. Unless noted otherwise, allfunctions evaluated at (x , y).

L(x + dx , y + dy) =F + dF

=F + Fxdx + Fydy

=[P + Pxdx + Pydy

]i

+[Q + Qxdx + Qydy

]j




Let’s parameterize the edges of a ∆x by ∆y rectangle. For each segment,0 ≤ t ≤ 1.

(x + t∆x)i + y j

(x + t∆x)i + (y + ∆y)j

x i + (y + t∆y)j (x + ∆x)i + (y + t∆y)j

(x , y) (x + ∆x , y)

(x + ∆x , y + ∆y)(x , y + ∆y)




When we replace F with L and compute the line integral, we getconvenient cancellation. Here are the top and bottom segments.

∫ 1

0L(x + t∆x , y) · (∆x i)dt −

∫ 1

0L(x + t∆x , y + ∆y) · (∆x i)dt

= ∆x

[∫ 1

0P + Px(t∆x)dt −

∫ 1

0P + Px(t∆x) + Py (∆y)dt

]= ∆x

∫ 1

0−Py∆ydt

= −Py∆y∆x

Similarly the left and right segments sum to Qx∆x∆y .




Finally we return to our original limit approximation. Note ∆y∆x is thearea of a ∆x by ∆y rectangle, so our expression conforms to the limitdefinition of a double integral.

∫C

F · dr = lim∆x ,∆y→0

n∑i=1

∫Ci

F · dri

= lim∆x ,∆y→0

n∑i=1

(Qx − Py )∆y∆x

= lim∆x ,∆y→0

n∑i=1

(Qx − Py )∆A

=

∫∫D

(Qx − Py )dA



Example 1

Let F(x , y) = (y2 − 3x)i + xy j. Let C be the path that travels along theline segment from (2, 4) to (−1, 1) and then back to (2, 4) along theparabola y = x2.

Compute∫C F · dr.



Exercise

Let F(x , y) = (3y − ex)i + (2x − sin y)j. Let C be a circle of radius 3traveling counterclockwise once around the origin.

1 Set up the line integral∫C F · dr as a single-variable integral of t.

2 If F were conservative what would the value of this integral be? Is Fconservative?

3 How would you apply Green’s theorem to the integral? What is itsvalue?

4 What would∫C F · dr be if C traveled clockwise instead?



Example 2

Let C be a semicircle from (2, 0) to (−2, 0) above the x-axis. Compute∫C

(x2 − y3)dx + (x3 + ey2)dy



Steps for Line Integrals

First decide whether you’re taking the line integral of a function: f (x , y)or a vector field: F(x , y).

1 Function f

a Parameterize C and set up the integral, replacing dx , dy , ds with theappropriate differential.

b Evaluate the integral.

2 Vector Field Fa Is F conservative? Use FTLI.b Can you draw the curve C? Is it closed? Use Green’s.c If neither works, parameterize C and set up the integral, replacing dr

with r′(t)dt, and evaluate.



Summary

What kind of integrals can we evaluate with Green’s theorem?

Why would we ever want to replace a single integral with a doubleintegral?


Section 16.5 Curl and Divergence

Section 16.5

Curl and Divergence

Goals:

Compute the curl and divergence of a vector field.

Interpret curl and divergence geometrically.



Derivatives of Vector Fields

If we compare the fundamental theorem of calculus to the fundamentaltheorem of line integrals,

f (b)− f (a) =

∫ b

af ′(x)dx

f (B)− f (A) =

∫C∇f · dr

we see that ∇f fills the role of a derivative of the multivariable function fin this context. In this section we try to define some derivative-likeoperations of vector fields.



The ∇ Operator

Notation

We define the gradient operator ∇ (“del”), which behaves in some wayslike a vector. Depending on our choice of dimension we can have

∇ =

⟨∂

∂x,∂

∂y

⟩∇ =

⟨∂

∂x,∂

∂y,∂

∂z

⟩



The Gradient of a Function

Given a function f (x , y), we can reexamine ∇f in terms of the gradientoperator:

∇f =

⟨∂

∂x,∂

∂y

⟩f

=

⟨∂

∂xf ,

∂

∂yf

⟩=

⟨∂f

∂x,∂f

∂y

⟩



Divergence of a Vector Field

The ∇ operator is more exciting when we apply it to vector fields.

