Integration On Manifolds

Diffeomorphism

U, V \subseteq ℝ^{n}

are open, a diffeomorphism

h : U \to V

is a bijection such that both

h

and its inverse function

h^{- 1}

are

C^{\infty}

k

-Dimensional Manifold in Euclidean Space

A subset

M \subseteq ℝ^{n}

is a

k

-dimensional manifold if for every

x \in M

, there are open sets

U, V \subseteq ℝ^{n}

with

x \in U

, and a diffeomorphism

h : U \to V

, such that

h (U \cap M) = V \cap (ℝ^{k} \times {0}) .

Regular Level Set is a Manifold

Let

A \subseteq ℝ^{n}

be open and let

g : A \to ℝ^{p}

C^{\infty}

. If the Jacobian matrix

g^{'} (x)

has rank

p

whenever

g (x) = 0

, then

g^{- 1} (0)

is an

(n - p)

-dimensional manifold in

ℝ^{n}

Let

x \in g^{- 1} (0)

, and write the coordinate variables in

ℝ^{n}

z_{1}, \dots, z_{n}

. The Jacobian matrix

g^{'} (x)

is a

p \times n

matrix whose

j

-th column is the derivative of

g

in the

z_{j}

-direction.

Since $g^{'} (x)$ has rank $p$ , its $n$ columns span a $p$ -dimensional subspace of $ℝ^{p}$ . Therefore some $p$ of those $n$ columns form a basis of $ℝ^{p}$ . Choose their indices and call them $j_{1}, \dots, j_{p}$ .

Let $v = (z_{j_{1}}, \dots, z_{j_{p}})$ , and let $u$ be the remaining $n - p$ coordinate variables. With the coordinates ordered as $(u, v) \in ℝ^{n - p} \times ℝ^{p}$ , the $p \times p$ matrix ${[\frac{\partial g^{i}}{\partial v_{j}} (x)]}_{1 \leq i, j \leq p}$ has those selected columns as its columns. Since those columns form a basis of $ℝ^{p}$ , this matrix is invertible.

By the implicit function theorem, after restricting to a small neighborhood, the equation $g (u, v) = 0$ uniquely determines $v$ as a $C^{\infty}$ function $v = φ (u)$ . Define $h (u, v) = (u, v - φ (u)) .$ Then $h$ is a local diffeomorphism, and it sends the local set $g^{- 1} (0)$ exactly to $ℝ^{n - p} \times {0}$ . This is the local model required in the definition of a manifold, so $g^{- 1} (0)$ is an $(n - p)$ -dimensional manifold.

Unit Sphere is a 2-Dimensional Manifold

The unit sphere

S^{2} = {(x, y, z) \in ℝ^{3} : x^{2} + y^{2} + z^{2} = 1}

is a 2-dimensional manifold in Euclidean space.

Define

g : ℝ^{3} \to ℝ

g (x, y, z) = x^{2} + y^{2} + z^{2} - 1.

Then

S^{2} = g^{- 1} (0)

, and

g

C^{\infty}

. Its Jacobian matrix is

g^{'} (x, y, z) = [\begin{matrix} 2 x & 2 y & 2 z \end{matrix}] .

(x, y, z) \in S^{2}

, then

x^{2} + y^{2} + z^{2} = 1

, so not all of

x, y, z

are zero. Therefore

g^{'} (x, y, z)

has rank

1

g^{- 1} (0)

. By the regular level set theorem,

S^{2}

is a

(3 - 1)

-dimensional manifold in

ℝ^{3}

, so it is a 2-dimensional manifold.

Coordinate System on a Manifold

Let

M \subseteq ℝ^{n}

be a

k

-dimensional manifold. A coordinate system around

x \in M

is the function

f : W \to ℝ^{n}

, where

W \subseteq ℝ^{k}

is the coordinate domain and

f (W)

is the patch of

M

being described. The function

f

must satisfy:

$W$ is open in $ℝ^{k}$ , and there is an open set $U \subseteq ℝ^{n}$ such that $f (W) = M \cap U$ and $x \in f (W)$ .
$f$ is one-to-one and $C^{\infty}$ .
The Jacobian matrix $f^{'} (y)$ has rank $k$ for every $y \in W$ .
The inverse $f^{- 1} : f (W) \to W$ is continuous.

A point

y \in W

is a coordinate value, and

f (y) \in M

is the corresponding point on the manifold.

