Motivation: Fundamental Theorems of Vector

Our goal as we close out the semester is to give several “Fundamental Theorem of Calculus”-type theorems which relate volume of on a given domain to line and surface integrals about the boundary of the domain.

The general form of these theorems, which we collectively call the fundamental theorems of , is the following:

The of a “-type object” on a given domain D may be computed using only the values along the boundary of D. 2. Fundamental Theorem for Conservative Vector Fields: A of the form R C ∇Vds may be computed by evaluating V at the boundary points of the 1-dimensional parametrized curve domain C.

Motivation: Fundamental Theorems of Vector Calculus

As an overview, we will roughly and informally summarize the content of the fundamental theorems of vector calculus. First let’s start with the ones we have already seen:

1. FTC: An area integral of the form

R b 0 a f (x)dx may be computed by evaluating f at the boundary points a and b of the 1-dimensional domain interval [a, b]. Motivation: Fundamental Theorems of Vector Calculus

As an overview, we will roughly and informally summarize the content of the fundamental theorems of vector calculus. First let’s start with the ones we have already seen:

1. FTC: An area integral of the form

R b 0 a f (x)dx may be computed by evaluating f at the boundary points a and b of the 1-dimensional domain interval [a, b].

2. Fundamental Theorem for Conservative Vector Fields: A line integral of the form R C ∇Vds may be computed by evaluating V at the boundary points of the 1-dimensional parametrized curve domain C. 2. Stokes’ Theorem: A of the form RR ~ S (F )dS

may be computed by line integrating F~ along the boundary of the 2-dimensional domain S.

Note: The “curl” of F~ is a derivative-type object we will define later.

3. Theorem: A volume integral of the form RRR ~ W div(F )d(x, y, z)

may be computed by surface integrating F~ along the boundary of the 3-dimensional domain W.

Note: The ”divergence” of F~ is another soon-to-be-introduced derivative-type object.

Fundamental Theorems of Vector Calculus, contd.

Now here are the new ones.

1. Green’s Theorem: A volume integral of the form

ZZ  δF2 δF1  − d(x, y) D δx δy

may be computed by line integrating F~ = (F1, F2) along the boundary of the 2-dimensional domain D. 3. : A volume integral of the form RRR ~ W div(F )d(x, y, z)

may be computed by surface integrating F~ along the boundary of the 3-dimensional domain W.

Note: The ”divergence” of F~ is another soon-to-be-introduced derivative-type object.

Fundamental Theorems of Vector Calculus, contd.

Now here are the new ones.

1. Green’s Theorem: A volume integral of the form

ZZ  δF2 δF1  − d(x, y) D δx δy

may be computed by line integrating F~ = (F1, F2) along the boundary of the 2-dimensional domain D.

2. Stokes’ Theorem: A surface integral of the form RR ~ S curl(F )dS

may be computed by line integrating F~ along the boundary of the 2-dimensional domain S.

Note: The “curl” of F~ is a derivative-type object we will define later. Fundamental Theorems of Vector Calculus, contd.

Now here are the new ones.

1. Green’s Theorem: A volume integral of the form

ZZ  δF2 δF1  − d(x, y) D δx δy

may be computed by line integrating F~ = (F1, F2) along the boundary of the 2-dimensional domain D.

2. Stokes’ Theorem: A surface integral of the form RR ~ S curl(F )dS

may be computed by line integrating F~ along the boundary of the 2-dimensional domain S.

Note: The “curl” of F~ is a derivative-type object we will define later.

3. Divergence Theorem: A volume integral of the form RRR ~ W div(F )d(x, y, z)

may be computed by surface integrating F~ along the boundary of the 3-dimensional domain W.

Note: The ”divergence” of F~ is another soon-to-be-introduced derivative-type object. Suppose C may be parametrized by a continuous one-to-one R2-valued function ~c with domain [a, b], where ~c(a) = ~c(b). Then we call C a simple closed curve. If δD is a simple closed curve, then we choose to orient δD in the counterclockwise direction. This is called the boundary orientation.

Definitions and Terminology

Definition Let D be a region in R2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δD. We call δD the boundary of D. If δD is a simple closed curve, then we choose to orient δD in the counterclockwise direction. This is called the boundary orientation.

Definitions and Terminology

Definition Let D be a region in R2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δD. We call δD the boundary of D.

