円柱座標変換
= \int_{z=-z}_{0}^{z=z}_{0}{\int_{x=-}^{x=r}{\int_{y=-\sqrt{r}_{0}^{2-x}^{2}^{y=\ \sqrt{r}_{0}^{2-x}^{2}{\ f(x,y,z)\ dxdydz}}  (5-1-3)
また、式(3-1-2)の直方体L に対しM = Φ(L )(式(3-1-3))であることと、ヤコビアン(式(2-1-5))に注意して、積分の変数変換公式を用いると、
\begin{align}\int_{M}^{f\ dxdydz}
&= \int_{L} {(f\circ\Phi)\cdot (\det J\Phi )\ drd\theta d\zeta } \\
&= {\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }\int_{\zeta =-z}_{0}^{\zeta =z}_{0}{\ r\cdot \left( f\circ \Phi  \right)(r,\theta ,\zeta )\ drd\theta d\zeta }}
\end{align}  (5-1-4)
が分かる。
以上、まとめると、
\begin{align}
\int_{M} {f\ dxdydz}
&= \int_{z=-z_0}^{z=z_0} {\int_{x=-r_0}^{x=r}{\int_{y=-\sqrt{r_0^2-x^2}^{y=\ \sqrt{r_0^2-x^2}{\ f(x,y,z)\ dxdydz}} \\
&= \int_{r=0}^{r=r_0}{\int_{\theta =0}^{\theta =2\pi}{\int_{\zeta =-z_0}^{\zeta =z_0}{\ r\cdot \left( f\circ \Phi  \right)(r,\theta ,\zeta )\ dr d\theta d\zeta }}
\end{align} (5-1-5)
がわかる。
ここでは、ベクトル場の円柱表面∂M 上での面積分の計算方法を説明する。
x -y -z 空間で定義されたベクトル場X に対して、円柱面∂M 上の面積分を
\int_{\partial \mathbf{M}{\mathbf{X} = \int_{\Delta }_{1}{\mathbf{X}+\int_{\Delta }_{2}{\mathbf{X}+\int_{\Delta }_{3}{\mathbf{X}   (5-2-1)
で定める。但し、右辺の各項は
\int_{\Delta }_{1}{\mathbf{X}=\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{\left( \mathbf{X}\cdot \left( \frac{\partial \mathbf{I_{1}}{\partial r}\times \frac{\partial \mathbf{I_{1}}{\partial \theta } \right) \right)}\ d\theta dr   (5-2-2)
\int_{\Delta }_{2}{\mathbf{X}=\int_{\zeta =0}^{\zeta =\zeta }_{0}{\int_{\theta =0}^{\theta =2\pi }{\left( \mathbf{X}\cdot \left( \frac{\partial \mathbf{I_{2}}{\partial \theta }\times \frac{\partial \mathbf{I_{2}}{\partial \zeta } \right) \right)}\ d\theta d\zeta   (5-2-3)
\int_{\Delta }_{3}{\mathbf{X}=\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{\left( \mathbf{X}\cdot \left( \frac{\partial \mathbf{I_{3}}{\partial r}\times \frac{\partial \mathbf{I_{3}}{\partial \theta } \right) \right)}\ d\theta dr   (5-2-4)
である。ここで、
\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)
同様に、
\mathbf{X}\cdot \left( \frac{\partial \mathbf{I_{2}}{\partial \theta }\times \frac{\partial \mathbf{I_{2}}{\partial \zeta } \right) = \cdot X}_{r   (5-2-6)
\mathbf{X}\cdot \left( \frac{\partial \mathbf{I_{3}}{\partial \theta }\times \frac{\partial \mathbf{I_{3}}{\partial \zeta } \right) = -r\cdot X}_{\zeta    (5-2-7)
なので、
\int_{\Delta }_{1}{\mathbf{X}=\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{r\cdot X}_{\zeta (r,\theta ,\zeta }_{0)}\ d\theta dr   (5-2-8)
\int_{\Delta }_{2}{\mathbf{X}=\int_{\theta =0}^{\theta =2\pi }{r}_{0\cdot X}_{r(,\theta ,\zeta )}\ d\theta d\zeta   (5-2-9)
\int_{\Delta }_{3}{\mathbf{X}=\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{-r\cdotX}_{\zeta (r,\theta ,0)}\ d\theta dr   (5-2-10)
