Open Access

The curl theorem of a triangular integral

Advances in Difference Equations20122012:23

https://doi.org/10.1186/1687-1847-2012-23

Received: 30 September 2011

Accepted: 2 March 2012

Published: 2 March 2012

Abstract

As the foundation of double integral, we propose a triangular integral, which is an antisymmetric double integral by single limit of double dependent sums of triangularly divided areas. Extending integrand from scalar function to tensor one, we derive the curl theorem based on this triangular double integral. It is derived by substituting the total differentials in the transformation lemma, which is based on this triangular double integral. We may thus infer that this triangular integral is the inverse operation of the total differential.

1 Introduction

The variational principle of the 2D theory is conventionally given as
δ D d x d y = 0 ,
(1.1)
where the integrand = (X, Y, X x , X y , Y x , Y y ) is a scalar functional and D is a domain. Here, X = X(x, y), Y = Y(x, y), X x X x , X y X y , Y x Y x and Y y Y y . The double integral in (1.1) is conventionally defined as
D d x d y = lim n i = 1 n lim k j = 1 k ( x i , y j ) Δ x i Δ y j ,
(1.2)
where Δx i x i - xi-1and Δy j y j - yj-1. According to the conventional method of the perpendicularly combined form of the Riemann and the Lebesgue integrals [1, 2], the area of double integral demands double limits at infinities, k → ∞ and n → ∞, of double independent sums, j = 1,2, . . . , k and i = 1,2, . . . , n, of rectangularly divided areas as shown in (1.2). Based on this definition of the conventional rectangular double integral (1.2), the curl theorem on the 2D plane is formulated as
D ( X d x + Y d y ) = D Y x - X y d x d y ,
(1.3)

where ∂D is an integral path.

Meanwhile, the total differential is widely used even in the exterior derivative [3]. However, it is not known how to derive the curl theorem (1.3) by substituting the total differentials in an integral formula based on the conventional rectangular double integral method. Extending integrand from scalar function to tensor one, we may derive the curl theorem by substituting the total differentials in an integral formula. It depends on how to define a new kind of double integral. We extend the variational principle (1.1) to
δ D L α β d x α d x β = 0 ,
(1.4)
where the integrand L αβ = L αβ (X μ , X μ,v ) is a tensor functional and indices are summed over α, β = 1, 2. Here, X μ = X μ (x λ ) and X μ , v X μ x v ( λ , μ , v = 1 , 2 ) . A new type of double integral in (1.4) is defined as
D L α β d x α d x β = lim n k = 1 n j = 1 k L α β ( x λ ) k , j Δ ( x α ) k Δ ( x β ) j ,
(1.5)
where Δ(x α ) k ≡ (x α ) k - (x α )k-1, Δ(x β ) j ≡ (x β ) j - (x β )j-1and indices are summed over α,β = 1,2. It makes possible to introduce a new kind of triangular double integral by the following two properties:
  1. 1.

    replacing rectangular area by triangular one;

     
  2. 2.

    replacing double limits of independent double sums by single limit of dependent double sums.

     

We propose an antisymmetric triangular double integral. It demands only single limit at infinity n → ∞ of double dependent sums, j = 1, 2, . . . , k and k = 1, 2, . . . , n, of triangularly divided areas as shown in Definition 1. We succeed to define a new kind of triangular integral method, which may derive the curl theorem by substituting the total differentials in an integral formula. In this article, we formulate the curl theorem based on a new kind of integral formula (1.5). We name it as the curl theorem of a triangular integral on the 2D plane as shown in the main theorem (4.6). In detail, we derive (4.1) by substituting the total differentials (4.3) and (4.4) in (4.2). The curl theorem of a triangular integral on the 2D plane (4.6) is finally derived from (4.1) and (4.5) under the condition of a closed curve (4.8). We may thus infer that this triangular integral is the inverse operation of the total differential.

There are three advantages of this theory. One is the conceptual coherence despite of its complicated procedure of calculation in the derivation of the curl theorem (see Section 4.1). Another one is that this theory is applicable for finite element method in the case of 1 < n < ∞ (see (2.12) and (2.18)). The other one is applicability to the integral in the variational principle of multiple variables in the case that the integrand is extended to tensor (see (1.4) and (1.5)).

This article is structured as follows. In Section 2, a triangularly divided area is introduced. This triangular double integral is defined by single limit of double sums of triangularly divided areas. In Section 3, the combination and the transformation lemmata are derived. In Section 4, the curl theorem on the 2D plane is derived by substituting the total differentials in the transformation lemma. In Section 5, the curl theorems of a triangular integral in the 3D space and the 4D hyper-space are presented.

2 Single limit of double sums

A triangular integral as the foundation of double integral is proposed in Section 2.1 and its example is shown in Section 2.2.

2.1 Sum of triangular areas

The sum of triangular areas is introduced as follows.

First of all, the respective triangular area is introduced. Let f(x, y) = 0 be a piecewise smooth curve of equation on the xy-plane, expressed in terms of the Cartesian coordinates (x, y) 2. Assume there are two fixed points of A(xA, yA) and B(xB, yB). Suppose there is a sequence of points {P k (x k , y k ) | k = 0, 1, 2, . . . , n} on f(x, y) = 0, where the initial and the terminal points are respectively P0 (x0, y0) = A(xA, yA) and P n (x n , y n ) = B(xB, yB). The respective triangular area ΔS k (k = 1, 2, . . . , n), which is surrounded by three vertices of O(0, 0), Pk-1(xk-1, yk-1) and P k (x k , y k ), is introduced as in Figure 1 by
Δ S k = 1 2 ( x k y k - 1 - y k x k - 1 ) ( k = 1 , 2 , , n ) .
(2.1)
Figure 1

Sum of triangularly divided areas.