Upper Hemisphere Coordinate System on the Sphere

Let

S^{2} = {(x, y, z) \in ℝ^{3} : x^{2} + y^{2} + z^{2} = 1}

and let

W = {(u, v) \in ℝ^{2} : u^{2} + v^{2} < 1} .

Show that

f : W \to ℝ^{3}

, defined by

f (u, v) = (u, v, \sqrt{1 - u^{2} - v^{2}}),

is a coordinate system around every point of the upper hemisphere

S^{2} \cap {z > 0}

The set

W

is open in

ℝ^{2}

. The map

f

C^{\infty}

W

, because

1 - u^{2} - v^{2} > 0

there and the square root function is smooth on positive real numbers.

The map $f$ is one-to-one: if $f (u_{1}, v_{1}) = f (u_{2}, v_{2})$ , then comparing the first two coordinates gives $u_{1} = u_{2}$ and $v_{1} = v_{2}$ .

Let $U = {(x, y, z) \in ℝ^{3} : z > 0}$ . Then $f (W) = S^{2} \cap U .$ Indeed, every $f (u, v)$ lies on $S^{2}$ and has positive $z$ -coordinate. Conversely, if $(x, y, z) \in S^{2} \cap U$ , then $x^{2} + y^{2} = 1 - z^{2} < 1$ , so $(x, y) \in W$ , and because $z > 0$ , $z = \sqrt{1 - x^{2} - y^{2}} .$ Thus $(x, y, z) = f (x, y)$ .

The inverse function on the image is $f^{- 1} : S^{2} \cap U \to W, f^{- 1} (x, y, z) = (x, y),$ which is continuous.

Finally, the Jacobian matrix is $f^{'} (u, v) = [\begin{matrix} 1 & 0 \\ 0 & 1 \\ \frac{- u}{\sqrt{1 - u^{2} - v^{2}}} & \frac{- v}{\sqrt{1 - u^{2} - v^{2}}} \end{matrix}] .$ Its first two rows form the $2 \times 2$ identity matrix, so its columns are linearly independent and the matrix has rank $2$ . Therefore $f$ satisfies every condition in the definition of a coordinate system around every point of $S^{2} \cap {z > 0}$ .

Standard Coordinate System on the

x

-Axis

Let

M = ℝ \times {0} \subseteq ℝ^{2} .

Show that

f : ℝ \to ℝ^{2}

, defined by

f (t) = (t, 0),

is a coordinate system around every point of

M

Let

x = (a, 0) \in M

. The domain

W = ℝ

is open in

ℝ

, and

f

is one-to-one, because

f (s) = f (t)

implies

(s, 0) = (t, 0)

, so

s = t

The map $f$ is $C^{\infty}$ , and its Jacobian matrix is $f^{'} (t) = [\begin{matrix} 1 \\ 0 \end{matrix}] .$ This matrix has rank $1$ for every $t \in ℝ$ .

Taking $U = ℝ^{2}$ , we have $f (W) = M = M \cap U$ . The inverse function on the image is $f^{- 1} : M \to ℝ, f^{- 1} (t, 0) = t,$ which is continuous. Therefore $f$ satisfies every condition in the definition of a coordinate system around $x$ . Since $x \in M$ was arbitrary, $f$ is a coordinate system around every point of $M$ .

Induced Volume of a Manifold Subset

Let

M \subseteq ℝ^{n}

be a

k

-dimensional manifold, and let

E \subseteq M

. Suppose first that

E \subseteq f (W)

for a coordinate system

f : W \to ℝ^{n}

. The induced $k$ -dimensional volume of

E

{vol}_{M} (E) = \int_{f^{- 1} (E)} \sqrt{\det ({f^{'} (u)}^{T} f^{'} (u))} d u,

whenever this integral exists. The matrix

f^{'} {(u)}^{T} f^{'} (u)

records the inner products of the coordinate tangent vectors, so its determinant gives the squared

k

-dimensional volume distortion of the coordinate map at

u

If $E$ is not contained in one coordinate image, decompose $E$ into countably many disjoint measurable pieces, each contained in a coordinate image, and add the above values. The change-of-variables formula makes the result independent of the chosen coordinate systems. When $k = 2$ , this induced volume is called the surface area of $E$ , and is denoted ${area}_{M} (E)$ .

Surface Area Measure

Let

M \subseteq ℝ^{n}

be a two-dimensional manifold. The surface area measure on

M

, denoted

A_{M}

, is the measure whose value on a measurable subset

E \subseteq M

is the surface area of

E

A_{M} (E) = {area}_{M} (E) .

