√ √

rameterization φ′ (r, t) = (’r cos t, r sin t, 1 ’ r 2 ). First note that φ′ ( 2/2, π/2) =

√ √

(0, 2/2, 2/2) and

√ √ √

‚φ′ 2 π ‚φ′ 2 π 2

, ) = ( 0, 1, ’1 ,

( , ), ( , 0, 0 )

‚r 2 2 ‚t 2 2 2

Now we check the value of ν when this pair is plugged in:

√ √

01

2 2

, 0, 0 ) = √2

dx § dy( 0, 1, ’1 , =’

0

2 2

2

The sign of this result is “’” so we need to use the negative sign in Equation 2

in order to use φ′ to evaluate the integral of ω over M.

‚φ′ ‚φ′

ω=’ (r, t) dr § dt

ωφ′ (r,t) (r, t),

‚r ‚t

M R

π

’ cos t sin t

(1 ’ r 2 )

=’ dr dt =

r sin t r cos t 2

R

Very often the surface that we are going to integrate over is given to us by a

parameterization. In this case there is a very natural choice of orientation. Just use

the “+” sign in Equation 2! We will call this the orientation of M induced by the

parameterization. In other words, if you see a problem phrased like this, “Calculate

the integral of the form ω over the manifold M given by parameterization φ with the

induced orientation,” then you should just go back to using Equation 1 and don™t

worry about anything else.

Exercise 3.8. Let M be the image of the parameterization, φ(a, b) = (a, a + b, ab),

where 0 ¤ a ¤ 1, and 0 ¤ b ¤ 1. Integrate the form ω = 2z dx§dz+y dy§dz’x dx§dz

over M using the orientation induced by φ. Answer: ’5/6

There is one subtle technical point here that should be addressed. The novice

reader may want to skip this for now. Suppose someone gives you a surface de¬ned by

a parameterization and tells you to integrate some 2-form over it, using the induced

4. INTEGRATING n-FORMS ON Rm 45

orientation. But you are clever, and you realize that if you change parameterizations

you can make the integral easier. Which orientation do you use? The problem is

that the orientation induced by your new parameterization may not be the same as

the one induced by the original parameterization.

To ¬x this we need to see how we can de¬ne a 2-form on some tangent space

Tp R3 , where p is a point of M, that yields an orientation of M that is consis-

tent with the one induced by a parameterization φ. This is not so hard. If dx §

dy ‚φ (xp , yp ), ‚φ (xp , yp) is positive then we simply let ν = dx § dy. If it is negative

‚x ‚y

then we let ν = ’dx§dy. In the unlikely event that dx§dy ‚φ (xp , yp ), ‚φ (xp , yp ) =

‚x ‚y

0 we can remedy things by either changing the point p or by using dx § dz instead

of dx § dy. Once we have de¬ned ν we know how to integrate M using any other

parameterization.

4. Integrating n-forms on Rm

In the previous sections we saw how to integrate a 2-form over a region in R2 , or

over a subset of R3 parameterized by a region in R2 . The reason that these dimensions

were chosen was because there is nothing new that needs to be introduced to move

to the general case. In fact, if the reader were to go back and look at what we did

he/she would ¬nd that almost nothing would change if we had been talking about

n-forms instead.

Before we jump to the general case, we will work one example showing how to

integrate a 1-form over a parameterized curve.

Example 3.5. Let C be the curve in R2 parameterized by

φ(t) = (t2 , t3 )

where 0 ¤ t ¤ 2. Let ν be the 1-form y dx + x dy. We calculate ν.

C

The ¬rst step is to calculate

dφ

= 2t, 3t2

dt

So, dx = 2t and dy = 3t . From the parameterization we also know x = t2 and

2

y = t3 . Hence, since ν = y dx + x dy, we have

46 3. DIFFERENTIAL FORMS

dφ

= (t3 )(2t) + (t2 )(3t2 ) = 5t4

νφ(t)

dt

Finally, we integrate:

2

dφ

ν= νφ(t) dt

dt

0

C

2

5t4 dt

=

0

2

= t5 0

= 32

Exercise 3.9. Let C be the curve in R3 parameterized by φ(t) = (t, t2 , 1 + t), where

0 ¤ t ¤ 2. Integrate the 1-form ω = y dx + z dy + xy dz over C using the induced

orientation. Answer: 16.

Exercise 3.10. Let M be the line segment in R2 which connects (0, 0) to (4, 6).