The respective triangular area ΔS k , which is surrounded by three vertices of O(0, 0), Pk-1(xk-1, yk-1) and P k (x k , y k ), is introduced as Δ S k = 1 2 x k y k - 1 - y k x k - 1 ( k = 1 , 2 , . . . , n ) .

The increments of x k and y k , i.e., Δx k and Δy k , are respectively denoted as
Δ x k x k - x k - 1 ( k = 1 , 2 , , n ) ,
(2.2)
Δ y k y k - y k - 1 ( k = 1 , 2 , , n ) .
(2.3)
Substituting Δx k and Δy k in ΔS k of (2.1), it is modified to be
Δ S k = 1 2 ( y k Δ x k - x k Δ y k ) ( k = 1 , 2 , , n ) .
(2.4)
Next, we introduce triangular sum S n (n = 1, 2, 3, . . .) as a sum of n triangular areas ΔS k , i.e.,
S n = k = 1 n Δ S k .
(2.5)
Substituting (2.4) in (2.5), it is modified to be
S n = 1 2 k = 1 n ( y k Δ x k - x k Δ y k ) .
(2.6)
Furthermore, the sum of triangular areas is modified to be the sum of a triangular area and double sums as follows. Using the notations of Δx k in (2.2) and Δy k in (2.3), x k and y k are respectively expressed as
x k = x 0 + j = 1 k Δ x j ( k = 1 , 2 , , n ) ,
(2.7)
y k = y 0 + j = 1 k Δ y j ( k = 1 , 2 , , n ) .
(2.8)
Substituting (2.7) and (2.8) in (2.6), it is finally modified to be
S n = 1 2 k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j + 1 2 ( x B y A - y B x A ) .
(2.9)
The cases of n = 1, 2, 3 and n → ∞ of
1 2 k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j
(2.10)
in (2.9) are shown in the following. Let the coordinates of point Q be (x n , y0).
  1. 1.
    In the case of n = 1, i.e., for two points of P 0(x 0, y 0) and P 1(x 1, y 1) = P n (x n , y n ), it holds
    1 2 k = 1 1 j = 1 k Δ x k Δ y j - Δ y k Δ x j = 0 .
    (2.11)
     
  2. 2.
    In the case of n = 2, i.e., for three points of P 0(x 0, y 0), P 1(x 1, y 1) and P 2(x 2, y 2) = P n (x n , y n ), as shown in Figure 2, the area of P 0 P 1 P 2 is
    1 2 k = 1 2 j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 2 ( Δ x 2 Δ y 1 - Δ y 2 Δ x 1 ) .
    (2.12)
     
Figure 2

Triangularly divided areas in the case of n = 2.

The area of triangle P0P1P2 is σ 2 = R 2 . 0 - R 2 . 1 - R 2 . 2 = 1 2 Δ x 2 Δ y 1 - Δ y 2 Δ x 1 .

Introducing R2.0 as the area of triangle P0QP n , we obtain
R 2 . 0 = 1 2 ( x 2 - x 0 ) ( y 0 - y 2 ) = - 1 2 ( Δ x 1 + Δ x 2 ) ( Δ y 1 + Δ y 2 ) .
(2.13)
Introducing R2.1 as the area of triangle P0P1Q, we obtain
R 2 . 1 = 1 2 ( x 2 - x 0 ) ( y 0 - y 1 ) = - 1 2 ( Δ x 1 + Δ x 2 ) Δ y 1 .
(2.14)
Introducing R2.2 as the area of triangle P1P2Q, we obtain
R 2 . 2 = 1 2 ( x 2 - x 1 ) ( y 0 - y 2 ) = - 1 2 Δ x 2 ( Δ y 1 + Δ y 2 ) .
(2.15)
We introduce σ2 as the area of triangle P0P1P2 as
σ 2 = R 2 . 0 - R 2 . 1 - R 2 . 2 .
(2.16)
Substituting (2.13), (2.14) and (2.15) in (2.16), it is modified to be
σ 2 = 1 2 ( Δ x 2 Δ y 1 - Δ y 2 Δ x 1 ) .
(2.17)
We thus see the coincidence of (2.12) and (2.17).
  1. 3.
    In the case of n = 3, i.e., for four points of P 0(x 0, y 0), P 1(x 1, y 1), P 2(x 2, y 2) and P 3(x 3, y 3) = P n (x n , y n ), as shown in Figure 3, the area of P 0 P 1 P 2 P 3 is
    1 2 k = 1 3 j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 2 ( Δ x 2 Δ y 1 - Δ y 2 Δ x 1 ) + 1 2 ( Δ x 3 Δ y 1 - Δ y 3 Δ x 1 ) + 1 2 ( Δ x 3 Δ y 2 - Δ y 3 Δ x 2 ) .
    (2.18)
     
Figure 3

Triangularly divided areas in the case of n = 3.

The area of P0P1P2P3 is σ 3 = R 3 . 0 - R 3 . 1 - R 3 . 2 - R 3 . 3 = 1 2 Δ x 2 Δ y 1 - Δ y 2 Δ x 1 + 1 2 Δ x 3 Δ y 1 - Δ y 3 Δ x 1 + 1 2 Δ x 3 Δ y 2 - Δ y 3 Δ x 2 .