Thus an expression such as

\int_{M} f (q) d A_{M} (q)

means integration with respect to the surface area measure

A_{M}

. The symbol

A_{M}

denotes a measure on subsets of

M

, not a scalar-valued function of the point

q

Surface Area Density in Coordinates

Let

M \subseteq ℝ^{3}

be a two-dimensional manifold, and let

f : U \to M, U \subseteq ℝ^{2},

be a coordinate system. The surface area density of $M$ in the coordinates $f$ is the function

ρ_{f} : U \to ℝ^{\geq 0}

defined by

ρ_{f} (u) = \sqrt{\det (f^{'} {(u)}^{T} f^{'} (u))} .

This is the factor that appears in the induced volume definition when

k = 2

. Thus locally,

d A = ρ_{f} (u) d u .

Surface Area Density Cross Product Formula

Let

M \subseteq ℝ^{3}

be a two-dimensional manifold, and let

f : U \to M

be a coordinate system with

U \subseteq ℝ^{2}

. Define the coordinate tangent vectors

f_{1} (u) = \frac{\partial f}{\partial u_{1}} (u), f_{2} (u) = \frac{\partial f}{\partial u_{2}} (u) .

Then the surface area density in the coordinates

f

ρ_{f} (u) = ‖ f_{1} (u) \times f_{2} (u) ‖ .

The Jacobian matrix

f^{'} (u)

has columns

f_{1} (u)

and

f_{2} (u)

. Therefore

f^{'} {(u)}^{T} f^{'} (u) = [\begin{matrix} f_{1} (u) \cdot f_{1} (u) & f_{1} (u) \cdot f_{2} (u) \\ f_{2} (u) \cdot f_{1} (u) & f_{2} (u) \cdot f_{2} (u) \end{matrix}] .

Its determinant is

(f_{1} \cdot f_{1}) (f_{2} \cdot f_{2}) - {(f_{1} \cdot f_{2})}^{2},

where the argument

u

is suppressed to reduce notation. By the component formula for the cross product and the definition of dot product,

{‖ f_{1} \times f_{2} ‖}^{2} = (f_{1} \cdot f_{1}) (f_{2} \cdot f_{2}) - {(f_{1} \cdot f_{2})}^{2} .

Hence

\det (f^{'} {(u)}^{T} f^{'} (u)) = {‖ f_{1} (u) \times f_{2} (u) ‖}^{2} .

Taking the nonnegative square root in the definition of surface area density gives

ρ_{f} (u) = ‖ f_{1} (u) \times f_{2} (u) ‖ .

Induced Volume of the Unit Square

Let

M = ℝ^{2} \times {0} \subseteq ℝ^{3}

and let

E = [0,1] \times [0,1] \times {0} \subseteq M .

Use the induced volume of a manifold subset to compute the surface area of

E

Use the coordinate system

f : ℝ^{2} \to ℝ^{3}

defined by

f (u, v) = (u, v, 0) .

Then

f^{- 1} (E) = [0,1] \times [0,1]

, and the Jacobian matrix is

f^{'} (u, v) = [\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix}] .

Therefore

f^{'} {(u, v)}^{T} f^{'} (u, v) = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}],

so its determinant is

1

. By the definition of induced volume,

{area}_{M} (E) = \int_{[0,1] \times [0,1]} \sqrt{1} d u d v = 1.

Thus the induced surface area of the unit square is

1

, as expected.

Induced Area of the Projected Unit Disk on the Sphere

Let

S^{2} = {(x, y, z) \in ℝ^{3} : x^{2} + y^{2} + z^{2} = 1} .

Project the open unit disk

W = {(u, v) \in ℝ^{2} : u^{2} + v^{2} < 1}

onto the top of the unit sphere by

f (u, v) = (u, v, \sqrt{1 - u^{2} - v^{2}}) .

Use the induced volume of a manifold subset to compute the surface area of

f (W)

The Jacobian matrix of

f

f^{'} (u, v) = [\begin{matrix} 1 & 0 \\ 0 & 1 \\ - \frac{u}{\sqrt{1 - u^{2} - v^{2}}} & - \frac{v}{\sqrt{1 - u^{2} - v^{2}}} \end{matrix}] .

Hence

f^{'} {(u, v)}^{T} f^{'} (u, v) = [\begin{matrix} 1 + \frac{u^{2}}{1 - u^{2} - v^{2}} & \frac{u v}{1 - u^{2} - v^{2}} \\ \frac{u v}{1 - u^{2} - v^{2}} & 1 + \frac{v^{2}}{1 - u^{2} - v^{2}} \end{matrix}] .

