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问题及解答

计算柱坐标, 球坐标变换的Jacobi

Posted by haifeng on 2012-06-04 14:21:08 last update 2014-08-04 20:02:34 | Edit | Answers (0)

(1) 柱坐标变换

\[\begin{cases}x=\rho\cos\theta\\ y=\rho\sin\theta\\ z=z\end{cases}\]

\[J=\frac{\partial(x,y,z)}{\partial(\rho,\theta,z)}=\begin{vmatrix}x'_\rho & x'_\theta & x'_z\\ y'_\rho & y'_\theta & y'_z\\ z'_\rho & z'_\theta & z'_z\end{vmatrix}=\begin{vmatrix}\cos\theta & -\rho\sin\theta & 0\\ \sin\theta &\rho\cos\theta & 0\\ 0 & 0 & 1\end{vmatrix}=\rho\]

如果令 $x=\rho\sin\theta$, $y=\rho\cos\theta$, 则算得的 $J=-\rho$.


(2) 球坐标变换

\[\begin{cases}x=r\sin\varphi\cos\theta\\ y=r\sin\varphi\sin\theta\\ z=r\cos\varphi\end{cases}\]

其中 $\varphi$ 是向量 $(x,y,z)$ 与 $z$ 轴正方向的夹角.

\[J=\frac{\partial(x,y,z)}{\partial(r,\varphi,\theta)}=\begin{vmatrix}x'_r & x'_\varphi & x'_\theta\\ y'_r & y'_\varphi & y'_\theta\\z'_r & z'_\varphi & z'_\theta \end{vmatrix}=\begin{vmatrix}\sin\varphi\cos\theta & r\cos\varphi\cos\theta & -r\sin\varphi\sin\theta\\ \sin\varphi\sin\theta &r\cos\varphi\sin\theta & r\sin\varphi\cos\theta\\ \cos\varphi & -r\sin\varphi & 0\end{vmatrix}=r^2\sin\varphi\]

若如下令

\[\begin{cases}x=r\cos\varphi\cos\theta\\ y=r\cos\varphi\sin\theta\\ z=r\sin\varphi\end{cases}\]

其中 $\varphi$ 是向量 $(x,y,z)$ 与 $xoy$ 平面的夹角.

\[J=\frac{\partial(x,y,z)}{\partial(r,\varphi,\theta)}=\begin{vmatrix}x'_r & x'_\varphi & x'_\theta\\ y'_r & y'_\varphi & y'_\theta\\z'_r & z'_\varphi & z'_\theta \end{vmatrix}=\begin{vmatrix}\cos\varphi\cos\theta & -r\sin\varphi\cos\theta & -r\cos\varphi\sin\theta\\ \cos\varphi\sin\theta &-r\sin\varphi\sin\theta & r\cos\varphi\cos\theta\\ \sin\varphi & r\cos\varphi & 0\end{vmatrix}=-r^2\cos\varphi\]