Introducing R3.0 as the area of triangle P0QP n , we obtain
R 3 . 0 = 1 2 ( x 3 - x 0 ) ( y 0 - y 3 ) = - 1 2 ( Δ x 1 + Δ x 2 + Δ x 3 ) ( Δ y 1 + Δ y 2 + Δ y 3 ) .
(2.19)
Introducing R3.1 as the area of triangle P0P1Q, we obtain
R 3 . 1 = 1 2 ( x 3 - x 0 ) ( y 0 - y 1 ) = - 1 2 ( Δ x 1 + Δ x 2 + Δ x 3 ) Δ y 1 .
(2.20)
Introducing R3.2 as the area of triangle P1P2Q, we obtain
R 3 . 2 = ( x 3 - x 1 ) ( y 0 - y 2 ) - 1 2 ( x 3 - x 1 ) ( y 0 - y 1 ) - 1 2 ( x 2 - x 1 ) ( y 1 - y 2 ) - 1 2 ( x 3 - x 2 ) ( y 0 - y 2 ) = - 1 2 Δ x 2 Δ y 1 - 1 2 Δ x 2 Δ y 2 - 1 2 Δ x 3 Δ y 2 .
(2.21)
Introducing R3.3 as the area of triangle P2P3Q, we obtain
R 3 . 3 = 1 2 ( x 3 - x 2 ) ( y 0 - y 3 ) = - 1 2 Δ x 3 ( Δ y 1 + Δ y 2 + Δ y 3 ) .
(2.22)
We introduce σ3 as the area of P0P1P2P3 as
σ 3 = R 3 . 0 - R 3 . 1 - R 3 . 2 - R 3 . 3 .
(2.23)
Substituting (2.19), (2.20), (2.21) and (2.22) in (2.23), it is modified to be
σ 3 = 1 2 ( Δ x 2 Δ y 1 - Δ y 2 Δ x 1 ) + 1 2 ( Δ x 3 Δ y 1 - Δ y 3 Δ x 1 ) + 1 2 ( Δ x 3 Δ y 2 - Δ y 3 Δ x 2 ) .
(2.24)
We thus see the coincidence of (2.18) and (2.24).
  1. 4.
    In the case of n → ∞, i.e., for a segment of A B ¯ and a piecewise smooth curve of equation f(x, y) = 0, as shown in Figure 4, the area σ surrounded by the segment and the curve is
    σ = 1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j .
    (2.25)
     
Figure 4

Single limit at infinity n → ∞ of double sums of triangularly divided areas.

The area surrounded by f(x, y) = 0 and A B ¯ is σ = 1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 2 f ( x , y ) 0 [ A , B ] ( d x d y - d y d x ) .

The area of the triangle OAB is 1 2 x B y A - y B x A .

Definition 1. (Definition of a triangular integral) The triangular double integral of 1 2 f ( x , y ) 0 [ A , B ] ( d x d y - d y d x ) is defined in the case of a piecewise smooth curve of equation f(x, y) = 0 by the formula,
1 2 f ( x , y ) 0 [ A , B ] ( d x d y - d y d x ) = 1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j ,
(2.26)

where dx' and dy' respectively correspond to increments of Δx j and Δy j , while dx and dy respectively correspond to those of Δx k and Δy k .

Suppose S is the limit of S n at infinity n → ∞, i.e.,
S = lim n S n .
(2.27)
Theorem 1. The area S of OAB surrounded by O A ¯ , O B ¯ and the graph of a piecewise smooth curve of equation f (x, y) = 0 is expressed as
S = 1 2 f ( x , y ) 0 [ A , B ] ( d x d y - d y d x ) + 1 2 ( x B y A - y B x A ) ,
(2.28)

where O(0,0), A(xA, yA), B(xB, yB), f(xA, yA) = 0 and f(xB, yB) = 0.

Proof. As n → ∞ in (2.9), we obtain S as
S = 1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j + 1 2 ( x B y A - y B x A )
(2.29)

as shown in Figure 4. Q.E.D.

Remark. The antisymmetric double integral here introduced demands only single limit of double sums of triangularly divided areas.

2.2 Two kinds of solutions of an example of a parabola

An example that the integrand is constant is shown in Example 1. An example that the integrand is not constant is shown in Section 4.2. Two kinds of solutions of the problem of a parabola are shown in the following. The first and the second solutions are respectively given by the arithmetic and the geometric sequences.

Example 1. An area surrounded by a parabola of
y = - x 2 + 9
(2.30)
and a segment of A B ¯ , where A(xA, yA) = (1,8) and B(xB, yB) = (2,5).
  1. 1.

    Integration by the arithmetic sequence (The first kind of solution)

     
The arithmetic sequences x j and x k are respectively
x j = 1 + j n ( j = 0 , 1 , 2 , . . . , k ) ,
(2.31)
x k = 1 + k n ( k = 0 , 1 , 2 , . . . , n ) .
(2.32)
Using (2.2), the increments of the arithmetic sequences x j and x k , i.e., Δx = Δx j = Δx k , are
Δ x = 1 n .
(2.33)
Substituting (2.30) in (2.3), the increments of the arithmetic sequences y j and y k , i.e., Δy j and Δy k , are respectively
Δ y j = - 2 n - 2 j - 1 n 2 ( j = 0 , 1 , 2 , , k ) ,
(2.34)
Δ y k = - 2 n - 2 k - 1 n 2 ( k = 0 , 1 , 2 , , n ) .
(2.35)
Thus, the antisymmetric double increment of the arithmetic sequence (2.31) and (2.32) is
1 2 ( Δ x k Δ y j - Δ y k Δ x j ) = k - j n 3 ( j = 1 , 2 , , k ) , ( k = 1 , 2 , , n ) .
(2.36)
The double dependent sums of (2.36) for j = 1,2, . . . , k and k = 1,2,. . . , n is
1 2 k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 n 3 k = 1 n j = 1 k ( k - j ) = 1 6 1 - 1 n 2 .
(2.37)
As n → ∞ in (2.37), we obtain
1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 6 lim n 1 - 1 n 2 = 1 6 .
(2.38)
  1. 2.