である。従って、
\int_{\partial \mathbf{M}{\mathbf{X}=\int_{r=0}^{r=}{\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}_{0X}_{r(,\theta ,\zeta )}\ d\theta d\zeta   (5-2-11)
が分かる。
x -y -z 空間で定義されたベクトル場X に対して、
\int_{\partial M}{\mathbf{X}=\int_{M}{\operatorname{div}\textbf{X}\ dxdydz}   (5-3-1)
が成立する。この事実を、円柱面におけるガウスの発散定理という。ここで、
\operatorname{div}\mathbf{X} = \left( \frac{\partial X}_{x}{\partial x} \right)+\left( \frac{\partial X}_{y}{\partial y} \right)+\left( \frac{\partial X}_{z}{\partial z} \right)   (5-3-2)
はベクトル場X の発散である。以下、証明を行う。
(5-1-5)、(5-2-2)、(5-2-3)、(5-2-4)より、
\int_{\zeta =0}^{\zeta =\zeta }_{0}{\int_{\theta =0}^{\theta =2\pi }{r}_{0X}_{r(,\theta ,\zeta )}\ d\theta d\zeta
= \int_{M}{\left( \left( \frac{\partial X}_{x}{\partial x} \right)+\left( \frac{\partial X}_{y}{\partial y} \right)\  \right)dxdydz}   (5-3-3)
\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{r\left( X}_{\varsigma (r,\theta ,\zeta }_{0)-X}_{\zeta (r,\theta ,0) \right)}\ \ d\theta dr
= \int_{M}{\left( \frac{\partial X}_{z}{\partial z} \right)dxdydz}   (5-3-4)
を示せばよいことがわかる。
まず、式(5-3-3)を示す。
x -y 平面上の曲線c (t ) を、
c(t)=\left( \begin{matrix}
\cos t  \\
\sin t  \\
\end{matrix} \right)   (5-3-5)
とする。また、変数z を固定して考えることで、x -y 平面上の二次元ベクトル場
\mathbf{A}\left( x,y \right)=\left( \begin{matrix}
a\left( x,y \right)  \\
b\left( x,y \right)  \\
\end{matrix} \right)=\left( \begin{matrix}
-X}_{y(x,y,z)  \\
X}_{x(x,y,z)  \\
\end{matrix} \right)   (5-3-6)
を考える。また、平面曲線と、二次元ベクトル場に対しては、(5-3-6)に対するグリーンの定理:
\int_{c}^{A}=\int_{x=-}^{x=}{\int_{-\sqrt{r}_{0^{2}-x}^{2}^{\sqrt{r}_{0^{2}-x}^{2}{\left( \frac{\partial b}{\partial x}-\frac{\partial a}{\partial y} \right)dxdy}   (5-3-7)
が成立することは、既知とする。
\begin{align}
\int_{\theta =0}^{\theta =2\pi}{r_0 X_r(r_0,\theta ,\zeta )}\ d\theta
&= \int_{\theta =0}^{\theta =2\pi}{\left((X_x \circ \Phi )r_0\cos\theta + (X_y\circ \Phi)r_0\sin\theta \right)}\ d\theta \\
&= \int_{\theta =0}^{\theta =2\pi}{\left(X_x(r_0 \cos\theta, r_0 \sin\theta, \zeta)r_0 \cos\theta +X_y(r_0\cos\theta, r_0\sin\theta, \zeta)r_0\sin\theta  \right)}\ d\theta \\
&= \int_{t=0}^{t=2\pi}{\left( \begin{matrix}
-X_y(r_0 \cos t,r_0 \sin t,\zeta )  \\
X_x(r_0 \cos t,r_0 \sin t,\zeta )  \\
\end{matrix} \right)}\ \cdot \left( \begin{matrix}
-r_0 \sin t  \\
r_0 \cos t  \\
\end{matrix} \right)dt \\
&= \int_{t=0}^{t=2\pi }{\left( \begin{matrix}
a\circ c(t)  \\
b\circ c(t)  \\
\end{matrix} \right)}\ \cdot \left( \frac{dc}{dt} \right)dt \\
&= \int_{x=-}^{x=}{\int_{y=-\sqrt{r}_{0^{2}-x}^{2}^{y=\sqrt{r}_{0^{2}-x}^{2}{\left( \frac{\partial b}{\partial x}-\frac{\partial a}{\partial y} \right)dydx} \\
