Let ( S , Ξ΅ ) (S,\varepsilon) ( S , Ξ΅ ) Ο \omega Ο Oriented k-Sub-Rectangle of Euclidean Space , so that
S = Ξ» R ( R ) , R = [ a 1 , b 1 ] Γ β― Γ [ a k , b k ] , S=\lambda_R(R), \qquad R=[a_1,b_1]\times\cdots\times[a_k,b_k], S = Ξ» R β ( R ) , R = [ a 1 β , b 1 β ] Γ β― Γ [ a k β , b k β ] , with active coordinate indices
1 β€ i 1 < β― < i k β€ n . 1\le i_1<\cdots<i_k\le n. 1 β€ i 1 β < β― < i k β β€ n . Write Ο \omega Ο Coordinate Expansion of Differential Forms on Euclidean Open Sets as
Ο = β 1 β€ j 1 < β― < j k β 1 β€ n a j 1 β¦ j k β 1 β d x j 1 β§ β― β§ d x j k β 1 . \omega=\sum_{1\le j_1<\cdots<j_{k-1}\le n} a_{j_1\dots j_{k-1}}\,dx_{j_1}\wedge\cdots\wedge dx_{j_{k-1}}. Ο = 1 β€ j 1 β < β― < j k β 1 β β€ n β β a j 1 β β¦ j k β 1 β β d x j 1 β β β§ β― β§ d x j k β 1 β β . For each r β { 1 , β¦ , k } r\in\{1,\dots,k\} r β { 1 , β¦ , k } I r ^ I^{\hat r} I r ^ ( k β 1 ) (k-1) ( k β 1 ) ( i 1 , β¦ , i k ) (i_1,\dots,i_k) ( i 1 β , β¦ , i k β ) i r i_r i r β a I r ^ a_{I^{\hat r}} a I r ^ β
By Active Coordinate Coefficient Formula for the Exterior Derivative , the coefficient of
d x i 1 β§ β― β§ d x i k dx_{i_1}\wedge\cdots\wedge dx_{i_k} d x i 1 β β β§ β― β§ d x i k β β in the coordinate expansion of d Ο d\omega d Ο
c = β r = 1 k ( β 1 ) r β 1 β a I r ^ β x i r . c=\sum_{r=1}^k (-1)^{r-1}\frac{\partial a_{I^{\hat r}}}{\partial x_{i_r}}. c = r = 1 β k β ( β 1 ) r β 1 β x i r β β β a I r ^ β β . Therefore, by Integral of a Differential Form over an Oriented k-Sub-Rectangle in Euclidean Space ,
β« ( S , Ξ΅ ) d Ο = Ξ΅ β« a k b k β― β« a 1 b 1 c ( Ξ» R ( t 1 , β¦ , t k ) ) β d t 1 β― d t k . \int_{(S,\varepsilon)} d\omega
=\varepsilon\int_{a_k}^{b_k}\cdots\int_{a_1}^{b_1} c(\lambda_R(t_1,\dots,t_k))\,dt_1\cdots dt_k. β« ( S , Ξ΅ ) β d Ο = Ξ΅ β« a k β b k β β β― β« a 1 β b 1 β β c ( Ξ» R β ( t 1 β , β¦ , t k β )) d t 1 β β― d t k β . Substituting the formula for c c c
β« ( S , Ξ΅ ) d Ο = Ξ΅ β r = 1 k ( β 1 ) r β 1 β« a k b k β― β« a 1 b 1 β a I r ^ β x i r ( Ξ» R ( t 1 , β¦ , t k ) ) β d t 1 β― d t k . \int_{(S,\varepsilon)} d\omega
=\varepsilon\sum_{r=1}^k (-1)^{r-1}
\int_{a_k}^{b_k}\cdots\int_{a_1}^{b_1}
\frac{\partial a_{I^{\hat r}}}{\partial x_{i_r}}(\lambda_R(t_1,\dots,t_k))\,dt_1\cdots dt_k. β« ( S , Ξ΅ ) β d Ο = Ξ΅ r = 1 β k β ( β 1 ) r β 1 β« a k β b k β β β― β« a 1 β b 1 β β β x i r β β β a I r ^ β β ( Ξ» R β ( t 1 β , β¦ , t k β )) d t 1 β β― d t k β . Fix r β { 1 , β¦ , k } r\in\{1,\dots,k\} r β { 1 , β¦ , k } t r t_r t r β