Its determinant is

(1 + \frac{u^{2}}{1 - u^{2} - v^{2}}) (1 + \frac{v^{2}}{1 - u^{2} - v^{2}}) - {(\frac{u v}{1 - u^{2} - v^{2}})}^{2} = \frac{1}{1 - u^{2} - v^{2}} .

Therefore

{area}_{S^{2}} (f (W)) = \int_{W} \frac{1}{\sqrt{1 - u^{2} - v^{2}}} d u d v .

The integrand becomes unbounded as

u^{2} + v^{2} \to 1

, so we first compute the area on smaller disks that stay away from the boundary circle. These smaller disks exhaust

W

, and the final area is obtained by taking their limit using monotone convergence.

For $0 < ρ < 1$ , let $W_{ρ} = {(u, v) \in ℝ^{2} : u^{2} + v^{2} < ρ^{2}} .$

Apply the polar coordinate change of variables on

W_{ρ}

with

q (u, v) = \frac{1}{\sqrt{1 - u^{2} - v^{2}}} .

Since

q (r \cos θ, r \sin θ) = \frac{1}{\sqrt{1 - r^{2}}},

the corollary gives

{area}_{S^{2}} (f (W_{ρ})) = \int_{0}^{2 π} \int_{0}^{ρ} \frac{r}{\sqrt{1 - r^{2}}} d r d θ

To evaluate the inner integral, use the substitution

s = 1 - r^{2}

, so

d s = - 2 r d r

. Therefore

\int_{0}^{ρ} \frac{r}{\sqrt{1 - r^{2}}} d r = - \frac{1}{2} \int_{1}^{1 - ρ^{2}} s^{- 1 / 2} d s = {[- \sqrt{1 - r^{2}}]}_{0}^{ρ} = 1 - \sqrt{1 - ρ^{2}},

where the bracket evaluation uses the fundamental theorem of calculus. Hence

{area}_{S^{2}} (f (W_{ρ})) = \int_{0}^{2 π} (1 - \sqrt{1 - ρ^{2}}) d θ = 2 π (1 - \sqrt{1 - ρ^{2}}) .

ρ \to 1^{-}

, the sets

W_{ρ}

increase to

W

. Since

q \geq 0

, the functions

q 𝟏_{W_{ρ}}

increase pointwise to

q 𝟏_{W}

. By the monotone convergence theorem,

\int_{W} q (u, v) d u d v = \lim_{ρ \to 1^{-}} \int_{W_{ρ}} q (u, v) d u d v .

Therefore taking the limit as

ρ \to 1^{-}

gives

{area}_{S^{2}} (f (W)) = 2 π .

Thus the projected unit disk covers the upper hemisphere, whose induced surface area is

2 π

Coordinate Characterization of Manifolds

A subset

M \subseteq ℝ^{n}

is a

k

-dimensional manifold if and only if every point of

M

has a coordinate system around it.

Suppose first that

M

is a

k

-dimensional manifold. Choose a diffeomorphism

h : U \to V

giving the local manifold model. Let

W = {a \in ℝ^{k} : (a, 0) \in h (M \cap U)}

and define

f : W \to ℝ^{n}

f (a) = h^{- 1} (a, 0)

. This gives

f (W) = M \cap U

, and the inverse coordinate map is continuous. If

H

denotes the first

k

components of

h

, then

H \circ f

is the identity on

W

, so

H^{'} (f (y)) f^{'} (y) = I

. Hence

f^{'} (y)

has rank

k

Conversely, suppose $f : W \to ℝ^{n}$ satisfies the coordinate condition near $x = f (y)$ . Choose $k$ component functions of $f$ whose $k \times k$ derivative matrix has nonzero determinant. After reordering coordinates, define $G (a, b) = f (a) + (0, b), b \in ℝ^{n - k} .$ Then $G^{'} (y, 0)$ is invertible. By the inverse function theorem, $G$ has a local differentiable inverse $h$ . Since $f^{- 1}$ is continuous on $f (W)$ , shrinking the neighborhoods gives $h (V \cap M) = V^{'} \cap (ℝ^{k} \times {0}) .$ This is exactly the local model for a $k$ -dimensional manifold.

Half-Space

The half-space

H^{k} \subseteq ℝ^{k}

H^{k} = {x \in ℝ^{k} : x^{k} \geq 0} .