&= \int_{x=-}^{x=}{\int_{y=-\sqrt{r}_{0^{2}-x}^{2}^{y=\sqrt{r}_{0^{2}-x}^{2}{\left( \frac{\partial X}_{x(x,y,z)}{\partial x}+\frac{\partial X}_{y(x,y,z)}{\partial y} \right)dydx}
\end{align}   (5-3-8)
が成り立つ。従って、
\int_{\zeta =0}^{\zeta =\zeta }_{0}{\int_{\theta =0}^{\theta =2\pi }{X}_{r(,\theta ,\zeta )}\ d\theta d\zeta
= \int_{z=0}^{z=\zeta }_{0}{\int_{\theta =0}^{\theta =2\pi }{X}_{r(,\theta ,z)}\ d\theta dz
\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)
次に、
\int_{r=0}^{r=}{\int_{\theta =0}^{\theta =2\pi }{r\left( X}_{\varsigma (r,\theta ,\zeta }_{0)-X}_{\zeta (r,\theta ,0) \right)}\ \ d\theta dr
= \int_{M}{\left( \frac{\partial X}_{z}{\partial z} \right)dxdydz}   (5-3-10)
を示す。
\int_{z=0}^{z=\zeta }_{0}{\frac{\partial X}_{\zeta (r,\theta ,\zeta )\ }{\partial \zeta }d\zeta }
=X}_{\zeta (r,\theta ,\zeta }_{0)-X}_{\zeta \ (r,\theta ,0)   (5-3-11)
なので、
\begin{align}
\int_{r=0}^{r=r_0}{\int_{\theta =0}^{\theta =2\pi}{r\left(X_\varsigma(r,\theta ,\zeta_0)-X_\zeta(r,\theta ,0) \right)\ \ d\theta}dr}
&= \int_{r=0}^{r=r_0}{r\left(\int_{\theta =0}^{\theta =2\pi}{\left(X_\varsigma(r,\theta,\zeta_0)-X_\zeta(r,\theta,0) \right)d\theta}\right)\ \ dr}\\
&= \int_{r=0}^{r=r_0}{r\left(\int_{\theta =0}^{\theta =2\pi}{\int_{\zeta =0}^{\zeta =\zeta_0}{\frac{\partial X_\zeta(r,\theta,\zeta)}{\partial \zeta}d\zeta}d\theta}\right)\ dr} \\
& =\int_{r=0}^{r=r_0}{\int_{\theta =0}^{\theta =2\pi}{\int_{\zeta =0}^{\zeta =\zeta_0}{\left(\frac{\partial X_\zeta(r,\theta,\zeta)}{\partial \zeta} \right)r d\zeta}d\theta}dr} \\
&= \int_{r=0}^{r=r_0}{\int_{\theta =0}^{\theta =2\pi}{\int_{\zeta =0}^{\zeta =\zeta_0}{\left( \left( \frac{\partial X_z}{\partial z} \right){}^\circ \Phi (r,\theta ,\zeta ) \right)r d\zeta}d\theta}dr} \\
&= \int_{r=0}^{r=r_0}{\int_{\theta =0}^{\theta =2\pi}{\int_{\zeta =0}^{\zeta =\zeta_0}{\left( \left( \frac{\partial X_z}{\partial z} \right){}^\circ \Phi (r,\theta ,\zeta ) \right)\left( \det (J\Phi (r,\theta ,\zeta )) \right)d\zeta} d\theta}dr} \\
&= \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_z(x,y,z)}{\partial z} \right)dy}dx}dz}
\end{align}   (5-3-12)
■
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:
\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)
\left( \frac{\partial f}{\partial y} \right)(\Phi (r,\theta ,\zeta ))
= \sin\theta \left( \left( \frac{\partial (f\circ \Phi )}{\partial r} \right)(r,\theta ,z) \right)
+ \frac{\cos\theta}{r}\left( \left( \frac{\partial (f\circ \Phi )}{\partial \theta } \right)(r,\theta ,z) \right) (6-1-2)
\left( \frac{\partial f}{\partial z} \right)(\Phi (r,\theta ,z))
