这里考虑三维欧氏空间中的正则曲线 $c: I\rightarrow\mathbb{R}^3$, 弧长微元 $ds$ 有下面这几种表示.
\[ds^2=|dc|^2=dx^2+dy^2+dz^2.\]
若使用柱坐标系,
\[x=\rho\cos\theta,\quad y=\rho\sin\theta,\quad z=z\]
则
\[ds^2=d\rho^2+\rho^2d\theta^2+dz^2;\]
若采用球坐标系
\[
\begin{cases}
x=\rho\sin\varphi\cos\theta,\\
y=\rho\sin\varphi\sin\theta,\\
z=\rho\cos\varphi.
\end{cases}
\]
则
\[ds^2=d\rho^2+\rho^2d\varphi^2+\rho^2\sin^2\varphi d\theta^2.\]
1
对于柱坐标系, 是很好验证的,
\[
\begin{cases}
dx=d(\rho\cos\theta)=\cos\theta d\rho-\rho\sin\theta d\theta,\\
dy=d(\rho\sin\theta)=\sin\theta d\rho+\rho\cos\theta d\theta,\\
dz=dz.
\end{cases}
\]
因此
\[
\begin{split}
dx^2+dy^2&=\cos^2\theta d\rho^2-2\rho\sin\theta\cos\theta d\rho d\theta+\rho^2\sin^2\theta d\theta^2\\
&=\sin^2\theta d\rho^2+2\rho\sin\theta\cos\theta d\rho d\theta+\rho^2\cos^2\theta d\theta^2\\
&=d\rho^2+\rho^2 d\theta^2.
\end{split}
\]
2
对于球坐标系,
\[
\begin{cases}
dx=d(\rho\sin\varphi\cos\theta)&=\sin\varphi\cos\theta d\rho+\rho\cos\varphi\cos\theta d\varphi-\rho\sin\varphi\sin\theta d\theta,\\
dy=d(\rho\sin\varphi\sin\theta)&=\sin\varphi\sin\theta d\rho+\rho\cos\varphi\sin\theta d\varphi+\rho\sin\varphi\cos\theta d\theta,\\
dz=d(\rho\cos\varphi)&=\cos\varphi d\rho-\rho\sin\varphi d\varphi.
\end{cases}
\]
于是
\[
\begin{split}
dx^2+dy^2+dz^2&=\sin^2\varphi\cos^2\theta d\rho^2+\rho^2\cos^2\varphi\cos^2\theta d\varphi^2+\rho^2\sin^2\varphi\sin^2\theta d\theta^2\\
&\quad +2\rho\sin\varphi\cos\varphi\cos^2\theta d\rho d\varphi-2\rho\sin^2\varphi\sin\theta\cos\theta d\rho d\theta-2\rho^2\sin\varphi\cos\varphi\sin\theta\cos\theta d\varphi d\theta\\
&\ +\sin^2\varphi\sin^2\theta d\rho^2+\rho^2\cos^2\varphi\sin^2\theta d\varphi^2+\rho^2\sin^2\varphi\cos^2\theta d\theta^2\\
&\quad +2\rho\sin\varphi\cos\varphi\sin^2\theta d\rho d\varphi+2\rho\sin^2\varphi\sin\theta\cos\theta d\rho d\theta+2\rho^2\sin\varphi\cos\varphi\sin\theta\cos\theta d\varphi d\theta\\
&=d\rho^2+\rho^2 d\varphi^2+\rho^2\sin^2\varphi d\theta^2.
\end{split}
\]