円柱座標変換
Δ 3 = { ( x y z ) | x 2 + y 2 ≤ r 0 z = z 0 } {\displaystyle {{\Delta }_{3}}=\left\{\left.\left({\begin{matrix}x\\y\\z\\\end{matrix}}\right)\ \left|{\begin{matrix}{{x}^{2}}+{{y}^{2}}\leq {{r}_{0}}\\z={{z}_{0}}\\\end{matrix}}\right.\right\}\right.} (3-2-3)
Δ1 , Δ2 , Δ3 をそれぞれ、下面、側面、上面という。
逆に言うと、Δ1 , Δ2 , Δ3 を貼り合わせたものが∂M である。集合演算を用いると、
∂ M = Δ 1 ∪ Δ 2 ∪ Δ 3 {\displaystyle \partial M={\Delta }_{1}\cup {\Delta }_{2}\cup {\Delta }_{3}} (3-2-4)
のようになる。
次に、Δ1 , Δ2 , Δ3 を、円柱座標変換を利用してパラメトライズすることを考える。D1 , D2 , D3 を以下のように定める。
D 1 = { ( r θ ) | 0 ≤ r ≤ r 0 0 ≤ θ ≤ 2 π } {\displaystyle {{D}_{1}}=\left\{\left.\left({\begin{matrix}r\\\theta \\\end{matrix}}\right)\ \left|{\begin{matrix}0\leq r\leq {{r}_{0}}\\0\leq \theta \leq 2\pi \\\end{matrix}}\right.\right\}\right.} (3-2-5)
∫ Δ 3 X = ∫ r = 0 r = r 0 ∫ θ = 0 θ = 2 π − r ⋅ X ζ ( r , θ , 0 ) d θ d r {\displaystyle \int _{{\Delta }_{3}}{\mathbf {X} }=\int _{r=0}^{r={{r}_{0}}}{\int _{\theta =0}^{\theta =2\pi }{-r\cdot {{X}_{\zeta }}(r,\theta ,0)}}\ d\theta dr} (5-2-10)
である。従って、
∫ ∂ M X = ∫ r = 0 r = r 0 ∫ θ = 0 θ = 2 π r ( X ζ ( r , θ , ζ 0 ) − X ζ ( r , θ , 0 ) ) d θ d r + ∫ ζ = 0 ζ = ζ 0 ∫ θ = 0 θ = 2 π r 0 X r ( r 0 , θ , ζ ) d θ d ζ {\displaystyle \int _{\partial \mathbf {M} }{\mathbf {X} }=\int _{r=0}^{r={{r}_{0}}}{\int _{\theta =0}^{\theta =2\pi }r{\left({{X}_{\zeta }}(r,\theta ,{{\zeta }_{0}})-{{X}_{\zeta }}(r,\theta ,0)\right)d\theta dr}}\ \ +\int _{\zeta =0}^{\zeta ={{\zeta }_{0}}}{\int _{\theta =0}^{\theta =2\pi }{{{r}_{0}}{{X}_{r}}({{r}_{0}},\theta ,\zeta )}}\ d\theta d\zeta } (5-2-11)
が分かる。
N θ ( r , θ , ζ ) = ( ∂ Φ ∂ ζ ( r , θ , ζ ) ) × ( ∂ Φ ∂ r ( r , θ , ζ ) ) ‖ ( ∂ Φ ∂ ζ ( r , θ , ζ ) ) × ( ∂ Φ ∂ r ( r , θ , ζ ) ) ‖ = ( − sin  θ cos  θ 0 ) {\displaystyle {{\mathbf {N} }_{\theta }}(r,\theta ,\zeta )={\frac {\left({\frac {\partial \Phi }{\partial \zeta }}(r,\theta ,\zeta )\right)\times \left({\frac {\partial \Phi }{\partial r}}(r,\theta ,\zeta )\right)}{\left\|\left({\frac {\partial \Phi }{\partial \zeta }}(r,\theta ,\zeta )\right)\times \left({\frac {\partial \Phi }{\partial r}}(r,\theta ,\zeta )\right)\right\|}}=\left({\begin{matrix}-\sin \theta \\\cos \theta \\0\\\end{matrix}}\right)} (2-2-2)
{ x = r cos  θ y = r sin  θ z = z {\displaystyle \left\{{\begin{array}{l}x=r\cos \theta \\y=r\sin \theta \\z=z\\\end{array}}\right.} (8-1-4)
の両辺を時刻t に関する関数と考えること、つまり、