Suppose C may be parametrized by a continuous one-to-one R2-valued function ~c with domain [a, b], where ~c(a) = ~c(b). Then we call C a simple closed curve. Definitions and Terminology

Definition Let D be a region in R2. Recall that a point (x, y) is called a boundary point of D if every open disk about (x, y) intersects both D and the exterior of D. Denote the set of all boundary points of D by δD. We call δD the boundary of D.

Suppose C may be parametrized by a continuous one-to-one R2-valued function ~c with domain [a, b], where ~c(a) = ~c(b). Then we call C a simple closed curve. If δD is a simple closed curve, then we choose to orient δD in the counterclockwise direction. This is called the boundary orientation. Green’s Theorem

Theorem (Green’s Theorem) Let D be a domain whose boundary δD is a simple closed curve. 2 Let F = (F1, F2) be a vector field over R . Then

I ZZ δF δF  F~ · ds = 2 − 1 d(x, y). δD D δx δy 2 If (F1, F2) = (xy , x), then

δF2 δF1 2 δx − δy = 1 − 2xy ,

and hence we are being asked to show that

H 2 RR C(xy , x) · ds = R (1 − 2xy)d(x, y),

where R is the interior of the unit circle.

Example H 2 Verify Green’s theorem for the line integral C(xy , x) · ds about the unit circle C. Example H 2 Verify Green’s theorem for the line integral C(xy , x) · ds about the unit circle C.

2 If (F1, F2) = (xy , x), then

δF2 δF1 2 δx − δy = 1 − 2xy , and hence we are being asked to show that

H 2 RR C(xy , x) · ds = R (1 − 2xy)d(x, y), where R is the interior of the unit circle. We start with the line integral on the left. Parametrize C by

~c(t) = (cos t, sin t) for 0 ≤ t ≤ 2π.

We have ~c 0(t) = (− sin t, cos t). Now compute:

I Z 2π (xy 2, x) · ds = (cos t sin2 t, sin t) · (− sin t, cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0  1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π.

Solution: The Line Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) Now compute:

I Z 2π (xy 2, x) · ds = (cos t sin2 t, sin t) · (− sin t, cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0  1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π.

Solution: The Line Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) We start with the line integral on the left. Parametrize C by

~c(t) = (cos t, sin t) for 0 ≤ t ≤ 2π.

We have ~c 0(t) = (− sin t, cos t). Solution: The Line Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) We start with the line integral on the left. Parametrize C by

~c(t) = (cos t, sin t) for 0 ≤ t ≤ 2π.

We have ~c 0(t) = (− sin t, cos t). Now compute:

I Z 2π (xy 2, x) · ds = (cos t sin2 t, sin t) · (− sin t, cos t)dt C 0 Z 2π = (− cos t sin3 t + cos2 t)dt 0  1 1 1 2π = − sin4 t + t + sin(2t) 4 2 4 0 = (0 + π + 0) − (0 + 0 + 0) = π. Now we compute the integral on the right and hope we get π:

√ ZZ Z 1 Z 1−x2 (1 − 2xy)d(x, y) = √ (1 − 2xy)dydx R −1 − 1−x2 1 √ Z 2 = [y − xy 2] 1−√x 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π.

So Green’s theorem is true in this case!

Solution: The Volume Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) H 2 Known: C(xy , x) · ds = π √ ZZ Z 1 Z 1−x2 (1 − 2xy)d(x, y) = √ (1 − 2xy)dydx R −1 − 1−x2 1 √ Z 2 = [y − xy 2] 1−√x 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π.

So Green’s theorem is true in this case!

Solution: The Volume Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) H 2 Known: C(xy , x) · ds = π Now we compute the integral on the right and hope we get π: Solution: The Volume Integral

H 2 RR Goal: C(xy , x) · ds = R (1 − 2xy)d(x, y) H 2 Known: C(xy , x) · ds = π Now we compute the integral on the right and hope we get π:

√ ZZ Z 1 Z 1−x2 (1 − 2xy)d(x, y) = √ (1 − 2xy)dydx R −1 − 1−x2 1 √ Z 2 = [y − xy 2] 1−√x 2 −1 y=− 1−x Z 1 p = 2 1 − x 2dx −1 1 = [2 arcsin x]−1 = π.

So Green’s theorem is true in this case! Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green’s theorem instead.