Definition

The divergence of a vector field F = P i + Qj at a point (x0, y0) measuresthe extent to which F points away from (x0, y0). The divergence functionis

div F = ∇ · F =∂

∂xP +

∂

∂yQ

Divergence is defined analogously for 3-dimensional vector fields.

Notice that ∇ · F(x0, y0) is a number, not a vector.



Recognizing Divergence Visually

It is easiest to recognize divergence when F(x0, y0) is the zero vector.

∇ · F > 0 ∇ · F < 0 ∇ · F ≈ 0



Divergence when F(x , y) 6= 0

When F(x0, y0) is not zero, it can be harder to tell whether F is expandingaway from (x0, y0) or accelerating toward it. If F(x0, y0) = v we cansubtract the constant vector v from F .

∇ · F = ∇ · (F− v)



Exercise

If F = xz i + xyz j− y2k, compute ∇ · F at (−2, 2, 1). What does it mean?



Green’s Theorem

Recall that ∂Q∂x −

∂P∂y measured the amount that a vector field curled around

a point. Green’s theorem related this to the line integral of a curve aroundthat point. This is a special case of our other “derivative,” the curl of F.



Curl F

Definition

The curl is defined for a three-dimensional vector field F = 〈P,Q,R〉. It isdenoted ∇× F and computed as follows:

curl F = ∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣=

(∂R

∂y− ∂Q

∂z

)i−(∂R

∂x− ∂P

∂z

)j +

(∂Q

∂x− ∂P

∂y

)k



Geometric Properties of ∇× F

Notice ∇× F(x0, y0, z0) is a vector.

∇× F(x0, y0, z0) is related to the line integrals of F around the point(x0, y0, z0).

The length indicates how large such integrals can be, as a multiple ofthe area they enclose.

The direction is perpendicular to the plane in which these lineintegrals are maximized, or around which F curls most strongly.



Exercise

If F = xz i + xyz j− y2k, compute ∇×F at (−2, 2, 1). What does it mean?



Vector Version of Green’s Theorem

Green’s theorem is two dimensional, so we assumeF(x , y , z) = P(x , y)i + Q(x , y)j + 0k. Most of the terms in the curl arezero. Specifically,

∇× F =

(∂Q

∂x− ∂P

∂y

)k.

Theorem (Green’s Theorem)

Suppose D is a simply connected, bounded region in the plane z = 0 andr(t) defines C , a piecewise smooth curve that traces the boundary of Dcounterclockwise. If F = 〈P,Q, 0〉 is a vector field, then∫

CF · dr =

∫∫D

(∇× F) · k dA



Composing ∇’s

We can also compose ∇ operators together. Here are some examples

Example

∇2f = ∇ · (∇f ) takes the divergence of the gradient vector field.

Example

∇ · (∇× F) computes the divergence of the curl of F.

Theorem

A vector field G on a simply connected 3-dimensional domain is equal to∇× F for some F, if and only if ∇ · G(x , y , z) = 0 for all (x , y , z).



Summary Questions

How do you compute divergence and curl?

How do you interpret divergence geometrically?

On what vector fields can you compute curl? Divergence?

If someone handed you two functions and tells you one is the curl of avector field and the other is the divergence, how could you tell whichis which?


Section 16.6 Parametric Surfaces and Their Areas

Section 16.6

Parametric Surfaces and Their Areas

Goals:

Parameterize a surface.

Compute tangent vectors to a parametric surface.



Parametric Surfaces

A parametric surface is the set of points defined by two-variableparametric equations, usually in three-space.

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

Where u and v are both parameters. Like with curves, we write this as avector function so we can do calculus, but we visualize it as a set of points.