Manifold with Boundary

A subset

M \subseteq ℝ^{n}

is a

k

-dimensional manifold with boundary if every point of

M

has a local diffeomorphic model either on

ℝ^{k} \times {0}

or on

H^{k}

\times {0}

. Points modeled on the boundary

{x^{k} = 0}

form the boundary of the manifold, denoted

\partial M

Boundary of a Manifold with Boundary is a Manifold

M

is a

k

-dimensional manifold with boundary, then

\partial M

is a

(k - 1)

-dimensional manifold and

M ∖ \partial M

is a

k

-dimensional manifold.

Let

x \in M

. By the definition of manifold with boundary, a neighborhood of

x

is locally diffeomorphic either to

ℝ^{k} \times {0}

or to

H^{k} \times {0}

If $x \notin \partial M$ , then in local coordinates $M$ is modeled on the interior of $H^{k}$ , which is locally $ℝ^{k}$ . Thus $M ∖ \partial M$ is locally a $k$ -dimensional manifold.

If $x \in \partial M$ , then a boundary chart carries $\partial M$ locally to ${y \in H^{k} : y^{k} = 0} \times {0},$ which is diffeomorphic to $ℝ^{k - 1} \times {0}$ . Therefore $\partial M$ is a $(k - 1)$ -dimensional manifold.

Tangent Space of a Manifold

Let

M \subseteq ℝ^{n}

be a

k

-dimensional manifold and let

x = f (a)

for a coordinate system

f : W \to ℝ^{n}

. The tangent space of

M

x

M_{x} = D f (a) (ℝ^{k}) .

Here

D f (a)

is the derivative linear transformation of

f

a

, and

D f (a) (ℝ^{k})

means the image of the set

ℝ^{k}

under that linear transformation:

D f (a) (ℝ^{k}) = {D f (a) (v) : v \in ℝ^{k}} .

This subspace is independent of the coordinate system.

Vector Field on a Manifold

A vector field on a manifold

M

assigns to each

x \in M

a vector

F (x) \in

M_{x}

. It is differentiable if its coordinate representation is differentiable in every coordinate system.

Differential Form on a Manifold

p

-form on a manifold

M

assigns to each

x \in M

an alternating tensor in

Λ^{p} (M_{x})

, where

M_{x}

is the tangent space of

M

x

. It is differentiable if its pullback by every coordinate system is differentiable.

Exterior Derivative on Manifolds

There is a unique

(p + 1)

-form

d ω

M

such that for every coordinate system

f : W \to ℝ^{n}

f^{*} (d ω) = d (f^{*} ω) .

Let

f : W \to ℝ^{n}

be a coordinate system with

x = f (a)

, and let

v_{1}, \dots, v_{p + 1} \in M_{x}

. Since

f^{'} (a)

identifies

ℝ^{k}

with the tangent space

M_{x}

, there are unique

w_{1}, \dots, w_{p + 1} \in ℝ^{k}

such that

f^{'} (a) w_{i} = v_{i}

Define $d ω (x) (v_{1}, \dots, v_{p + 1}) = d (f^{*} ω) (a) (w_{1}, \dots, w_{p + 1}) .$ The right side uses the exterior derivative on the coordinate domain. The compatibility of coordinate changes shows that this value does not depend on the coordinate system $f$ , so $d ω$ is well-defined.

This construction gives $f^{*} (d ω) = d (f^{*} ω)$ in every coordinate system. It is also forced by that equation, so the form $d ω$ is unique.

Orientable Manifold

A manifold

M

is orientable if its tangent spaces

M_{x}

can be assigned orientations consistently across coordinate systems. A manifold together with such a choice is an oriented manifold.

Induced Boundary Orientation

M

is an oriented manifold with boundary, the induced orientation on

\partial M

is determined by requiring that

[n (x), v_{1}, \dots, v_{k - 1}]

have the orientation of

M_{x}

, where

n (x)

is the outward unit normal and

v_{1}, \dots, v_{k - 1} \in {(\partial M)}_{x}

Integral of a Form over a Manifold

M

is an oriented

k

-dimensional manifold and

ω

is a

k

-form on

M

, the integral

\int_{M} ω

is defined by pulling

ω

back to orientation-preserving coordinate systems and assembling the local integrals with a partition of unity.

Stokes' Theorem on Manifolds

M

is a compact oriented

k

-dimensional manifold with boundary and

ω

is a

(k - 1)

-form on

M

, then

\int_{M} d ω = \int_{\partial M} ω,

where

\partial M

has the induced orientation.