    Integration by the geometric sequence (The second kind of solution)

     
The geometric sequences x j and x k are respectively
x j = 2 j n ( j = 0 , 1 , 2 , , k ) ,
(2.39)
x k = 2 k n ( k = 0 , 1 , 2 , , n ) .
(2.40)
The increments of the geometric sequences x j , x k , y j and y k , i.e., Δx j , Δx k , Δy j and Δy k , are respectively
Δ x j = 2 j n ( 1 - 2 - 1 n ) ( j = 0 , 1 , 2 , , k ) ,
(2.41)
Δ y j = 2 2 j n 2 - 2 n - 1 ( j = 0 , 1 , 2 , , k ) ,
(2.42)
Δ x k = 2 k n ( 1 - 2 - 1 n ) ( k = 0 , 1 , 2 , , n ) ,
(2.43)
Δ y k = 2 2 k n ( 2 - 2 n - 1 ) ( k = 0 , 1 , 2 , , n ) .
(2.44)
Thus, the antisymmetric double increment of the geometric sequence (2.39) and (2.40) is
1 2 Δ x k Δ y j - Δ y k Δ x j = 1 2 1 - 2 - 1 n 2 - 2 n - 1 2 k n 2 2 j n - 2 2 k n 2 j n ( j = 1 , 2 , , k ) , ( k = 1 , 2 , , n ) .
(2.45)
The double dependent sums of (2.45) for j = 1,2, . . . , k and k = 1, 2, . . . , n is
1 2 k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 2 k = 1 n j = 1 k 1 - 2 - 1 n 2 - 2 n - 1 2 k n 2 2 j n - 2 2 k n 2 j n .
(2.46)
As n → ∞ in (2.46), we obtain
1 2 lim n k = 1 n j = 1 k Δ x k Δ y j - Δ y k Δ x j = 1 2 lim n k = 1 n j = 1 k 1 - 2 - 1 n 2 - 2 n - 1 2 k n 2 2 j n - 2 2 k n 2 j n = 1 6 .
(2.47)

3 The combination lemma and the transformation lemma

The combination lemma is shown in Section 3.1 and the transformation lemma is shown in Section 3.2.

Let y = f(x) be a differentiable function on the xy-plane. Assume there are two fixed points of A = (xA, yA) and B = (xB, yB) on y = f(x), where yA = f(xA) and yB = f(xB). An ordinary integral is denoted in accordance with the notation of double integral denoted in Definition 1. New notation for an ordinary integral of y = f(x) in x [xA, xB] is
y = f ( x ) [ A , B ] y d x x A x B f ( x ) d x .
(3.1)

3.1 The combination lemma on the 2D plane

The combination lemma shows that the sum of an integral along x-axis and that along y-axis between [A, B] is equal to the subtraction of two rectangles, x B y B and x A y A .

Lemma 1. (The combination lemma) Assume y = f(x) is a differentiable function on the xy-plane, and x = f-1(y) is also a differentiable one, where f-1is the inverse function of f. Between A(xA, yA) = (x0, y0) and B(xB, yB) = (x n , y n ), it holds
y = f ( x ) [ A , B ] y d x + x = f - 1 ( y ) [ A , B ] x d y = x B y B - x A y A ,
(3.2)

where yA = f(xA) and yB = f(xB).

We name (3.2) as the combination lemma.

Proof. It is divided into the following cases.
  1. 1.
    For a monotonically increasing function y = f(x) in x [x A, x B] or x = f -1(y) in y [y A, y B], it is indicated that the first term of the left-hand side of (3.2) is an integral of y = f(x) along x-axis i.e.,
    y = f ( x ) [ A , B ] y d x = lim n k = 1 n y k Δ x k
    (3.3)
     
and the second term of the left-hand side of (3.2) is an integral of x = f-1(y) along y-axis, i.e.,
x = f - 1 ( y ) [ A , B ] x d y = lim n k = 1 n x k Δ y k
(3.4)
as shown in Figure 5.
Figure 5

Two kinds of integrals for a monotonically increasing function.

An integral along x-axis is y = f ( x ) [ A , B ] y d x = lim n k = 1 n y k Δ x k

and that along y-axis is x = f - 1 ( y ) [ A , B ] x d y = lim n k = 1 n x k Δ y k .

It holds y = f ( x ) [ A , B ] y d x + x = f - 1 ( y ) [ A , B ] x d y = x B y B - x A y A ,

where yA = f(xA) and yB = f(xB).

  1. 2.
    For a monotonically decreasing function y = f(x) in x [x A, x B] or x = f -1(y) in y [y B, y A], it holds
    x = f - 1 y B , A x d y - x A y A - y B = y = f x A , B y d x - y B x B - x A
    (3.5)
     
as shown in Figure 6. We thus obtain (3.2). Q.E.D.
Figure 6

Two kinds of integrals for a monotonically decreasing function.

An integral along x-axis is y = f ( x ) [ A , B ] y d x = lim n k = 1 n y k Δ x k

and that along y-axis is x = f - 1 ( y ) [ B , A ] x d y = - lim n k = 1 n x k Δ y k .

It holds x = f - 1 ( y ) [ B , A ] x d y - x A ( y A - y B ) = y = f ( x ) [ A , B ] y d x - y B ( x B - x A ) ,

where yA = f(xA) and yB = f(xB).

3.2 The transformation lemma on the 2D plane

The transformation lemma transforms an integrand to the second integral variable. It transforms a single integral to an antisymmetric double integral of which integrand is constant.