= \left( \frac{\partial (f\circ \Phi )}{\partial \zeta } \right)(r,\theta ,z) (6-1-3)
上記の3式は、しばしば略記的に以下のように表記される:
\left\{ \begin{array}{*{35}{l}
\left( \frac{\partial f}{\partial x} \right)=\cos\theta \left( \frac{\partial f}{\partial r} \right)-\frac{\sin\theta}{r}\left( \frac{\partial f}{\partial \theta } \right)  \\
\left( \frac{\partial f}{\partial y} \right)=\sin\theta \left( \frac{\partial f}{\partial r} \right)+\frac{\cos\theta}{r}\left( \frac{\partial f}{\partial \theta } \right)  \\
\left( \frac{\partial f}{\partial z} \right)=\left( \frac{\partial f}{\partial \zeta } \right)  \\
\end{array} \right. (6-1-4)
勾配 grad :
\begin{align}
(\operatorname{grad}f)(x,y,z) &= \left( \begin{matrix}
(\partial f/\partial x)(x,y,z)  \\
(\partial f/\partial y)(x,y,z)  \\
(\partial f/\partial z)(x,y,z)  \\
\end{matrix} \right) \\
&= \left( \left( \frac{\partial f}{\partial x} \right)(x,y,z) \right)\left( \begin{matrix}
1  \\
0  \\
0  \\
\end{matrix} \right) + \left( \left( \frac{\partial f}{\partial y} \right)(x,y,z) \right)\left( \begin{matrix}
0  \\
1  \\
0  \\
\end{matrix} \right) + \left( \left( \frac{\partial f}{\partial z} \right)(x,y,z) \right)\left( \begin{matrix}
0  \\
0  \\
1  \\
\end{matrix} \right)\end{align}         (6-2-1)
...
以下、証明を行う。(grad f )(Φ(r , θ, ζ)) と、Nr (r , θ, ζ) の内積を取ると、
\begin{align}
\left[ \left(\operatorname{grad}f \right)(\Phi (r,\theta ,\zeta )) \right]\cdot \left( \mathbf{N_{r}(r,\theta ,\zeta ) \right)
=& \left( \left( \frac{\partial f}{\partial x} \right)(\Phi (r,\theta ,\zeta )),\left( \frac{\partial f}{\partial y} \right)(\Phi (r,\theta ,\zeta )),\left( \frac{\partial f}{\partial z} \right)(\Phi (r,\theta ,\zeta )) \right)\left( \begin{matrix}
\cos\theta   \\
\sin\theta   \\
0  \\
\end{matrix} \right) \\
=& \cos\theta \left( \left( \frac{\partial f}{\partial x} \right)(\Phi (r,\theta ,\zeta )) \right)+\sin\theta \left( \left( \frac{\partial f}{\partial y} \right)(\Phi (r,\theta ,\zeta )) \right) \\
=& \cos\theta \left[ \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) \right]\\
&+ \sin\theta \left[ \sin\theta \left( \left( \frac{\partial (f\circ \Phi )}{\partial r} \right)(r,\theta ,z) \right)+\frac{\cos\theta}{r}\left( \left( \frac{\partial (f\circ \Phi )}{\partial \theta } \right)(r,\theta ,z) \right) \right]\\
=& \left( \frac{\partial (f\circ \Phi )}{\partial r} \right)(r,\theta ,z)
\end{align} (6-2-2)
であり、同様に、Nθ(r , θ, ζ) , Nζ(r , θ, ζ) についても、
\left[ \left(\operatorname{grad}f \right)(\Phi (r,\theta ,\zeta )) \right]\cdot \left( \mathbf{N_{\theta }(r,\theta ,\zeta ) \right)=\frac{1}{r}\left( \frac{\partial (f\circ \Phi )}{\partial \theta } \right)(r,\theta ,z)        (6-2-3)