3 3

Exercise 3.12. Let E be the region in R2 parameterized by Ψ(u, v) = (u2 + v 2 , 2uv),

where 0 ¤ u ¤ 1 and 0 ¤ v ¤ 1. Evaluate

1

√ dA

x’y

E

Answer: 4

Exercise 3.13. Let R be the region of the xy-plane bounded by the ellipse 9x2 + 4y 2 =

36. Integrate the 2-form ω = x2 dx § dy over R. (Hint: Use the parameterization

φ(u, v) = (2u, 3v).) Answer: 6π.

Example 3.8. Often in multivariable calculus classes we integrate functions

f (x, y) over regions R bounded by the graphs of equations y = g1 (x) and

y = g2 (x), and by the lines x = a and x = b, where g1 (x) < g2 (x) for all

x ∈ [a, b]. We show here that such problems can always be translated into inte-

grals over rectangular regions.

The region R described above is parameterized by

Ψ(u, v) = (u, (1 ’ v)g1 (u) + vg2 (u))

50 3. DIFFERENTIAL FORMS

where a ¤ u ¤ b and 0 ¤ v ¤ 1. The partials of this parameterization are

‚Ψ dg1(u) dg2 (u)

1, (1 ’ v)

= +v

‚u du du

‚Ψ

= 0, ’g1 (u) + g2 (u)

‚v

Hence,

1 (u) 2 (u)

1 (1 ’ v) dgdu + v dgdu

dx § dy = = ’g1 (u) + g2 (u)

’g1 (u) + g2 (u)

0

We conclude with the identity

b g2 (u) 1

b

f (u, (1 ’ v)g1 (u) + vg2 (u))(g2(u) ’ g1 (u)) dv du

f (x, y) dy dx =

0

a g1 (u) a

1

b

g2 (u)f (u, (1 ’ v)g1 (u) + vg2(u)) dv du

=

0

a

1

b

’ g1 (u)f (u, (1 ’ v)g1 (u) + vg2 (u)) dv du

0

a

This may be of more theoretical importance than practical (see Example 5.1).

Example 3.9. Let V = {(r, θ, z)|1 ¤ r ¤ 2, 0 ¤ z ¤ 1}. (V is the region

between the cylinders of radii 1 and 2 and between the planes z = 0 and z = 1.)

We will calculate

z(x2 + y 2 ) dx § dy § dz

V

The region V is best parameterized using cylindrical coordinates:

Ψ(r, θ, z) = (r cos θ, r sin θ, z),

where 1 ¤ r ¤ 2, 1 ¤ θ ¤ 2π, and 0 ¤ z ¤ 1.

Computing the partials:

5. INTEGRATING n-FORMS ON PARAMETERIZED SUBSETS OF Rn 51

‚Ψ

= cos θ, sin θ, 0

‚r

‚Ψ

’r sin θ, r cos θ, 0

=

‚θ

‚Ψ

= 0, 0, 1

‚z

Hence,

cos θ sin θ 0

’r sin θ r cos θ 0

dx § dy § dz = =r

0 01

Also,

z(x2 + y 2 ) = z(r 2 cos2 θ + r 2 sin2 θ) = zr 2

So we have

1 2π 2

z(x2 + y 2) dx § dy § dz = (zr 2 )(r) dr dθ dz

0 0 1

V

1 2π 2

zr 3 dr dθ dz

=

0 0 1

1 2π

15

= z dθ dz

4

0 0

1

15π

= z dz

2

0

15π

=

4

Exercise 3.14. Integrate the 3-form ω = x dx § dy § dz over the region of R3 in the

¬rst octant bounded by the cylinders x2 + y 2 = 1 and x2 + y 2 = 4, and the plane z = 2.

Answer: 143

52 3. DIFFERENTIAL FORMS

Exercise 3.15. Let R be the region in the ¬rst octant of R3 bounded by the spheres

x2 + y 2 + z 2 = 1 and x2 + y 2 + z 2 = 4. Integrate the 3-form ω = dx § dy § dz over R.