g r ( t ) = a I r ^ ( Ξ» R ( t 1 , β¦ , t r β 1 , t , t r + 1 , β¦ , t k ) ) g_r(t)=a_{I^{\hat r}}(\lambda_R(t_1,\dots,t_{r-1},t,t_{r+1},\dots,t_k)) g r β ( t ) = a I r ^ β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , t , t r + 1 β , β¦ , t k β )) on [ a r , b r ] [a_r,b_r] [ a r β , b r β ] Derivative of a Coordinate Slice of a C^1 Function on a Euclidean Open Set , the function g r g_r g r β [ a r , b r ] [a_r,b_r] [ a r β , b r β ] ( a r , b r ) (a_r,b_r) ( a r β , b r β )
g r β² ( t ) = β a I r ^ β x i r ( Ξ» R ( t 1 , β¦ , t r β 1 , t , t r + 1 , β¦ , t k ) ) . g_r'(t)=\frac{\partial a_{I^{\hat r}}}{\partial x_{i_r}}(\lambda_R(t_1,\dots,t_{r-1},t,t_{r+1},\dots,t_k)). g r β² β ( t ) = β x i r β β β a I r ^ β β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , t , t r + 1 β , β¦ , t k β )) . Thus g r g_r g r β [ a r , b r ] [a_r,b_r] [ a r β , b r β ] t r t_r t r β Fundamental Theorem of Calculus, Part II in One Dimension with I = [ a r , b r ] I=[a_r,b_r] I = [ a r β , b r β ]
β« a r b r β a I r ^ β x i r ( Ξ» R ( t 1 , β¦ , t k ) ) β d t r = a I r ^ ( Ξ» R ( t 1 , β¦ , t r β 1 , b r , t r + 1 , β¦ , t k ) ) β a I r ^ ( Ξ» R ( t 1 , β¦ , t r β 1 , a r , t r + 1 , β¦ , t k ) ) . \int_{a_r}^{b_r}\frac{\partial a_{I^{\hat r}}}{\partial x_{i_r}}(\lambda_R(t_1,\dots,t_k))\,dt_r
= a_{I^{\hat r}}(\lambda_R(t_1,\dots,t_{r-1},b_r,t_{r+1},\dots,t_k))
- a_{I^{\hat r}}(\lambda_R(t_1,\dots,t_{r-1},a_r,t_{r+1},\dots,t_k)). β« a r β b r β β β x i r β β β a I r ^ β β ( Ξ» R β ( t 1 β , β¦ , t k β )) d t r β = a I r ^ β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , b r β , t r + 1 β , β¦ , t k β )) β a I r ^ β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , a r β , t r + 1 β , β¦ , t k β )) . Substituting this identity into the iterated integral gives
β« ( S , Ξ΅ ) d Ο = β r = 1 k ( E r + + E r β ) , \int_{(S,\varepsilon)} d\omega
=\sum_{r=1}^k (E_r^+ + E_r^-), β« ( S , Ξ΅ ) β d Ο = r = 1 β k β ( E r + β + E r β β ) , where
E r + = Ξ΅ ( β 1 ) r β 1 β« a k b k β― β« a r b r ^ β― β« a 1 b 1 a I r ^ ( Ξ» R ( t 1 , β¦ , t r β 1 , b r , t r + 1 , β¦ , t k ) ) β d t 1 β― d t r ^ β― d t k E_r^+=\varepsilon(-1)^{r-1}\int_{a_k}^{b_k}\cdots\widehat{\int_{a_r}^{b_r}}\cdots\int_{a_1}^{b_1}
a_{I^{\hat r}}(\lambda_R(t_1,\dots,t_{r-1},b_r,t_{r+1},\dots,t_k))\,dt_1\cdots \widehat{dt_r}\cdots dt_k E r + β = Ξ΅ ( β 1 ) r β 1 β« a k β b k β β β― β« a r β b r β β β β― β« a 1 β b 1 β β a I r ^ β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , b r β , t r + 1 β , β¦ , t k β )) d t 1 β β― d t r β β β― d t k β and