Parameterizing the Graph of an Explicit Function

The graph of an explicit function z = f (x , y) can be parameterized bysubstituting

x = u y = v z = f (u, v)

r(u, v) = ui + v j + f (u, v)k



Parameterizing a Plane

Formula

If a plane p contains the point (x0, y0, z0) and the vectors a and b, then aparametrization of p is

r(u, v) = 〈x0, y0, z0〉+ ua + vb



Parameterizations from Other Coordinate Systems

Another source of parameterizations comes from coordinate systems we’velearned. Constant multiples and constant terms stretch and shift thesurface.

r(u, v) = (a cos u sin v + x0)i + (b sin u sin v + y0)j + (c cos v + z0)k0 ≤ u ≤ 2π 0 ≤ v ≤ π



Exercise

Describe the surfaces with the following parametric equations.

1 r(u, v) = 3 cos ui + 3 sin uj + vk 0 ≤ u ≤ 2π, 0 ≤ v ≤ 5

2 r(u, v) = (3− 3u − 3v)i + (6u + 2v)j + (2− 9v)k

3 r(u, v) = u cos π4 sin v i + u sin π4 sin v j + u cos vk

0 ≤ u ≤ 5, 0 ≤ v ≤ π



Derivatives of a Parametric Surface

Definitions

The partial derivatives ru(u0, v0) and rv (u0, v0) are tangent vectors tothe surface S .

We can use them to produce a linearization of S at r(u0, v0).

L(u, v) = r(u0, v0) + ru(u0, v0)(u − u0) + rv (u0, v0)(v − v0)

Some algebra shows that this is the equation of a tangent plane.



Summary Questions

How do we parameterize a plane?

How do we parametrize the graph z = f (x , y)?

How do we parametrize a sphere or a cylinder?

What is the relationship between the tangent vectors and the tangentplane of a surface?


Section 16.7 Surface Integrals

Section 16.7

Surface Integrals

Goals:

Understand the geometric significance of the different surfaceintegrals.

Set up and evaluate surface integrals.



Integrating on Parametric Surfaces

Just like with line integrals, we’d like to find ways of integrating a functionf (x , y , z) on a surface S that do not depend on the choice ofparameterization.

Definition

An integral dS is computed with respect to the geometric area on thesurface. We divide S into regions and let

∆Si be the area of the i th region.

(x∗i , y∗i , z∗i ) be a test point in the i th region.

D be the largest diameter of any of the regions (the longest distancebetween two points in the region).

We then define the surface integral:∫∫Sf (x , y , z)dS = lim

D→0

∑i

f (x∗i , y∗i , z∗i )∆Si



Computing a Surface Integral

It’s easier to choose subregions defined by a change in u and a change inv . Still, we may not know the area of such a region, so we use our oldtrick and linearize r(u, v) at a point and use the area of the parallelogramgiven by ∆u and ∆v . The area is thus

dS = |ru × rv |dudv



Example 1∫∫S

1dS computes the area of a surface. Compute the surface area of a

sphere of radius R.



Exercise

Let L be the surface

r(u, v) = 3 cos ui + 3 sin uj + vk 0 ≤ u ≤ 2π, 0 ≤ v ≤ 5.

Set up (but do not evaluate) the surface integral∫∫Lx2zdS .



Motivational Question

Suppose water, traveling at 3 meters per second is passing through acircular opening of radius 4m.

1 How much water flows through the opening per second?

2 What if the water is not flowing perpendicular to the circle?



Motivational Question

If the water is flowing with velocity v and n is normal to the opening Swith length equal to the area of S , then the flow rate is v · n.



Flux Integrals

3 What if the velocity of the water is not constant, but depends on thelocation where it is measured?

4 What if the opening isn’t a flat shape, but a surface in 3 dimensions?



Flux Integrals

Definition

The flux integral of F through S is denoted∫∫S

F · dS.

For a parameterization r(u, v) we define dS = (ru × rv )dudv . F · dSmeasures the flow of F through the parallelogram dS .



Example 2

Let F(x , y , z) = x i− y j be a flow of water and let N be a net given byN = {(x , y , z) : z = x2 − y2, x2 + y2 ≤ 1}.

