x

xi y

x

Figure 4. Using φ to integrate a 2-form.

1 2

f (xi , yj ) dx § dy(Vi,j , Vi,j ),

At the conclusion of Step 4 we have two sums,

i j

1 2

ωφ(xi ,yj ) (Vi,j , Vi,j ).

and In order for these to be equal we must have:

i j

1 2 1 2

f (xi , yj ) dx § dy(Vi,j , Vi,j ) = ωφ(xi ,yj ) (Vi,j , Vi,j )

And so,

1 2

ωφ(xi ,yj ) (Vi,j , Vi,j )

f (xi , yj ) = 1 2

dx § dy(Vi,j , Vi,j )

1 2

But, since we are using a rectangular lattice in R we know dx § dy(Vi,j , Vi,j ) =

1 2 1 2

Area(Vi,j , Vi,j ) = |Vi,j | · |Vi,j |. We now have

1 2

ωφ(xi ,yj ) (Vi,j , Vi,j )

f (xi , yj ) = 1 2

|Vi,j | · |Vi,j |

Using the bilinearity of ω this reduces to

1 2

Vi,j Vi,j

f (xi , yj ) = ωφ(xi ,yj ) ,2

1

|Vi,j | |Vi,j |

But, as the distance between adjacent points of our partition tends toward 0,

1

Vi,j φ(xi+1 , yj ) ’ φ(xi , yj ) φ(xi+1 , yj ) ’ φ(xi , yj ) ‚φ

’

= = (xi , yj )

1

|Vi,j | |(xi+1 , yj ) ’ (xi , yj )| |xi+1 ’ xi | ‚x

2

Vi,j ‚φ

Similarly, converges to (xi , yj ).

2

|Vi,j | ‚y

40 3. DIFFERENTIAL FORMS

Let™s summarize what we have so far. We have de¬ned f (x, y) so that

1 2 1 2

ωφ(xi ,yj ) (Vi,j , Vi,j ) = f (xi , yj ) dx § dy(Vi,j , Vi,j )

i j i j

1 2

Vi,j Vi,j 1 2

dx § dy(Vi,j , Vi,j )

= ωφ(xi ,yj ) ,2

1

|Vi,j | |Vi,j |

i j

We have also shown that when we take the limit as the distance between adjacent

partition point tends toward 0 this sum converges to the sum

‚φ ‚φ 1 2

(x, y) dx § dy(Vi,j , Vi,j )

ωφ(x,y) (x, y),

‚x ‚y

i j

Hence, it must be that

‚φ ‚φ

(x, y), (x, y) dx § dy

(1) ω= ωφ(x,y)

‚x ‚y

M R

At ¬rst glance, this seems like a very complicated formula. Let™s break it down

by examining the integrand on the right. The most important thing to notice is that

this is just a di¬erential 2-form on R, even though ω is a 2-form on R3 . For each

pair of numbers, (x, y), the function ωφ(x,y) ‚φ (x, y), ‚φ (x, y) just returns some real

‚x ‚y

number. Hence, the entire integrand is of the form g dx § dy, where g : R ’ R.

The only way to really convince oneself of the usefulness of this formula is to

actually use it.

Example 3.3. Let M denote the top half of the unit sphere in R3 . Let ω =

z 2 dx § dy be a di¬erential 2-form on R3 . Calculating ω directly by setting up

M

a Riemann sum would be next to impossible. So we employ the parameterization

√

φ(r, t) = (r cos t, r sin t, 1 ’ r 2 ), where 0 ¤ t ¤ 2π and 0 ¤ r ¤ 1.

2. INTEGRATING DIFFERENTIAL 2-FORMS 41

‚φ ‚φ

(r, t) dr § dt

ω= ωφ(r,t) (r, t),

‚r ‚t

M R

’r

cos t, sin t, √ , ’r sin t, r cos t, 0 dr § dt

= ωφ(r,t)

2

1’r

R

cos t sin t

(1 ’ r 2 ) dr § dt

=

’r sin t r cos t

R

(1 ’ r 2 )(r)dr § dt

=

R

2π 1

π

r ’ r 3 dr dt =

=

2

0 0

‚φ

(r, t), ‚φ (r, t)

Notice that as promised, the term ωφ(r,t) in the second integral

‚r ‚t

3

above simpli¬ed to a function from R to R, r ’ r .

Exercise 3.3. Integrate the 2-form

1 1

ω = dy § dz ’ dx § dz

x y

over the top half of the unit sphere using the following parameterizations from rectangular,

cylindrical, and spherical coordinates:

(1) (x, y) ’ (x, y, 1 ’ x2 ’ y 2 ), where x2 + y 2 ¤ 1.

√

(2) (r, θ) ’ (r cos θ, r sin θ, 1 ’ r 2 ), where 0 ¤ θ ¤ 2π and 0 ¤ r ¤ 1.

(3) (θ, φ) ’ (sin φ cos θ, sin φ sin θ, cos φ), where 0 ¤ θ ¤ 2π and 0 ¤ φ ¤ π .