Proposition 1. (Integral along x-axis) For a piecewise smooth curve of equation f(x, y) = 0, it holds
f x , y = 0 A , B y d x = f x , y 0 A , B d x d y + x B y A - x A y A .
(3.6)
Proof. Substituting (2.8) into (3.3), it is modified to be
f x , y = 0 A , B y d x = lim n k = 1 n j = 1 k Δ y j Δ x k + y A x B - x A .
(3.7)

Q.E.D.

Proposition 2. (Integral along y-axis) For a piecewise smooth curve of equation f(x, y) = 0, it holds
f x , y = 0 B , A x d y = - f x , y 0 A , B d y d x + x A y A - x A y B .
(3.8)
Proof. Substituting (2.7) into (3.4), it is modified to be
f x , y = 0 B , A x d y = - lim n k = 1 n j = 1 k Δ x j Δ y k + x A y A - y B .
(3.9)

Q.E.D.

Lemma 2. (The transformation lemma) For a piecewise smooth curve of equation f(x, y) = 0, it holds
f x , y = 0 A , B y d x = 1 2 f x , y 0 A , B d x d y - d y d x + 1 2 x B - x A y A + y B .
(3.10)
Proof. (Combining integral along x-axis with that along y-one) Combining (3.6) with (3.8), we obtain
f x , y = 0 A , B y d x + f x , y = 0 B , A x d y = f x , y 0 A , B d x d y - d y d x + x B y A - x A y B .
(3.11)

Substituting (3.2) into (3.11), we obtain the lemma. Q.E.D.

4 The curl theorem of a triangular integral on the 2D plane

The curl theorem of the conventional rectangular integral (1.3) is modified to be that of a new kind of triangular integral (4.6) in Section 4.1 and its example is shown in Section 4.2.

4.1 Proof of the curl theorem on the 2D plane

We present two lemmata in the following before the proof of Theorem 2.

Lemma 3. Let X = X(x, y) be a partially differentiable function with respect to x and y. It holds
f x , y = 0 A , B X d x = 1 2 f x , y 0 A , B X y d x d y - X y d y d x + 1 2 f x , y 0 A , B X x - X x d x d x + 1 2 ( x B - x A ) ( X A + X B )
(4.1)

for a piecewise smooth curve of equation f(x, y) = 0 between A(xA, yA) and B(xB, yB), where XA = X(xA, yB) and XB = X(xB, yB).

Proof. Rewriting the transformation lemma (3.10) for X = X(x, y) and f(x, y) = 0, it is expressed as
f x , y = 0 A , B X d x = 1 2 f x , y 0 A , B ( d x d X - d X d x ) + 1 2 ( x B - x A ) ( X A + X B ) .
(4.2)
Substituting two kinds of the total differentials of X, i.e., dX and dX',
d X = X x d x + X y d y ,
(4.3)
d X = X x d x + X y d y
(4.4)

in (4.2), it is modified to be (4.1). Q.E.D.

Remark. Precise modifications in the proof of Lemma 3 are shown in Appendix 1.

Lemma 4. Let Y = Y(x, y) be a partially differentiable function with respect to x and y. It holds
f x , y = 0 A , B Y d y = 1 2 f x , y 0 A , B Y x d y d x - Y x d x d y + 1 2 f x , y 0 A , B Y y - Y y d y d y + 1 2 y B - y A Y A + Y B
(4.5)

for a piecewise smooth curve of equation f(x, y) = 0 between A(xA, yA) and B(xB, yB), where YA = Y(xA, yA) and YB = Y(xB, yB).

Proof. In the similar procedure as Lemma 3, we obtain (4.5). Q.E.D.

Theorem 2. (The curl theorem of a triangular integral on the 2D plane) LetD be a piecewise smooth curve of equation f(x, y) = 0 on the xy-plane, which is expressed in terms of the Cartesian coordinates (x, y) 2. Let D be the region inside and onD. Let X1 = X = X(x, y) and X2 = Y = Y(x, y) be partially differentiable functions with respect to x1 = x and x2 = y in D. It holds
D X α d x α = 1 2 D X α x β - X β x α d x α d x β ,
(4.6)

where indices are summed over α, β = 1,2.

Proof. Combining (4.1) with (4.5), we obtain
f x , y = 0 A , B X d x + Y d y = 1 2 f x , y 0 A , B X y - Y x d x d y + Y x - X y d y d x + 1 2 f x , y 0 A , B X x - X x d x d x + Y y - Y y d y d y + 1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) .
(4.7)
For an integral on a closed curve, the initial point A(xA, yA) coincides with the terminal one B(xB, yB), i.e., A(xA, yA) = B(xB, yB). It then holds
1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) = 0
(4.8)

regardless of values of XA, XB, YA and YB. We thus obtain (4.6). Q.E.D.

Remark. X x and X x must be rigorously distinguished as integrands. See (5.20) and (5.19) in detail. An inequality holds for X x and X x in an integral form, i.e.,
D X x - X x d x d x = lim n k = 1 n j = 1 k Δ X [ Δ x k ] Δ x k - Δ X [ Δ x j ] Δ x j Δ x k Δ x j 0 .
(4.9)
Here, ΔXx k ] is denoted as ΔXx k ] ≡ X(x k , y k )-X(xk-1, y k ). See an example in (5.23). However, they coincide as explicit forms of derivatives. An equality holds for X x and X x in arbitrary derivative form, i.e.,
X x , y x - X x , y x = lim Δ x k 0 Δ X Δ x k Δ x k - lim Δ x j 0 Δ X Δ x j Δ x j = 0 .
(4.10)

Similar formulae hold for X y and X y , for Y x and Y x and for Y y and Y y .