First suppose

ω

is supported in the image of a single orientation-preserving singular

k

-cube

c

lying in

M ∖ \partial M

. By the definition of exterior derivative on a manifold and the cube version of Stokes' theorem,

\int_{M} d ω = \int_{c} d ω = \int_{\partial c} ω .

Since

ω

is zero on

\partial c

and also zero on

\partial M

, both sides of the theorem are zero.

Next suppose $ω$ is supported in one singular $k$ -cube $c$ whose only face in $\partial M$ is $c_{(k, 0)}$ . The induced boundary orientation was chosen so that the sign of this face agrees with the boundary integral. Therefore the cube Stokes theorem gives $\int_{M} d ω = \int_{\partial M} ω .$

In the general case, choose an open cover and a partition of unity $Φ$ so that each $φ ω$ , for $φ \in Φ$ , has one of the two supported forms above. Since $M$ is compact, only finitely many terms contribute. Also, $\sum_{φ \in Φ} d φ \land ω = 0$ because $\sum_{φ \in Φ} φ = 1$ . Hence $\int_{M} d ω = \sum_{φ \in Φ} \int_{M} d (φ ω) = \sum_{φ \in Φ} \int_{\partial M} φ ω = \int_{\partial M} ω .$

Volume Element on a Manifold

M

is an oriented

k

-dimensional manifold with the inner product inherited from

ℝ^{n}

, the volume element

d V

is the

k

-form assigning to each

x \in M

the volume element of the oriented inner product space

M_{x}

. The volume of

M

\int_{M} d V

whenever this integral exists.

Surface Area Element in

ℝ^{3}

M \subseteq ℝ^{3}

is an oriented surface with unit normal

n

, then its area element

d A

satisfies

d A (v, w) = (v \times w, n)

for positively oriented tangent vectors

v, w \in M_{x}

x \in M

, the volume element on the two-dimensional tangent space

M_{x}

is the oriented area form. If

v, w \in M_{x}

are positively oriented tangent vectors, then

v \times w

is parallel to the chosen unit normal

n

The scalar $(v \times w, n)$ is the signed area of the parallelogram spanned by $v$ and $w$ , positive exactly when $v, w$ have the orientation determined by $n$ . Therefore the oriented area element satisfies $d A (v, w) = (v \times w, n) .$

Green's Theorem

Let

M \subseteq ℝ^{2}

be a compact oriented two-dimensional manifold with boundary, with the usual orientation. If

α, β : M \to ℝ

are differentiable, then

\int_{\partial M} α d x + β d y = \int_{M} (D_{1} β - D_{2} α) d x \land d y .

Apply Stokes' theorem on manifolds to the

1

-form

ω = α d x + β d y .

Using the definition of exterior derivative and the antisymmetry of the wedge product,

d ω = D_{1} β d x \land d y + D_{2} α d y \land d x = (D_{1} β - D_{2} α) d x \land d y .

Substituting this into Stokes' theorem gives the stated formula.

Divergence Theorem

Let

M \subseteq ℝ^{3}

be a compact oriented three-dimensional manifold with boundary, and let

n

be the outward unit normal on

\partial M

. If

F

is a differentiable vector field on

M

, then

\int_{M} div F d V = \int_{\partial M} (F, n) d A .

Write

F = (P, Q, R)

and associate to

F

the

2

-form

ω = P d y \land d z + Q d z \land d x + R d x \land d y .

By the definition of exterior derivative,

d ω = (D_{1} P + D_{2} Q + D_{3} R) d x \land d y \land d z = div F d V .

On the boundary, the pullback of

ω

(F, n) d A

, by the definition of the outward normal and the surface area element. Applying Stokes' theorem gives the divergence theorem.

Classical Stokes' Theorem

Let

M \subseteq ℝ^{3}

be a compact oriented surface with boundary, let

n

be the unit normal determined by the orientation, and let

T

be the positively oriented unit tangent field on

\partial M

. If

F

is a differentiable vector field near

M

, then

\int_{M} (\nabla \times F, n) d A = \int_{\partial M} (F, T) d s .

Associate to

F = (P, Q, R)

the

1

-form

ω = P d x + Q d y + R d z .

Its exterior derivative is the

2

-form corresponding to the curl:

d ω = (\nabla \times F, n) d A

when restricted to the oriented surface

M

. On the boundary curve, the pullback of

ω

(F, T) d s,

where

T

is the positively oriented unit tangent. Applying Stokes' theorem on manifolds to

ω

gives

\int_{M} (\nabla \times F, n) d A = \int_{\partial M} (F, T) d s .