E r β = β Ξ΅ ( β 1 ) r β 1 β« a k b k β― β« a r b r ^ β― β« a 1 b 1 a I r ^ ( Ξ» R ( t 1 , β¦ , t r β 1 , a r , t r + 1 , β¦ , t k ) ) β d t 1 β― d t r ^ β― d t k . E_r^-=-\varepsilon(-1)^{r-1}\int_{a_k}^{b_k}\cdots\widehat{\int_{a_r}^{b_r}}\cdots\int_{a_1}^{b_1}
a_{I^{\hat r}}(\lambda_R(t_1,\dots,t_{r-1},a_r,t_{r+1},\dots,t_k))\,dt_1\cdots \widehat{dt_r}\cdots dt_k. E r β β = β Ξ΅ ( β 1 ) r β 1 β« a k β b k β β β― β« a r β b r β β β β― β« a 1 β b 1 β β a I r ^ β ( Ξ» R β ( t 1 β , β¦ , t r β 1 β , a r β , t r + 1 β , β¦ , t k β )) d t 1 β β― d t r β β β― d t k β . Now the face S r + S_r^+ S r + β R r R_r R r β ( Ξ» R β ΞΌ r + ) (\lambda_R\circ\mu_r^+) ( Ξ» R β β ΞΌ r + β )
i 1 , β¦ , i r β 1 , i r + 1 , β¦ , i k . i_1,\dots,i_{r-1},i_{r+1},\dots,i_k. i 1 β , β¦ , i r β 1 β , i r + 1 β , β¦ , i k β . Hence, by the definition Integral of a Differential Form over an Oriented k-Sub-Rectangle in Euclidean Space of the integral over an oriented sub-rectangle,
β« ( S r + , Ξ΅ ( β 1 ) r β 1 ) Ο = E r + . \int_{(S_r^+,\varepsilon(-1)^{r-1})}\omega=E_r^+. β« ( S r + β , Ξ΅ ( β 1 ) r β 1 ) β Ο = E r + β . Likewise, since S r β S_r^- S r β β R r R_r R r β ( Ξ» R β ΞΌ r β ) (\lambda_R\circ\mu_r^-) ( Ξ» R β β ΞΌ r β β ) β Ξ΅ ( β 1 ) r β 1 -\varepsilon(-1)^{r-1} β Ξ΅ ( β 1 ) r β 1
β« ( S r β , β Ξ΅ ( β 1 ) r β 1 ) Ο = E r β . \int_{(S_r^-,-\varepsilon(-1)^{r-1})}\omega=E_r^-. β« ( S r β β , β Ξ΅ ( β 1 ) r β 1 ) β Ο = E r β β . Therefore
β« ( S , Ξ΅ ) d Ο = β r = 1 k ( β« ( S r + , Ξ΅ ( β 1 ) r β 1 ) Ο + β« ( S r β , β Ξ΅ ( β 1 ) r β 1 ) Ο ) . \int_{(S,\varepsilon)} d\omega
=\sum_{r=1}^k \left(\int_{(S_r^+,\varepsilon(-1)^{r-1})}\omega+\int_{(S_r^-,-\varepsilon(-1)^{r-1})}\omega\right). β« ( S , Ξ΅ ) β d Ο = r = 1 β k β ( β« ( S r + β , Ξ΅ ( β 1 ) r β 1 ) β Ο + β« ( S r β β , β Ξ΅ ( β 1 ) r β 1 ) β Ο ) . Since
β ( S , Ξ΅ ) = { ( S r + , Ξ΅ ( β 1 ) r β 1 ) , ( S r β , β Ξ΅ ( β 1 ) r β 1 ) : r β { 1 , β¦ , k } } , \partial(S,\varepsilon)=\{(S_r^+,\varepsilon(-1)^{r-1}),(S_r^-,-\varepsilon(-1)^{r-1}): r\in\{1,\dots,k\}\}, β ( S , Ξ΅ ) = {( S r + β , Ξ΅ ( β 1 ) r β 1 ) , ( S r β β , β Ξ΅ ( β 1 ) r β 1 ) : r β { 1 , β¦ , k }} , this is exactly
β« ( S , Ξ΅ ) d Ο = β ( T , Ξ· ) β β ( S , Ξ΅ ) β« ( T , Ξ· ) Ο . \int_{(S,\varepsilon)} d\omega=\sum_{(T,\eta)\in\partial(S,\varepsilon)} \int_{(T,\eta)}\omega. β« ( S , Ξ΅ ) β d Ο = ( T , Ξ· ) β β ( S , Ξ΅ ) β β β« ( T , Ξ· ) β Ο . This proves the theorem.