Corollary 1. In the case of which it holds
D X α d x α = - 1 2 D E α β d x α d x β ,
(4.11)
where indices are summed over α, β = 1, 2, a sufficient condition to hold (4.11) is
- E α β = X α x β - X β x α ( α , β = 1 , 2 ) .
(4.12)
Proof. Using (4.6), (4.11) is modified to be
1 2 D X α x β - X β x α d x α d x β = - 1 2 D E α β d x α d x β ,
(4.13)

where indices are summed over α, β = 1, 2. A sufficient condition to hold (4.13) is (4.12). Q.E.D.

Corollary 2. In the case of (4.11), we obtain an antisymmetric property of E αβ as
- E α β = E β α ( α , β = 1 , 2 ) .
(4.14)
Proof. Interchanging α and β in (4.12), it also holds
E β α = X α x β - X β x α α , β = 1 , 2 .
(4.15)

Comparing (4. 12) with (4. 15), we obtain (4. 14). Q.E.D.

4.2 An example of the curl theorem on the 2D plane

An example of Theorem 2 is shown in Example 2. The integral path of this problem is an ellipse of
x = a cos θ ,
(4.16)
y = b sin θ .
(4.17)
Definition 2. (Definition of elliptical sequence) Using (4.16) and (4.17), we may respectively introduce elliptical sequences x j , y j , x k and y k as
x j = a cos θ j , y j = b sin θ j ( j = 0 , 1 , 2 , . . . , k ) ,
(4.18)
x k = a cos θ k , y k = b sin θ k ( k = 0 , 1 , 2 , . . . , n ) .
(4.19)
The increments of x j and y j in (4.18), i.e., Δx j and Δy j , of the angular arithmetic sequence θ j are respectively
Δ x j = a cos θ j - cos θ j - 1 ( j = 0 , 1 , 2 , . . . , k ) ,
(4.20)
Δ y j = b sin θ j - sin θ j - 1 ( j = 0 , 1 , 2 , . . . , k ) .
(4.21)
The increments of x k and y k in (4.19), i.e., Δx k and Δy k , of the angular arithmetic sequence θ k are respectively
Δ x k = a ( cos θ k - cos θ k - 1 ) ( k = 0 , 1 , 2 , . . . , n ) ,
(4.22)
Δ y k = b ( sin θ k - sin θ k - 1 ) ( k = 0 , 1 , 2 , . . . , n ) .
(4.23)

Example 2. We calculate both of the line integral and the area integral in the counterclockwise direction. The example is the curl theorem of a triangular integral on the 2D plane in the case of X = X(x, y) = -x2y and Y = Y(x, y) = xy2, where the closed curve is an ellipse of (4.16) and (4.17).

The arithmetic sequences of θ j and θ k are respectively
θ j = j 2 π n ( j = 0 , 1 , 2 , . . . , k ) ,
(4.24)
θ k = k 2 π n k = 0 , 1 , 2 , . . . , n .
(4.25)
The increments of θ j and θ k , i.e., Δθ = Δθ j = Δθ k , are
Δ θ = 2 π n .
(4.26)
  1. 1.

    Calculation by line integral

     
The line integral is calculated to be
D X d x + Y d y = x 2 a 2 + y 2 b 2 = 1 - x 2 y d x + x y 2 d y = a b a 2 + b 2 0 2 π sin 2 θ cos 2 θ d θ .
(4.27)
The integral in (4.27) is executed in the formula of
0 2 π sin 2 θ cos 2 θ d θ = lim n k = 1 n sin 2 θ k cos 2 θ k Δ θ = lim n 2 π n k = 1 n sin 2 k 2 π n cos 2 k 2 π n = π 4 .
(4.28)
Substituting (4.28) into (4.27), we finally obtain
x 2 a 2 + y 2 b 2 = 1 ( X d x + Y d y ) = π 4 a b a 2 + b 2 .
(4.29)
  1. 2.

    Calculation by area integral

     
The area double integral in the region of x 2 a 2 + y 2 b 2 1 is calculated to be
1 2 x 2 a 2 + y 2 b 2 1 X y - Y x d x d y + 1 2 x 2 a 2 + y 2 b 2 1 Y x - X y d y d x + 1 2 x 2 a 2 + y 2 b 2 1 X x - X x d x d x + 1 2 x 2 a 2 + y 2 b 2 1 Y y - Y y d y d y = π 4 a b a 2 + b 2 .
(4.30)

Calculations of the respective terms of (4.30) are shown in Appendix 2. We hence conclude that Theorem 2 holds for this problem.

Remark. It is very complicated to execute this triangular integral only by human's brain. Computer and formula manipulation software are recommended to verify these calculations in the case of n → ∞. This theory is also applicable for numerical calculation in the case of 1 < n < ∞.

5 The curl theorem of a triangular integral in the higher dimensions

The curl theorem of a triangular integral in the 3D space is derived in Section 5.1 and that in the 4D hyper-space is derived in Section 5.2.

5.1 The curl theorem in the 3D space

We extend the curl theorem of a triangular integral in the 3D space. We present three lemmata in the following before the proof of Theorem 3.

Lemma 5. Let X = X(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
g ( x , y , z ) = 0 [ A , B ] X d x = 1 2 g ( x , y , z ) 0 [ A , B ] X y d x d y - X y d y d x + X z d x d z - X z d z d x + 1 2 g ( x , y , z ) 0 [ A , B ] X x - X x d x d x + 1 2 ( x B - x A ) ( X A + X B )
(5.1)

for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where XA = X(xA, yA, zA) and XB = X(xB, yB, zB).

Proof. Rewriting the transformation lemma (3.10) for X = X(x, y, z) and g(x, y, z) = 0, it is expressed as
g ( x , y , z ) = 0 [ A , B ] X d x = 1 2 g ( x , y , z ) 0 [ A , B ] ( d x d X - d X d x ) + 1 2 ( x B - x A ) ( X A + X B ) .
(5.2)
Substituting two kinds of the total differentials of X, i.e., dX and dX',
d X = X x d x + X y d y + X z d z ,
(5.3)
d X = X x d x + X y d y + X z d z
(5.4)

in (5.2), it is modified to be (5.1). Q.E.D.