1 Compute

∫∫N

F · dS.

2 What does the sign of your answer mean?



Exercises

1 Let L be the surface

r(u, v) = 3 cos ui + 3 sin uj + vk 0 ≤ u ≤ 2π, 0 ≤ v ≤ 5.

Let F(x , y , z) = xy i + z j + exk. Set up, but do not evaluate the fluxintegral:

∫S F · dS.

2 An experimental solar car has solar panels in the shape of the graph

z =√x − y2

4 for z ≥ 0 and x ≤ 16. The direction and intensity of thesunlight is given by the constant vector field: F = c(2i + 3j− 2k). Setup a flux integral to compute the energy collected by the car.



Orientation

Depending on our choice of parameterization, ru × rv could point in one oftwo normal directions. If a surface has two sides, then choosing a normaldirection defines an orientation of the surface.

Note that in general not all surfaces have two sides. Surfaces without twosides are non-orientable.



The Unit Normal Vector

We can connect surface integrals to flux integrals via the followingdefinition

Definition

Given a surface S , the unit normal vector n to S at the point P is thenormal vector of length 1 at P whose direction agrees with the orientationof the surface.

Notice that at the point r(u, v), n = ru×rv|ru×rv | so

∫∫S

F·dS =

∫∫F(r)·(ru×rv )dudv =

∫∫F(r)·n|ru×rv |dudv =

∫∫S

F·ndS

This gives us an alternative notation for flux integrals.



Summary Questions

What are the two kinds of surface integrals? What do they compute?

What expression do we substitute for the differentials dS and dS?

How many different orientations can a connected surface have? Doesthis change if the surface consists of two or more disconnected parts?


Section 16.8 Stokes’ Theorem

Section 16.8

Stokes’ Theorem

Use Stokes’ Theorem to evaluate integrals.



Line Integrals that Bound a Surface

Green’s Theorem related a line integral around C to a double integral of aregion bounded by C . In three dimensions, a curve C might bound asurface S . We can attempt to apply the method of Green’s to thissituation.



Subdividing and Linearizing a Surface

Like in two dimensions, given a subdivision of S into smaller pieces withperimeter curves ri , we have∫

CF · dr =

∑i

∫Ci

F · dri

We approximate the Ci with parallelograms from the linearization of S .This lets us write the line integrals in terms of the parameterization of S .



Line Integrals Around the Parallelograms

Let s(u, v) be a parameterization of S . We will parameterize the edges ofparallelogram that results from a change of ∆u and ∆v in the linearizationof s(u, v). The domains are all 0 ≤ t ≤ 1.

r′(t)︷︸︸︷

s + tsu∆u

s + sv∆v + tsu∆u

r(t) = s + tsv∆v s + su∆u + tsv∆v

F + Fut∆u

F + Fut∆u + Fv∆v

F + Fu∆u + Fv t∆vF(r(t)) = F + Fv t∆v

s s + su∆u

s + su∆u + sv∆vs + sv∆v

We approximate F(ri (t)) by L(u, v) = F + Fudu + Fvdv .Multivariable Calculus 100 / 130