{ x ( r ( t ) , θ ( t ) , z ( t ) ) = r ( t ) cos  θ ( t ) y ( r ( t ) , θ ( t ) , z ( t ) ) = r ( t ) sin  θ ( t ) z ( r ( t ) , θ ( t ) , z ( t ) ) = z ( t ) {\displaystyle \left\{{\begin{array}{l}x(r(t),\theta (t),z(t))=r(t)\cos \theta (t)\\y(r(t),\theta (t),z(t))=r(t)\sin \theta (t)\\z(r(t),\theta (t),z(t))=z(t)\\\end{array}}\right.} (8-1-5)
と考えることと同じである。
∫ ζ = 0 ζ = ζ 0 ∫ θ = 0 θ = 2 π X r ( r 0 , θ , ζ ) d θ d ζ = ∫ z = 0 z = ζ 0 ∫ θ = 0 θ = 2 π X r ( r 0 , θ , z ) d θ d z {\displaystyle \int _{\zeta =0}^{\zeta ={{\zeta }_{0}}}{\int _{\theta =0}^{\theta =2\pi }{{{X}_{r}}({{r}_{0}},\theta ,\zeta )}}\ d\theta d\zeta =\int _{z=0}^{z={{\zeta }_{0}}}{\int _{\theta =0}^{\theta =2\pi }{{{X}_{r}}({{r}_{0}},\theta ,z)}}\ d\theta dz}
∫ z = 0 z = ζ 0 ∫ x = − r x = r ∫ y = − r 2 − x 2 y = r 2 − x 2 ( ∂ X x ( x , y , z ) ∂ x + ∂ X y ( x , y , z ) ∂ y ) d y d x d z {\displaystyle \int _{z=0}^{z={{\zeta }_{0}}}{\int _{x=-r}^{x=r}{\int _{y=-{\sqrt {{{r}^{2}}-{{x}^{2}}}}}^{y={\sqrt {{{r}^{2}}-{{x}^{2}}}}}{\left({\frac {\partial {{X}_{x}}(x,y,z)}{\partial x}}+{\frac {\partial {{X}_{y}}(x,y,z)}{\partial y}}\right)dydx}}}dz} (5-3-9)
である。ここで、
X ⋅ ( ∂ I 1 ∂ r × ∂ I 1 ∂ θ ) = X ⋅ ( ∂ I 1 ∂ r × ∂ I 1 ∂ θ ) ‖ ∂ I 1 ∂ r × ∂ I 1 ∂ θ ‖ ‖ ∂ I 1 ∂ r × ∂ I 1 ∂ θ ‖ = ( X ⋅ N ζ ) ‖ ∂ I 1 ∂ r × ∂ I 1 ∂ θ ‖ = r ⋅ X ζ {\displaystyle \mathbf {X} \cdot \left({\frac {\partial {{\mathbf {I} }_{1}}}{\partial r}}\times {\frac {\partial {{\mathbf {I} }_{1}}}{\partial \theta }}\right)=\mathbf {X} \cdot {\frac {\left({\frac {\partial {{\mathbf {I} }_{1}}}{\partial r}}\times {\frac {\partial {{\mathbf {I} }_{1}}}{\partial \theta }}\right)}{\left\|{\frac {\partial {{\mathbf {I} }_{1}}}{\partial r}}\times {\frac {\partial {{\mathbf {I} }_{1}}}{\partial \theta }}\right\|}}\left\|{\frac {\partial {{\mathbf {I} }_{1}}}{\partial r}}\times {\frac {\partial {{\mathbf {I} }_{1}}}{\partial \theta }}\right\|=\left(\mathbf {X} \cdot {{\mathbf {N} }_{\zeta }}\right)\left\|{\frac {\partial {{\mathbf {I} }_{1}}}{\partial r}}\times {\frac {\partial {{\mathbf {I} }_{1}}}{\partial \theta }}\right\|=r\cdot {{X}_{\zeta }}} (5-2-5)
f を、x -y -z 空間で定義されたスカラー値関数とする。このとき、r ≠ 0 をみたす任意の (r , θ, ζ) に対して以下の(6-1-1)~(6-1-3)の等式が成立する:
( ∂ f ∂ x ) ( Φ ( r , θ , ζ ) ) = cos  θ ( ( ∂ ( f ∘ Φ ) ∂ r ) ( r , θ , z ) ) − sin  θ r ( ( ∂ ( f ∘ Φ ) ∂ θ ) ( r , θ , z ) ) {\displaystyle \left({\frac {\partial f}{\partial x}}\right)(\Phi (r,\theta ,\zeta ))=\cos \theta \left(\left({\frac {\partial (f\circ \Phi )}{\partial r}}\right)(r,\theta ,z)\right)-{\frac {\sin \theta }{r}}\left(\left({\frac {\partial (f\circ \Phi )}{\partial \theta }}\right)(r,\theta ,z)\right)} (6-1-1)