2

Answer: 4π.

Exercise 3.4. Let S be the surface in R3 parameterized by

Ψ(θ, z) = (cos θ, sin θ, z)

1

where 0 ¤ θ ¤ π and 0 ¤ z ¤ 1. Let ω = xyz dy § dz. Calculate ω. Answer: 3

S

Exercise 3.5. Let ω be the di¬erential 2-form on R3 given by

ω = xyz dx § dy + x2 z dy § dz ’ y dx § dz

42 3. DIFFERENTIAL FORMS

(1) Let P be the portion of the plane 3 = 2x + 3y ’ z in R3 which lies above the

ω. Answer: ’ 17 .

square {(x, y)|0 ¤ x ¤ 1, 0 ¤ y ¤ 1}. Calculate 12

P

(2) Let M be the portion of the graph of z = x + y 2 in R3 which lies above the

2

ω. Answer: ’ 29 .

rectangle {(x, y)|0 ¤ x ¤ 1, 0 ¤ y ¤ 2}. Calculate 6

M

Exercise 3.6. Let ω = f (x, y, z) dx § dy be a di¬erential 2-form on R3 . Let D be

some region in the xy-plane. Let M denote the portion of the graph of z = g(x, y) that

lies above D. Show that

ω= f (x, y, g(x, y)) dx dy

M D

Exercise 3.7. Let S be the surface obtained from the graph of z = f (x) = x3 , where

0 ¤ x ¤ 1, by rotating around the z-axis. Integrate the 2-form ω = y dx § dz over S.

3π

(Hint: use cylindrical coordinates to parameterize S.) Answer: .

5

3. Orientations

What would have happened in Example 3.3 if we had used the parameterization

√

′

φ (r, t) = (’r cos t, r sin t, 1 ’ r 2 ) instead? We leave it to the reader to check that

we end up with the answer ’π/2 rather than π/2. This is a problem. We de¬ned

ω before we started talking about parameterizations. Hence, the value which we

M

calculate for this integral should not depend on our choice of parameterization. So

what happened?

To analyze this completely, we need to go back to the de¬nition of ω from

M

the previous section. We noted at the time that a choice was made to calculate

1 2 1 2

ωpi,j (Vi,j , Vi,j ) instead of ωpi,j (’Vi,j , Vi,j ). But was this choice correct? The answer

is a resounding maybe! We are actually missing enough information to tell. An

orientation is precisely some piece of information about M which we can use to

make the right choice. This way we can tell a friend what M is, what ω is, and what

the orientation on M is, and they are sure to get the same answer. Recall Equation

1:

‚φ ‚φ

(x, y) dx § dy

ω= ωφ(x,y) (x, y),

‚x ‚y

M R

3. ORIENTATIONS 43

Depending on the speci¬ed orientation of M, it may be incorrect to use Equation

1. Sometimes we may want to use:

‚φ ‚φ

ωφ(x,y) ’ (x, y) dx § dy

ω= (x, y),

‚x ‚y

M R

Both ω and are linear. This just means the negative sign in the integrand on

the right can come all the way outside. Hence, we can write both this equation and

Equation 1 as

‚φ ‚φ

ω=± (x, y) dx § dy

(2) ωφ(x,y) (x, y),

‚x ‚y

M R

We now de¬ne:

Definition. An orientation of M is any piece of information that can be used

to decide, for each choice of parameterization φ, whether to use the “+” or “’” sign

in Equation 2, so that the integral will always yield the same answer.

We will see several ways to specify an orientation on M. The ¬rst way is to

simply pick a point p of M and choose any 2-form ν on Tp R3 such that ν(Vp1 , Vp2 ) = 0

whenever Vp1 and Vp2 are vectors tangent to M, and V1 is not a multiple of V2 . Don™t

confuse this 2-form with the di¬erential 2-form, ω, of Equation 2. The 2-form ν is

only de¬ned at the single tangent space Tp R3 , whereas ω is de¬ned everywhere.

Let™s see now how we can use ν to decide whether to use the “+” or “’” sign in

Equation 2. All we must do is calculate ν ‚φ (xp , yp ), ‚φ (xp , yp ) , where φ(xp , yp ) =

‚x ‚y

p. If the result is positive then we will use the “+” sign to calculate the integral in

Equation 2. If it™s negative then we use the “’” sign. Let™s see how this works with

an example.

Example 3.4. Let™s revisit Example 3.3. The problem was to integrate the form

z 2 dx § dy over M, the top half of the unit sphere. But no orientation was ever

given for M, so the problem wasn™t very well stated. Let™s pick an easy point,

√ √

p, on M: (0, 2/2, 2/2). The vectors 1, 0, 0 p and 0, 1, ’1 p in Tp R3 are

both tangent to M. To give an orientation on M all we have to do is specify a

44 3. DIFFERENTIAL FORMS

2-form ν on Tp R3 such that ν( 1, 0, 0 , 0, 1, ’1 ) = 0. Let™s pick an easy one:

ν = dx § dy.