The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that

δF2 3 δF1 δx = 2xy and δy = 0.

So by Green’s theorem we get:

I ZZ (sin x, x2y 3) · ds = (2xy 3 − 0)d(x, y) δD D Z 2 Z x = 2xy 3dydx 0 0 Z 2 1 4 x = [ xy ]y=0dx 0 2 1 Z 2 = x5dx 2 0 1 1 16 = · · 26 = . 2 6 3

Example Compute the circulation of F~ = (sin x, x2y 3) about the path C = δD, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that

δF2 3 δF1 δx = 2xy and δy = 0.

So by Green’s theorem we get:

I ZZ (sin x, x2y 3) · ds = (2xy 3 − 0)d(x, y) δD D Z 2 Z x = 2xy 3dydx 0 0 Z 2 1 4 x = [ xy ]y=0dx 0 2 1 Z 2 = x5dx 2 0 1 1 16 = · · 26 = . 2 6 3

Example Compute the circulation of F~ = (sin x, x2y 3) about the path C = δD, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green’s theorem instead. So by Green’s theorem we get:

I ZZ (sin x, x2y 3) · ds = (2xy 3 − 0)d(x, y) δD D Z 2 Z x = 2xy 3dydx 0 0 Z 2 1 4 x = [ xy ]y=0dx 0 2 1 Z 2 = x5dx 2 0 1 1 16 = · · 26 = . 2 6 3

Example Compute the circulation of F~ = (sin x, x2y 3) about the path C = δD, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green’s theorem instead.

The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that

δF2 3 δF1 δx = 2xy and δy = 0. Example Compute the circulation of F~ = (sin x, x2y 3) about the path C = δD, where D is the triangle with vertices (0, 0), (2, 0), and (2, 2). Solution. Computing the circulation might be tedious in this case, as we would need to parametrize the three sides of the triangle separately. So we appeal to Green’s theorem instead.

The interior of the triangle is bounded below by y = 0 and above by y = x, and on the left and right by y = 0 and y = 2. Check that

δF2 3 δF1 δx = 2xy and δy = 0.

So by Green’s theorem we get:

I ZZ (sin x, x2y 3) · ds = (2xy 3 − 0)d(x, y) δD D Z 2 Z x = 2xy 3dydx 0 0 Z 2 1 4 x = [ xy ]y=0dx 0 2 1 Z 2 = x5dx 2 0 1 1 16 = · · 26 = . 2 6 3 Proof. ~ 1 δF2 δF1 1 1 Set F = (F1, F2) = 2 (−y, x) and check that δx − δy = 2 − (− 2 ) = 1. Now apply Green’s theorem: ZZ Area of D = 1d(x, y) D ZZ δF δF = ( 2 − 1 )d(x, y) D δx δy I = F~ · ds C 1 I = (−y, x) · ds. 2 C

Corollary (Area Formula) Let D be a region in R2 and assume C = δD is a simple closed curve. Then the area of D is equal to

1 H 2 C(−y, x) · ds. Corollary (Area Formula) Let D be a region in R2 and assume C = δD is a simple closed curve. Then the area of D is equal to

1 H 2 C(−y, x) · ds.

Proof. ~ 1 δF2 δF1 1 1 Set F = (F1, F2) = 2 (−y, x) and check that δx − δy = 2 − (− 2 ) = 1. Now apply Green’s theorem: ZZ Area of D = 1d(x, y) D ZZ δF δF = ( 2 − 1 )d(x, y) D δx δy I = F~ · ds C 1 I = (−y, x) · ds. 2 C Definition ~ 2 Let F = (F1, F2) be a vector field over R and let ~c(t) = (x(t), y(t)) be a smooth 2 parametrization of a curve in R with domain [a, b]. Define the x- and y-components of the vector field line integral of F~ over C to be, respectively,

R R b 0 C F1dx = a F1(~c(t))x (t)dt

R R b 0 C F2dy = a F2(~c(t))y (t)dt.

It follows from the definitions that

Z Z b 0 0 F~ · ds = (F1(~c(t)), F2(~c(t))) · (x (t), y (t))dt C a Z Z = F1dx + F2dy. C C

Lemma The values of the x- and y-components of a vector field line integral are independent of the choice of parametrization ~c.