Computations

∫Ci

F · dri

≈∫ 1

0(F + Fut∆u) · (su∆u)dt +

∫ 1

0(F + Fu∆u + Fv t∆v) · (sv∆v)dt

−∫ 1

0(F + Fv∆v + Fut∆u) · (su∆u)dt −

∫ 1

0(F + Fv t∆v) · (sv∆v)dt

=−∫ 1

0Fv∆v · (su∆u)dt +

∫ 1

0Fu∆u · (sv∆v)dt

=

∫ 1

0(Fu∆u) · (sv∆v)− (Fv∆v) · (su∆u)dt

=

∫ 1

0(Fu · sv − Fv · su)∆v∆udt

=(Fu · sv − Fv · su)∆v∆u



Computations II

(Fu · sv − Fv · su) =

[∂P

∂x

∂x

∂u+∂P

∂y

∂y

∂u+∂P

∂z

∂z

∂u

]∂x

∂v

+

[∂Q

∂x

∂x

∂u+∂Q

∂y

∂y

∂u+∂Q

∂z

∂z

∂u

]∂y

∂v

+

[∂R

∂x

∂x

∂u+∂R

∂y

∂y

∂u+∂R

∂z

∂z

∂u

]∂z

∂v

−[∂P

∂x

∂x

∂v+∂P

∂y

∂y

∂v+∂P

∂z

∂z

∂v

]∂x

∂u

−[∂Q

∂x

∂x

∂v+∂Q

∂y

∂y

∂v+∂Q

∂z

∂z

∂v

]∂y

∂u

−[∂R

∂x

∂x

∂v+∂R

∂y

∂y

∂v+∂R

∂z

∂z

∂v

]∂z

∂u



Computations III

(Fu · sv − Fv · su) =

(∂R

∂y− ∂Q

∂z

)(∂y

∂u

∂z

∂v− ∂z

∂u

∂y

∂v

)+

(∂P

∂z− ∂R

∂x

)(∂x

∂u

∂z

∂v− ∂z

∂u

∂x

∂v

)+

(∂Q

∂x− ∂P

∂y

)(∂x

∂u

∂y

∂v− ∂y

∂u

∂x

∂v

)= (∇× F) · (su × sv )



Taking a Limit

Using this computation we can see what happens as we let the size of oursubdivisions approach 0.∫

CF · dr =

∑i

∫Ci

F · dri

= lim∆u,∆v→0

∑i

(Fu · sv − Fv · su)∆v∆u

= lim∆u,∆v→0

∑i

(∇× F) · (su × sv )∆v∆u

=

∫∫S

(∇× F) · dS



Stokes’ Theorem

Theorem

If S is a smooth surface that is bounded by a simple closed boundarycurve C with positive orientation and F is a vector field with continuouspartial derivatives, then∫

CF · dr =

∫∫S

(∇× F) · dS

Positive orientation means the rotation of C and the direction of dS obeythe right hand rule. If the surface is negatively oriented, introduce a minussign.



Visualizing Stokes’ Theorem

Much like Green’s theorem, Stokes’ theorem is understood as adding upthe extent to which the vector field curls around each point to get thetotal work around the boundary. The dot product measures the extent towhich the overall curl of F takes place in the surface.



Example 1

Let C be the curve given by r(t) = cos(t)i + sin(t)j + (cos2(t)− sin2(t))k.Let F(x , y , z) = xyk. How does Stokes’ Theorem apply to∫

CF · dr?



Exercise

Suppose that F is a conservative vector field on R3 and C is a smoothcurve that bounds a positively-oriented surface S .

1 Can the Fundamental Theorem of Line Integrals compute∫C F · dr?

What is the value?

2 Use your characterization of F from 1 to compute ∇× F.



Example 2

Let F(x , y , z) be a smooth vector field on R3. Let S be a sphere. What is∫∫S(∇× F) · dS?



Physics Application of Stokes’ Theorem

Faraday’s law of induction says that the change in magnetic field througha surface S induces an electromotive force through a wire on its boundaryC . Physicists measure the induced voltage by integrating the change inmagnetic field dS.



Exercise

Suppose S is part of the paraboloid z = 16− x2 − y2 above the xy -plane,and C is its boundary. Is there an easier way to compute

∫∫S F · dS?



Summary Questions

What two types of integrals does Stokes’ Theorem equate?

What does positive orientation mean?


Section 16.9 The Divergence Theorem

Section 16.9

The Divergence Theorem

Use the Divergence Theorem to evaluate integrals.



Divergence of a Vector Field and Flux Integrals

A vector field with positive divergence expands outward. Let S be aclosed surface. Visually, we expect the flux integral through that surfaceto be related to this divergence.




Theorem

If E is a simple solid region, S is its boundary surface with outwardorientation, and F is a vector field with continuous partial derivatives.Then ∫∫

SF · dS =

∫∫∫E∇ · FdV

Outward orientation means that the normal vectors ru × rv point out of Erather than into E .