An orientation on M is speci¬ed by the 1-form ’dx on T(2,3) R2 . Integrate the form

2 3 2

cos 6 ’ 2 sin 4 ’

ω = sin y dx + cos x dy over M. Answer: 3 3

To proceed to the general case, we need to know what the integral of an n-form

over a region of Rn is. The steps to de¬ne such an object are precisely the same as

before, and the results are similar. If our coordinates in Rn are (x1 , x2 , ..., xn ) then an

n-form is always given by f (x1 , ..., xn )dx1 § dx2 § ... § dxn . Going through the steps

we ¬nd that the de¬nition of ω is exactly the same as the de¬nition we learned in

Rn

calculus for f dx1 dx2 ...dxn .

Rn

Exercise 3.11. Let „¦ be the cube in R3

{(x, y, z)| 0 ¤ x, y, z ¤ 1}

1

Let γ be the 3-form x2 z dx § dy § dz. Calculate γ. Answer: 6

„¦

4. INTEGRATING n-FORMS ON Rm 47

Moving on to integrals of n-forms over subsets of Rm parameterized by a region

in Rn we again ¬nd nothing surprising. Suppose we denote such a subset as M. Let

φ : R ‚ Rn ’ M ‚ Rm be a parameterization. Then we ¬nd that the following

generalization of Equation 2 must hold:

‚φ ‚φ

ω=± (x1 , ...xn ) dx1 § ... § dxn

(3) ωφ(x1 ,...,xn) (x1 , ...xn ), ...,

‚x1 ‚xn

M R

To decide whether or not to use the negative sign in this equation we must specify

an orientation. Again, one way to do this is to give an n-form ν on Tp Rm , where p is

some point of M. We use the negative sign when the value of

‚φ ‚φ

ν (x1 , ...xn ), ..., (x1 , ...xn )

‚x1 ‚xn

is negative, where φ(x1 , ...xn ) = p. If M was originally given by a parameterization

and we are instructed to use the induced orientation then we can ignore the negative

sign.

Example 3.6. Suppose φ(a, b, c) = (a + b, a + c, bc, a2 ), where 0 ¤ a, b, c ¤ 1.

Let M be the image of φ with the induced orientation. Suppose ω = dy § dz §

dw ’ dx § dz § dw ’ 2y dx § dy § dz. Then,

‚φ ‚φ ‚φ

(a, b, c) da § db § dc

ω= ωφ(a,b,c) (a, b, c), (a, b, c),

‚a ‚b ‚c

M R

ωφ(a,b,c) ( 1, 1, 0, 2a , 1, 0, c, 0 , 0, 1, b, 0 ) da § db § dc

=

R

1 0 2a 1 0 2a 110

0 c 0 ’ 1 c 0 ’ 2(a + c) 1 0 c da § db § dc

=

1b 0 0b 0 01b

R

1 1 1

7

2bc + 2c2 da db dc =

=

6

0 0 0

48 3. DIFFERENTIAL FORMS

5. Integrating n-forms on parameterized subsets of Rn

There is a special case of Equation 3 which is worth noting. Suppose φ is a

parameterization that takes some subregion, R, of Rn into some other subregion, M,

of Rn and ω is an n-form on Rn . At each point ω is just a volume form, so it can be

written as f (x1 , ..., xn ) dx1 § ... § dxn . If we let x = (x1 , ...xn ) then Equation 3 can

¯

be written as:

‚φ ‚φ

f (¯)dx1 ...dxn = ±

x f (φ(¯))

x (¯), ...,

x (¯) dx1 ...dxn

x

‚x1 ‚xn

M R

When n = 1 this is just the substitution rule for integration from Calculus. For

other n this is the general change of variables formula.

Example 3.7. We will use the parameterization Ψ(u, v) = (u, u2 + v 2 ) to eval-

uate

(x2 + y) dA

R

where R is the region of the xy-plane bounded by the parabolas y = x2 and

y = x2 + 4, and the lines x = 0 and x = 1.

The ¬rst step is to ¬nd out what the limits of integration will be when we change

coordinates.

y = x2 ’ u2 + v 2 = u2 ’ v = 0

y = x2 + 4 ’ u2 + v 2 = u2 + 4 ’ v = 2

x=0’u=0

x=1’u=1

Next, we will need the partial derivatives.

‚Ψ

=< 1, 2u >

‚u

‚Ψ

=< 0, 2v >

‚v

Finally, we can integrate.

5. INTEGRATING n-FORMS ON PARAMETERIZED SUBSETS OF Rn 49

(x2 + y) dA = (x2 + y) dx § dy

R R

2 1

1 2u

u2 + (u2 + v 2 )

= du dv

0 2v

0 0

2 1

4vu2 + 2v 3 du dv

=

0 0

2

4

v + 2v 3 dv

=

3

0

8 32