Some Quick Terminology We wish to give some indications toward the proof of Green’s Theorem. For this we need a little more notation and a lemma. It follows from the definitions that

Z Z b 0 0 F~ · ds = (F1(~c(t)), F2(~c(t))) · (x (t), y (t))dt C a Z Z = F1dx + F2dy. C C

Lemma The values of the x- and y-components of a vector field line integral are independent of the choice of parametrization ~c.

Some Quick Terminology We wish to give some indications toward the proof of Green’s Theorem. For this we need a little more notation and a lemma. Definition ~ 2 Let F = (F1, F2) be a vector field over R and let ~c(t) = (x(t), y(t)) be a smooth 2 parametrization of a curve in R with domain [a, b]. Define the x- and y-components of the vector field line integral of F~ over C to be, respectively,

R R b 0 C F1dx = a F1(~c(t))x (t)dt

R R b 0 C F2dy = a F2(~c(t))y (t)dt. Lemma The values of the x- and y-components of a vector field line integral are independent of the choice of parametrization ~c.

Some Quick Terminology We wish to give some indications toward the proof of Green’s Theorem. For this we need a little more notation and a lemma. Definition ~ 2 Let F = (F1, F2) be a vector field over R and let ~c(t) = (x(t), y(t)) be a smooth 2 parametrization of a curve in R with domain [a, b]. Define the x- and y-components of the vector field line integral of F~ over C to be, respectively,

R R b 0 C F1dx = a F1(~c(t))x (t)dt

R R b 0 C F2dy = a F2(~c(t))y (t)dt.

It follows from the definitions that

Z Z b 0 0 F~ · ds = (F1(~c(t)), F2(~c(t))) · (x (t), y (t))dt C a Z Z = F1dx + F2dy. C C Some Quick Terminology We wish to give some indications toward the proof of Green’s Theorem. For this we need a little more notation and a lemma. Definition ~ 2 Let F = (F1, F2) be a vector field over R and let ~c(t) = (x(t), y(t)) be a smooth 2 parametrization of a curve in R with domain [a, b]. Define the x- and y-components of the vector field line integral of F~ over C to be, respectively,

R R b 0 C F1dx = a F1(~c(t))x (t)dt

R R b 0 C F2dy = a F2(~c(t))y (t)dt.

It follows from the definitions that

Z Z b 0 0 F~ · ds = (F1(~c(t)), F2(~c(t))) · (x (t), y (t))dt C a Z Z = F1dx + F2dy. C C

Lemma The values of the x- and y-components of a vector field line integral are independent of the choice of parametrization ~c. Proof of a Special Case of Green’s Theorem

δF δF  Goal: Show H F~ · ds = RR 2 − 1 d(x, y). δD D δx δy We are unable to give a proof of Green’s theorem in its full generality, but we can prove it if we make the following very special simplifying assumptions:

I the boundary δD may be described as the union of two graphs of the form y = T (x) with B(x) ≤ T (x) for a ≤ x ≤ b; and

I δD may also be described as the union of two graphs of the form x = L(y) and x = R(y) with L(y) ≤ R(y) for c ≤ y ≤ d.

(Picture on whiteboard!) We begin by parametrizing the bottom C1 and top C2 of δD.

C1: ~c1(t) = (x1(t), y1(t)) = (t, B(t)) C2: ~c2(t) = (x2(t), y2(t)) = (t, T (t))

Note ~c1 traverses C1 along the boundary orientation but ~c2 goes clockwise, i.e. ~c2 parametrizes −C2.

Proof of a Special Case, contd.

δF δF  Goal: Show H F~ · ds = RR 2 − 1 d(x, y). δD D δx δy In order to observe Green’s theorem, we will break it up into two parts. H ~ H H Since δD F · ds = δD F1dx + δD F2dy, it suffices for us to show the following two equalities:

H RR δF1 δD F1dx = − D δy d(x, y) H RR δF2 δD F2dy = D δx d(x, y). Proof of a Special Case, contd.

δF δF  Goal: Show H F~ · ds = RR 2 − 1 d(x, y). δD D δx δy In order to observe Green’s theorem, we will break it up into two parts. H ~ H H Since δD F · ds = δD F1dx + δD F2dy, it suffices for us to show the following two equalities:

H RR δF1 δD F1dx = − D δy d(x, y) H RR δF2 δD F2dy = D δx d(x, y).