Example 1

Let S be the surface of the cube with vertices (±1,±1,±1), orientedoutward. Let F(x , y , z) = x3i + yz j + exzk. Compute

∫∫S F · dS.



Example 2

Let F = (2x − sin(y))i + (ez − 4y)j +−zk. Let L be given byr(u, v) = 3 cos(u)i + 3 sin uj + vk for 0 ≤ u ≤ 2π and 0 ≤ v ≤ 4. Compute∫∫

LF · dS.



Exercise

Let S be a sphere of radius 5 with outward orientation. LetF(x , y , z) = (x + 3y)i + (x sin(z))j− zk. What does the divergencetheorem tell you about the value of∫∫

SF · dS?



Application: Diffusion

This is a classical situation studied in a partial differential equations course.Suppose a function u(x , y , z , t) gives the density of smoke (or heat) at thepoint (x , y , z) at time t. Suppose F(x , y , z , t) gives the flow of the smoke.Then for any domain D with smooth boundary S we have

d

dt

∫∫∫DudV = −

∫∫S

F · dS

We can reverse the order of differentiation and integration on the left.

We can apply the Divergence Theorem on the right.∫∫∫D

∂u

∂tdV = −

∫∫∫D

(∇ · F) · dV

If this is true for all domains D, the integrands must be equal

∂u

∂t= −∇ · F



Application: Integration by Parts

The Divergence Theorem allows us to derive a multivariable integration byparts which is also used to solve partial differential equations.

Product Rule

If s is a function and F is a vector field then

∇ · (sF) = (∇s) · F + s(∇ · F)

Theorem (Integration by Parts)

If R is a region with boundary ∂R then∫∫∫R

(∇s) · F +

∫∫∫Rs(∇ · F) =

∫∫∂R

sF



The Divergence Theorem in One Dimension

Unlike Green’s and Stokes’ Theorems, the Divergence Theorem can begeneralized to other dimensions

The Fundamental Theorem of Calculus

For a vector field F(x) = P(x)i on a domain [a, b]. We have∇ · F = P ′(x). The outward normal vectors at a and b are −i and i. Sothe Divergence Theorem is∫

[a,b]P ′(x)dx = F(a) · (−i) + F(b) · i = −P(a) + P(b)



The Divergence Theorem in Two Dimension

Two Dimensions

For a closed curve C = (x(t), y(t)) traveling counterclockwise around adomain D, we have the outward normal vector:

n(t) =

∣∣∣∣ i jx ′(t) y ′(t)

∣∣∣∣The Divergence Theorem is

∫∫D∇ · FdA =

∫ b

aF(x(t), y(t)) · n(t)dt.



Summary Questions

What two types of integrals does the Divergence Theorem equate?

What needs to be true about the domain for the theorem to work?

Explain why the Divergence Theorem is more likely to be useful incomputation than Stokes’ Theorem.


Section 16.10 Summary of Integration Theorems

Section 16.10

Summary of Integration Theorems

Notice the parallels between the integration theorems we have learned.



The Fundamental Theorem of Calculus

Given a function f and an interval [a, b], we have

∫ b

af ′(x)dx = f (b)− f (a)



The Fundamental Theorem of Line Integrals

Given a function f and a vector function r(t) from A to B, we have

∫ b

a(∇f ) · dr = f (b)− f (a)



Green’s Theorem

Given a vector field F and a domain D in R2 bounded by a curve C , we

∫∫D

(∂Q

∂x− ∂P

∂y

)dA =

∫C

F · dr



Stokes’ Theorem

Given a vector field F and a smooth surface S in R3 that is bounded by asimple closed boundary curve C , we have

∫∫S

(∇× F) · dS =

∫C

F · dr




Given a vector field F and a region E with (outward oriented) boundarysurface S we have

∫∫∫E∇ · FdV =

∫∫S

F · dS



Summary Questions/Exercise

On each side of an integration theorem:

How are the functions related to each other?

How are the domains related to each other?


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