Lemma 6. Let Y = Y(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
g ( x , y , z ) = 0 [ A , B ] Y d y = 1 2 g ( x , y , z ) 0 [ A , B ] Y x d y d x - Y x d x d y + Y z d y d z - Y z d z d y + 1 2 g ( x , y , z ) 0 [ A , B ] Y y - Y y d y d y + 1 2 ( y B - y A ) ( Y A + Y B )
(5.5)

for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where YA = Y(xA, yA, zA) and YB = Y(xB, yB, zB).

Proof. In the similar procedure as Lemma 5, we obtain (5.5). Q.E.D.

Lemma 7. Let Z = Z(x, y, z) be a partially differentiable function with respect to x, y and z. It holds
g ( x , y , z ) = 0 [ A , B ] Z d z = 1 2 g ( x , y , z ) 0 [ A , B ] Z y d z d y - Z y d y d z + Z x d z d x - Z x d x d z + 1 2 g ( x , y , z ) 0 [ A , B ] Z z - Z z d z d z + 1 2 ( z B - z A ) ( Z A + Z B )
(5.6)

for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where ZA = Z(xA, yA, zA) and ZB = Z(xB, yB, zB).

Proof. In the similar procedure as Lemma 5, we obtain (5.6). Q.E.D.

Theorem 3. (The curl theorem of a triangular integral in the 3D space) Let D be a piecewise smooth surface in the xyz-space, which is expressed in terms of the Cartesian coordinates (x, y, z) 3. LetD be the boundary of D. Let X1 = X = X(x, y, z), X2 = Y = Y(x, y, z) and X3 = Z = Z(x, y, z) be partially differentiable functions with respect to x1 = x, x2 = y and x3 = z in D. It holds
D X α d x α = 1 2 D X α x β - X β x α d x α d x β ,
(5.7)

where indices are summed over α, β = 1,2,3.

Proof. Combining (5.1) with (5.5) and (5.6), we obtain
g ( x , y , z ) = 0 [ A , B ] X d x + Y d y + Z d z = 1 2 g ( x , y , z ) 0 [ A , B ] X y - Y x d x d y + Y x - X y d y d x + Y z - Z y d y d z + Z y - Y z d z d y + Z x - X z d z d x + X z - Z x d x d z + X x - X x d x d x + Y y - Y y d y d y + Z z - Z z d z d z + 1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) + 1 2 ( z B - z A ) ( Z A + Z B )
(5.8)
for a piecewise smooth curve of equation g(x, y, z) = 0 between A(xA, yA, zA) and B(xB, yB, zB), where XA = X(xA, yA, zA), XB = X(xB, yB, zB), YA = Y(xA, yA, zA), YB = Y(xB, yB, zB), ZA = Z(xA, yA, zA) and ZB = Z(xB, yB, zB). For an integral on a closed curve, the initial point A(xA, yA, zA) coincides with the terminal one B(xB, yB, zB), i.e., A(xA, yA, zA) = B(xB, yB, zB). It then holds
1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) + 1 2 ( z B - z A ) ( Z A + Z B ) = 0
(5.9)

regardless of the values of XA, XB, YA, YB, ZA and ZB. We thus obtain (5.7). Q.E.D.

Corollaries 1 and 2 also hold for α, β = 1,2,3.

5.2 The curl theorem in the 4D hyper-space

We extend the curl theorem of a triangular integral in the 4D hyper-space.

Theorem 4. (The curl theorem of a triangular integral in the 4D hyper-space) Let D be a piecewise smooth surface in the txyz-hyper-space, which is expressed in terms of the Cartesian coordinates (t, x, y, z) 4. LetD be the boundary of D. Let X0 = T = T(t, x, y, z), X1 = X = X(t, x, y, z), X2 = Y = Y(t, x, y, z) and X3 = Z = Z(t, x, y, z) be partially differentiable functions with respect to x0 = t, x1 = x, x2 = y and x3 = z in D. It holds
D X α d x α = 1 2 D X α x β - X β x α d x α d x β ,
(5.10)

where indices are summed over α, β = 0,1,2,3.

Proof. In the 4D, it holds
h ( t , x , y , z ) = 0 [ A , B ] T d t + X d x + Y d y + Z d z = 1 2 h ( t , x , y , z ) 0 [ A , B ] X y - Y x d x d y + Y x - X y d y d x + Y z - Z y d y d z + Z y - Y z d z d y + Z x - X z d z d x + X z - Z x d x d z + X x - X x d x d x + Y y - Y y d y d y + Z z - Z z d z d z + T t - T t d t d t + X t - T x d x d t + Y t - T y d y d t + Z t - T z d z d t + T x - X t d t d x + T y - Y t d t d y + T z - Z t d t d z + 1 2 ( t B - t A ) ( T A + T B ) + 1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) + 1 2 ( z B - z A ) ( Z A + Z B )
(5.11)
for a piecewise smooth curve of equation h(t, x, y, z) = 0 between A(tA, xA, yA, zA) and B(tB, xB, yB, zB), where TA = T(tA, xA, yA, zA), TB = T(tB, xB, yB, zB), XA = X(tA, xA, yA, zA), XB = X(tB, xB, yB, zB), YA = Y(tA, xA, yA, zA), YB = Y(tB, xB, yB, zB), ZA = Z(tA, xA, yA, zA) and ZB = Z(tB, xB, yB, zB). For an integral on a closed curve, the initial point A(tA, xA, yA, zA) coincides with the terminal one B(tB, xB, yB, zB), i.e., A(tA, xA, yA, zA) = B(tB, xB, yB, zB). It then holds
1 2 ( t B - t A ) ( T A + T B ) + 1 2 ( x B - x A ) ( X A + X B ) + 1 2 ( y B - y A ) ( Y A + Y B ) + 1 2 ( z B - z A ) ( Z A + Z B ) = 0
(5.12)

regardless of the values of TA, TB, XA, XB, YA, YB, ZA and ZB. We thus obtain (5.10). Q.E.D.