We begin by parametrizing the bottom C1 and top C2 of δD.

C1: ~c1(t) = (x1(t), y1(t)) = (t, B(t)) C2: ~c2(t) = (x2(t), y2(t)) = (t, T (t))

Note ~c1 traverses C1 along the boundary orientation but ~c2 goes clockwise, i.e. ~c2 parametrizes −C2. δF Now we compute RR 1 d(x, y). D δy

ZZ δF Z b Z T (x) δF 1 d(x, y) = 1 dydx D δy a B(x) δy Z b = [F1(x, T (x)) − F1(x, B(x))]dx a Z b Z b 0 0 = F1(c2(t))x2(t)dt − F1(c1(t))x1(t)dt a a Z Z = F1dx − F1dx −C2 C1 I = − F1dx. δD

This completes the proof of the mid-goal.

Proof of a Special Case, contd.

H RR δF1 Mid-Goal: Show that δD F1dx = − D δy d(x, y) Proof of a Special Case, contd.

H RR δF1 Mid-Goal: Show that δD F1dx = − D δy d(x, y) δF Now we compute RR 1 d(x, y). D δy

ZZ δF Z b Z T (x) δF 1 d(x, y) = 1 dydx D δy a B(x) δy Z b = [F1(x, T (x)) − F1(x, B(x))]dx a Z b Z b 0 0 = F1(c2(t))x2(t)dt − F1(c1(t))x1(t)dt a a Z Z = F1dx − F1dx −C2 C1 I = − F1dx. δD

This completes the proof of the mid-goal. Now let C3 and C4 denote the left and right boundaries, and parametrize again:

C3: ~c3(t) = (x3(t), y3(t)) = (L(t), t) C4: ~c4(t) = (x4(t), y4(t)) = (R(t), t).

This time ~c3 parametrizes −C3 but ~c4 parametrizes C4.

ZZ δF Z d Z R(y) δF 2 d(x, y) = 2 dxdy D δx c L(y) δx Z d Z d = F2(R(y), y)dy − F2(L(y), y)dy c c Z d Z d 0 0 = F2(~c4(t))y4(t)dt − F2(~c3(t))y3(t)dt c c Z Z = F2dy − F2dy C4 −C3 I = F2dy. δD

Proof of a Special Case, contd. H RR δF2 Mid-Goal: Show that δD F2dy = D δx d(x, y) ZZ δF Z d Z R(y) δF 2 d(x, y) = 2 dxdy D δx c L(y) δx Z d Z d = F2(R(y), y)dy − F2(L(y), y)dy c c Z d Z d 0 0 = F2(~c4(t))y4(t)dt − F2(~c3(t))y3(t)dt c c Z Z = F2dy − F2dy C4 −C3 I = F2dy. δD

Proof of a Special Case, contd. H RR δF2 Mid-Goal: Show that δD F2dy = D δx d(x, y)

Now let C3 and C4 denote the left and right boundaries, and parametrize again:

C3: ~c3(t) = (x3(t), y3(t)) = (L(t), t) C4: ~c4(t) = (x4(t), y4(t)) = (R(t), t).

This time ~c3 parametrizes −C3 but ~c4 parametrizes C4. Proof of a Special Case, contd. H RR δF2 Mid-Goal: Show that δD F2dy = D δx d(x, y)

Now let C3 and C4 denote the left and right boundaries, and parametrize again:

C3: ~c3(t) = (x3(t), y3(t)) = (L(t), t) C4: ~c4(t) = (x4(t), y4(t)) = (R(t), t).

This time ~c3 parametrizes −C3 but ~c4 parametrizes C4.

ZZ δF Z d Z R(y) δF 2 d(x, y) = 2 dxdy D δx c L(y) δx Z d Z d = F2(R(y), y)dy − F2(L(y), y)dy c c Z d Z d 0 0 = F2(~c4(t))y4(t)dt − F2(~c3(t))y3(t)dt c c Z Z = F2dy − F2dy C4 −C3 I = F2dy. δD Proof of a Special Case, contd.

We conclude by observing that, as promised:

I I I ~ F · ds = F1dx + F2dy δD δD δD ZZ δF ZZ δF = − 1 d(x, y) + 2 d(x, y) D δy D δx ZZ δF δF  = 2 − 1 d(x, y). D δx δy