Answer: ’7π 6

6. Summary: How to Integrate a Di¬erential Form

6.1. The Steps. To compute the integral of a di¬erential n-form, ω, over a

region, S, the steps are as follows:

(1) Choose a parameterization, Ψ : R ’ S, where R is a subset of Rn (see

Figure 5).

z

v

S

Ψ

R

u

y

x

Figure 5.

(2) Find all n vectors given by the partial derivatives of Ψ. Geometrically, these

are tangent vectors to S which span its tangent space (see Figure 6).

z

‚Ψ

‚u

y

x

‚Ψ

‚v

Figure 6.

(3) Plug the tangent vectors into ω at the point Ψ(u1 , u2, ..., un ).

(4) Integrate the resulting function over R.

6. SUMMARY: HOW TO INTEGRATE A DIFFERENTIAL FORM 53

6.2. Integrating 2-forms. The best way to see the above steps in action is to

look at the integral of a 2-form over a surface in R3 . In general, such a 2-form is

given by

ω = f1 (x, y, z) dx § dy + f2 (x, y, z) dy § dz + f3 (x, y, z) dx § dz

To integrate ω over S we now follow the steps:

(1) Choose a parameterization, Ψ : R ’ S, where R is a subset of R2 .

Ψ(u, v) = (g1 (u, v), g2(u, v), g3(u, v))

(2) Find both vectors given by the partial derivatives of Ψ.

‚Ψ ‚g1 ‚g2 ‚g3

= , ,

‚u ‚u ‚u ‚u

‚Ψ ‚g1 ‚g2 ‚g3

= , ,

‚v ‚v ‚v ‚v

(3) Plug the tangent vectors into ω at the point Ψ(u, v).

To do this, x, y, and z will come from the coordinates of Ψ. That

is, x = g1 (u, v), y = g2 (u, v), and z = g3 (u, v). Terms like dx § dy will

be determinants of 2 — 2 matrices, whose entries come from the vectors

computed in the previous step. Geometrically, the value of dx § dy will be

the area of the parallelogram spanned by the vectors ‚Ψ and ‚Ψ

(tangent

‚u ‚v

vectors to S), projected onto the dx-dy plane (see Figure 7).

The result of all this will be:

‚g1 ‚g2 ‚g2 ‚g3

f1 (g1 , g2 , g3 ) + f2 (g1 , g2 , g3)

‚u ‚u ‚u ‚u

‚g1 ‚g2 ‚g2 ‚g3

‚u ‚v ‚u ‚v

‚g1 ‚g3

+f3 (g1 , g2 , g3 ) ‚u ‚u

‚g1 ‚g3

‚u ‚v

Note that when you simplify this you get a function of u and v.

(4) Integrate the resulting function over R. In other words, if h(u, v) is the

function you ended up with in the previous step, then compute

h(u, v) du dv

R

54 3. DIFFERENTIAL FORMS

dz

‚Ψ

‚u

‚Ψ

‚v

dy

dx

‚Ψ ‚Ψ

Area=dx § dy ,

‚u ‚v

Figure 7. Evaluating dx § dy geometrically

In practice, the limits of integration will come from the shape of R, determined

in Step 1. They will all be constants only if R was a rectangle.

6.3. A sample 2-form. Let ω = (x2 + y 2 ) dx § dy + z dy § dz. Let S denote

the subset of the cylinder x2 + y 2 = 1 that lies between the planes z = 0 and z = 1.

(1) Choose a parameterization, Ψ : R ’ S.

Ψ(u, v) = (cos θ, sin θ, z)

Where R = {(θ, z)|0 ¤ θ ¤ 2π, 0 ¤ z ¤ 1}.

(2) Find both vectors given by the partial derivatives of Ψ.

‚Ψ

’ sin θ, cos θ, 0

=

‚θ