Corollaries 1 and 2 also hold for α, β = 0 1,2,3.

Appendix 1

Using Definition 1, the first term of the right-hand side of (4.2) in the proof of Lemma 3 is expressed as
1 2 f ( x , y ) 0 [ A , B ] ( d x d X - d X d x ) = 1 2 lim n k = 1 n j = 1 k ( Δ x k Δ X j - Δ X k Δ x j ) .
(5.13)
The total increments ΔX j and ΔX k in the right-hand side of (5.13) are respectively expressed as
Δ X j X ( x j , y j ) - X ( x j - 1 , y j - 1 ) = X ( x j , y j ) - X ( x j - 1 , y j ) x j - x j - 1 Δ x j + X ( x j - 1 , y j ) - X ( x j - 1 , y j - 1 ) y j - y j - 1 Δ y j ,
(5.14)
Δ X k X ( x k , y k ) - X ( x k - 1 , y k - 1 ) = X ( x k , y k ) - X ( x k - 1 , y k ) x k - x k - 1 Δ x k + X ( x k - 1 , y k ) - X ( x k - 1 , y k - 1 ) y k - y k - 1 Δ y k .
(5.15)
Substituting (5.14) and (5.15) in the right-hand side of (5.13), it is modified to be
1 2 lim n k = 1 n j = 1 k ( Δ x k Δ X j - Δ X k Δ x j ) = 1 2 lim n k = 1 n j = 1 k X x j - 1 , y j - X x j - 1 , y j - 1 y j - y j - 1 Δ x k Δ y j - 1 2 lim n k = 1 n j = 1 k X ( x k - 1 , y k ) - X ( x k - 1 , y k - 1 ) y k - y k - 1 Δ y k Δ x j + 1 2 lim n k = 1 n j = 1 k X x j , y j - X x j - 1 , y j x j - x j - 1 Δ x k Δ x j - 1 2 lim n k = 1 n j = 1 k X ( x k , y k ) - X ( x k - 1 , y k ) x k - x k - 1 Δ x k Δ x j .
(5.16)
Each term of (5.16) is respectively expressed as
1 2 f ( x , y ) 0 [ A , B ] X y d x d y = 1 2 lim n k = 1 n j = 1 k X x j - 1 , y j - X x j - 1 , y j - 1 y j - y j - 1 Δ x k Δ y j ,
(5.17)
1 2 f ( x , y ) 0 [ A , B ] X y d y d x = 1 2 lim n k = 1 n j = 1 k X ( x k - 1 , y k ) - X ( x k - 1 , y k - 1 ) y k - y k - 1 Δ y k Δ x j ,
(5.18)
1 2 f ( x , y ) 0 [ A , B ] X x d x d x = 1 2 lim n k = 1 n j = 1 k X x j , y j - X x j - 1 , y j x j - x j - 1 Δ x k Δ x j ,
(5.19)
1 2 f ( x , y ) 0 [ A , B ] X x d x d x = 1 2 lim n k = 1 n j = 1 k X ( x k , y k ) - X ( x k - 1 , y k ) x k - x k - 1 Δ x k Δ x j .
(5.20)

We thus obtain (4.1) in Lemma 3.

Appendix 2

Substituting (4.24) in (4.20) and (4.21), substituting (4.25) in (4.22) and (4.23), the respective terms of (4.30) are calculated as follows.
  1. 1.
    The first term of the left-hand side of (4.30) is
    1 2 x 2 a 2 + y 2 b 2 1 X y - Y x d x d y = - 1 2 x 2 a 2 + y 2 b 2 1 ( x ) 2 + y 2 d x d y = - 1 2 lim n k = 1 n j = 1 k ( x j ) 2 + ( y k ) 2 Δ x k Δ y j = - a b 2 lim n k = 1 n j = 1 k a 2 cos 2 θ j + b 2 sin 2 θ k ( cos θ k - cos θ k - 1 ) sin θ j - sin θ j - 1 = - a b 2 lim n k = 1 n j = 1 k a 2 cos 2 j 2 π n + b 2 sin 2 k 2 π n × cos k 2 π n - cos ( k - 1 ) 2 π n sin j 2 π n - sin ( j - 1 ) 2 π n = 3 8 π a b ( a 2 + b 2 ) .
    (5.21)
     
  2. 2.
    The second term of the lefthand side of (4.30) is
    1 2 x 2 a 2 + y 2 b 2 1 Y x - X y d y d x = 1 2 x 2 a 2 + y 2 b 2 1 ( y ) 2 + x 2 d y d x = 1 2 lim n k = 1 n j = 1 k ( y j ) 2 + ( x k ) 2 Δ y k Δ x j = a b 2 lim n k = 1 n j = 1 k b 2 sin 2 θ j + a 2 cos 2 θ k sin θ k - sin θ k - 1 cos θ j - cos θ j - 1 = a b 2 lim n k = 1 n j = 1 k b 2 sin 2 j 2 π n + a 2 cos 2 k 2 π n × sin k 2 π n - sin ( k - 1 ) 2 π n cos j 2 π n - cos ( j - 1 ) 2 π n = 3 8 π a b ( a 2 + b 2 )