1. Quiz 13
Let ω = f (x)dx1 ∧ · · · ∧ dxn be a smooth n-form on an open subset U of Rn . Here
xi : Rn → R is the i-th coordinate function on Rn , i.e.
xi (a1 , · · · , an ) = ai
for any (a1 , · · · , an ) ∈ Rn . We define the integral of ω over a Jordan measurable set S ⊆ U
by
Z
Z
f (x)dµ
ω=
S
S
where dµ denotes the Jordan measure on Rn . When n = 1, the definition is obvious. If
ω = f (x, y)dx ∧ dy is a smooth two form on R2 and S is a Jordan measurable set of R2 , we
define
ZZ
Z
f (x, y)dA.
ω=
S
S
If ω = f (x, y, z)dx ∧ dy ∧ dz is a smooth three form on R3 and S is Jordan measurable, then
Z
ZZZ
ω=
f (x, y, z)dV.
S
S
Remark. Since f (x, y)dy ∧ dx = −f (x, y)dx ∧ dy, we have
Z
ZZ
f (x, y)dy ∧ dx = −
f (x, y)dA.
S
S
Rn
If U and V are open subset of
and φ : U → V is a C 1 -diffeomorphism, then the
change of variable formula reads : for any smooth n-form ω on V, one has
Z
Z
(1.1)
ω=
φ∗ ω
φ(S)
S
where S is a Jordan measurable subset of U.
A singular 0 cube in Rn is a function c : {0} → Rn . A singular 0-cube in Rn is equivalent
to a point in Rn . A zero form on Rn is a function f : Rn → R. The integral of a zero form
f over a singular 0 cube c : {0} → Rn is defined to be
Z
f = f (c(0)).
c
Rn
Let k ≥ 1. A singular k cube in
is a function c : [0, 1]k → Rn such that there exists
k
an open subset U containing [0, 1] of Rn and a smooth function ψ : U → Rn such that
ψ|[0,1]k = c. If ω is a k-form on Rn , we define the integral of ω over c by
Z
Z
ω=
ψ ∗ ω.
c
[0,1]k
(Sometimes, we will use the notation c for ψ since ψ is an extension of c.) A singular 1-cube
is usually called a singular curve and a singular 2-cube is usually called a singular surface.
The integral of a one form over a singular one cube is called a line integral and the integral
of a two form over a singular two cube is called a surface integral.
1
2
(1) Evaluate
the following integral of differential forms.
Z
x
(a)
dx ∧ dy.
1
+
xy
Z[0,1]×[0,1]
(b)
(5 − 3x + 2y)dy ∧ dx.
[0,1]×[0,1]
(c) Let R = {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ 4, y ≥ 0}. Evaluate the integral
Z
(3x + 4y 2 )dx ∧ dy
R
via φ : [1, 2] × [0, 2π] → R2 , φ(r, θ) = (r cos θ, r sin θ) and by the change of
variable formula (1.1).
(2) Evaluate
the line integral
Z
xydx + 3y 2 dy where γ(t) = (11t4 , t3 ) for 0 ≤ t ≤ 1.
Z
xdx − zdy + ydz where γ(t) = (2t, 3t, −t2 ) for
(b) Evaluate the line integral
(a)
γ
γ
−1 ≤ t ≤ 1.
(3) Evaluate the surface integral
Z
xdy ∧ dz + ydz ∧ dx + zdx ∧ dy
(x2 + y 2 + z 2 )3/2
X
where X(θ, ψ) = (sin θ cos ψ, sin θ sin ψ, cos θ) for 0 ≤ θ ≤ π and 0 ≤ ψ ≤ 2π.
Let D be a compact Jordan measurable region on R2 whose boundary C = ∂D is a
curve. If ω = P (x, y)dx + Q(x, y)dy is a smooth one-form on an open subset U
of R2 so that D is contained in U. Green’s Theorem states that
I
ZZ
ω=
dω.
C 1 -oriented
C
D
It is easy to show that dω = (Qx − Py )dx ∧ dy. Hence the Green’s Theorem can be rewritten
as
I
ZZ
P (x, y)dx + Q(x, y)dy =
(Qx − Py )dx ∧ dy.
C
D
Use Green’s Theorem
to evaluate the following integrals.
I
(1) Evaluate
x4 dx+xydy where C is the triangle curve consisting of the line segments
C
from (0, 0)I to (1, 0), from (1, 0) to (0, 1) and from (0, 1) to (0, 0).
p
(2) Evaluate (3y − esin x )dx + (7x + y 4 + 1)dy where C is the circle x2 + y 2 = 9 with
C
positive orientation (counterclockwise).
y
x
(3) Let ω = − 2
dx + 2
dy be the smooth one form on R2 \ {(0, 0)} and
2
x +y
x + y2
I
2
2
2
C = {(x, y) ∈ R : 4x + 16y = 16} with positive orientation. Evaluate
ω.
C