문서의 임의 삭제는 제재 대상으로, 문서를 삭제하려면 삭제 토론을 진행해야 합니다. 문서 보기문서 삭제토론 스토크스 정리 (문단 편집) ==== 켈빈-스토크스 정리의 증명 ==== 헤르만 한켈(Hermann Hankel)이 1861년 그린 정리를 사용해 켈빈-스토크스 정리를 증명하였다.[*증명1 Zur allgemeinen Theorie der Bewegung der Flüssigkeiten, Preisschrift (공)저: Hermann Hankel §7 ,P34 [[https://books.google.co.kr/books/about/Zur_allgemeinen_Theorie_der_Bewegung_der.html?id=lHjRWuyMBSUC&redir_esc=y]]][*증명2 Vector Calculus, Michael Corral (Schoolcraft College) PDF LastEedition 2022(original 2008) GNU GFDL [[https://www.mecmath.net/]]][*증명3 The History of Stokes' Theorem, Author(s): Victor J. Katz, Source: Mathematics Magazine, Vol. 52, No. 3 (May, 1979), pp. 146-156, Published by: Mathematical Association of America, Stable URL: [[http://www.jstor.org/stable/2690275]] Accessed: 09-01-2017 23:04 UTC [[https://sites.math.washington.edu/~morrow/335_17/history of stokes thm.pdf]]] 발산정리(divergence theorem)의 [[연쇄 법칙]] [math(m(x,y,z))], [math(x(t))], [math(y(t))], [math(z(t))]에서 {{{#!wiki style="text-align:center" [math( \dfrac{\partial m}{\partial t} = \!\left( \dfrac{\partial m}{\partial x} \dfrac{\partial x}{\partial t} + \dfrac{\partial m}{\partial y} \dfrac{\partial y}{\partial t} + \dfrac{\partial m}{\partial z} \dfrac{\partial z}{\partial t} \right) \quad - \quad (1) )]}}} 1861년 한켈의 그린 정리 {{{#!wiki style="text-align:center" [math(\displaystyle \int ( \xi \,{\rm d}x +\eta \,{\rm d}y ) = \iint \biggl( \frac{{\rm d}\xi}{{\rm d}y} -\frac{{\rm d}\eta}{{\rm d}x} \biggr) {\rm d}x \,{\rm d}y \quad - \quad (2) )]}}} 헤르만 한켈의 켈빈-스토크스 정리 {{{#!wiki style="text-align:center" [math(\displaystyle \int ( \xi \,{\rm d}x +\eta \,{\rm d}y +\zeta \,{\rm d}z) \quad - \quad (3) )]}}} [math(z = z(x,y))], [math(x=x(t))], [math(y=y(t))]이므로 (1)을 사용하면 {{{#!wiki style="text-align:center" [math(\displaystyle \begin{aligned} \frac{{\rm d}z}{{\rm d}t} &= \frac{{\rm d}z}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} +\frac{{\rm d}z}{{\rm d}y} \frac{{\rm d}y}{{\rm d}t} \\ {\rm d}z &= \frac{{\rm d}z}{{\rm d}x} {\rm d}x +\frac{{\rm d}z}{{\rm d}y} {\rm d}y \quad - \quad (4) \end{aligned} )]}}} (3)에 (4)을 대입하면 {{{#!wiki style="text-align:center" [math(\displaystyle \begin{aligned} &\int \biggl\{ \xi \,{\rm d}x +\eta \,{\rm d}y +\zeta \biggl( \frac{{\rm d}z}{{\rm d}x} {\rm d}x +\frac{{\rm d}z}{{\rm d}y} {\rm d}y \biggr) \!\biggr\} \\ = &\int \biggl\{ \xi \,{\rm d}x +\eta \,{\rm d}y +\zeta \frac{{\rm d}z}{{\rm d}x} {\rm d}x +\zeta \frac{{\rm d}z}{{\rm d}y} {\rm d}y \biggr\} \\ = &\int \biggl\{ \!\biggl( \xi +\frac{{\rm d}z}{{\rm d}x} \zeta \biggr) {\rm d}x +\biggl( \eta +\frac{{\rm d}z}{{\rm d}y} \zeta \biggr) {\rm d}y \biggr\} \end{aligned} )]}}} 이것을 (2)에 대입하면 [math( \displaystyle \int \xi dx + \eta dy = \iint \left( \frac{d \xi}{d y} - \frac{d \eta}{dx} \right) dxdy )]이므로 [math( \displaystyle \int \begin{Bmatrix} \displaystyle \left( \xi + \frac{d z}{d x} \zeta \right)dx + \left( \eta + \frac{d z}{dy} \zeta \right)dy \end{Bmatrix} = \displaystyle \iint \begin{Bmatrix} \frac{d \left( \xi + \frac{d z}{d x} \zeta \right)}{d y} - \frac{d \left( \eta + \frac{d z}{dy} \zeta \right) }{dx} \end{Bmatrix} dxdy )] (1)을 [math( \xi , \eta , \zeta )]에 대입하면 [math( \displaystyle \frac{d \square}{d x} = \left( \dfrac{\partial \square}{\partial x}\dfrac{\partial x}{\partial x} + \dfrac{\partial \square}{\partial y}\dfrac{\partial y}{\partial x} + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial x} \right) = \left( \dfrac{\partial \square}{\partial x}1 + \dfrac{\partial \square}{\partial x}0 + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial x} \right) = \left( \dfrac{\partial \square}{\partial x} + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial x} \right) )] [math( \displaystyle \frac{d \square}{d y} = \left( \dfrac{\partial \square}{\partial x}\dfrac{\partial x}{\partial y} + \dfrac{\partial \square}{\partial y}\dfrac{\partial y}{\partial y} + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial y} \right) = \left( \dfrac{\partial \square}{\partial x}0 + \dfrac{\partial \square}{\partial y}1 + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial x} \right) = \left( \dfrac{\partial \square}{\partial y} + \dfrac{\partial \square}{\partial z}\dfrac{\partial z}{\partial y} \right) )] [math( \displaystyle \frac{d \left( \xi + \frac{d z}{d x} \zeta \right)}{d y} = \left( \frac{d \xi}{d y} + \frac{d \xi}{d z} \frac{d z}{d y} \right) + \frac{d z}{d y}\left(\frac{d \zeta}{d y} + \frac{d \zeta}{d z} \frac{d z}{d x}\right)+ \frac{d^2 z}{dx dy} \zeta )] [math( \displaystyle {} = \frac{d \xi}{d y} + \frac{d \xi}{d z} \frac{d z}{d y} + \frac{d \zeta}{d y} \frac{d z}{d x}+ \frac{d \zeta}{d z} \frac{d z}{d x}\frac{d z}{d y}+ \frac{d^2 z}{dx dy}\zeta )] [math( \displaystyle \frac{d \left( \eta + \frac{d z}{dy} \zeta \right) }{dx} = \left( \frac{d \eta}{d x} + \frac{d \eta}{d z} \frac{d z}{d x} \right) + \frac{d z}{d x} \left( \frac{d \zeta}{d x} + \frac{d \zeta}{d z} \frac{d z}{d y} \right) + \frac{d^2 z}{dy dx} \zeta )] [math( \displaystyle = \frac{d \eta}{d x} + \frac{d \eta}{d z} \frac{d z}{d x} + \frac{d \zeta}{d x} \frac{d z}{d y}+ \frac{d \zeta}{d z} \frac{d z}{d y}\frac{d z}{d x}+ \frac{d^2 z}{dy dx} \zeta )] 따라서 [math( \displaystyle \iint \begin{Bmatrix} \displaystyle \frac{d \left( \xi + \frac{d z}{d x} \zeta \right)}{d y} - \frac{d \left( \eta + \frac{d z}{dy} \zeta \right) }{dx} \end{Bmatrix} dxdy )] [math( = \displaystyle \iint \begin{Bmatrix} \displaystyle \left( \frac{d \xi}{d y} + \frac{d \xi}{d z} \frac{d z}{d y} + \frac{d \zeta}{d y} \frac{d z}{d x}+ \cancel{\frac{d \zeta}{d z} \frac{d z}{d x}\frac{d z}{d y}}+ \cancel{\frac{d^2 z}{dx dy}\zeta} \right)- \left( \frac{d \eta}{d x} + \frac{d \eta}{d z} \frac{d z}{d x} + \frac{d \zeta}{d x} \frac{d z}{d y}+ \cancel{\frac{d \zeta}{d z} \frac{d z}{d y}\frac{d z}{d x}}+ \cancel{\frac{d^2 z}{dy dx} \zeta} \right) \end{Bmatrix} dxdy )] [math( = \displaystyle \iint \begin{Bmatrix} \displaystyle \left( \frac{d \xi}{d y} - \frac{d \eta}{d x} \right)+\left( \frac{d \zeta}{d y} - \frac{d \eta}{d z} \right)\frac{d z}{d x} + \left( \frac{d \xi}{d z} - \frac{d \zeta}{d x}\right)\frac{d z}{d y} \end{Bmatrix} dxdy )] 따라서 [math( \displaystyle \int \left( \xi dx + \eta dy + \zeta dz \right) = \displaystyle \iint \begin{Bmatrix} \displaystyle \left( \frac{d \xi}{d y} - \frac{d \eta}{d x} \right)+\left( \frac{d \zeta}{d y} - \frac{d \eta}{d z} \right)\frac{d z}{d x} + \left( \frac{d \xi}{d z} - \frac{d \zeta}{d x}\right)\frac{d z}{d y} \end{Bmatrix} dxdy )]이고 [math(\begin{Bmatrix} \displaystyle \left( \frac{d \xi}{d y} - \frac{d \eta}{d x} \right)+\left( \frac{d \zeta}{d y} - \frac{d \eta}{d z} \right)\frac{d z}{d x} + \left( \frac{d \xi}{d z} - \frac{d \zeta}{d x}\right)\frac{d z}{d y} \end{Bmatrix} = \nabla \times S (curl \; S) \;,\, dxdy = dA )] [math( \displaystyle \int_{\partial S} S \cdot dC = \iint_{S} \left( \nabla \times S \right) \cdot dA )] 회전(curl)과 발산(divergence)으로부터 스토크스 정리를 조사할수있다.저장 버튼을 클릭하면 당신이 기여한 내용을 CC-BY-NC-SA 2.0 KR으로 배포하고,기여한 문서에 대한 하이퍼링크나 URL을 이용하여 저작자 표시를 하는 것으로 충분하다는 데 동의하는 것입니다.이 동의는 철회할 수 없습니다.캡챠저장미리보기