List of common coordinate transformations

${\displaystyle {\begin{aligned}x&=r\,\cos \theta \\y&=r\,\sin \theta \\{\frac {\partial (x,y)}{\partial (r,\theta )}}&={\begin{pmatrix}\cos \theta &-r\,\sin \theta \\\sin \theta &r\,\cos \theta \end{pmatrix}}\\\det {\frac {\partial (x,y)}{\partial (r,\theta )}}&=r\end{aligned}}}$

${\displaystyle {\begin{aligned}x&=e^{\rho }\cos \theta ,\\y&=e^{\rho }\sin \theta .\end{aligned}}}$

By using complex numbers ${\displaystyle (x,y)=x+iy'}$, the transformation can be written as

${\displaystyle x+iy=e^{\rho +i\theta }}$

I.e., it is given by the complex exponential function.

${\displaystyle {\begin{aligned}x&=a{\frac {\sinh \tau }{\cosh \tau -\cos \sigma }}\\y&=a{\frac {\sin \sigma }{\cosh \tau -\cos \sigma }}\end{aligned}}}$

${\displaystyle {\begin{aligned}x&={\frac {1}{4c}}\left(r_{1}^{2}-r_{2}^{2}\right)\\y&=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-(r_{1}^{2}-r_{2}^{2}+4c^{2})^{2}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}x&=\int \cos \left[\int \kappa (s)\,ds\right]ds\\y&=\int \sin \left[\int \kappa (s)\,ds\right]ds\end{aligned}}}$

${\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta ^{\prime }&=\arctan \left|{\frac {y}{x}}\right|\end{aligned}}}$

Note: solving for ${\displaystyle \theta ^{\prime }}$ returns the resultant angle in the first quadrant (${\displaystyle 0<\theta <{\frac {\pi }{2}}}$). To find ${\displaystyle \theta }$, one must refer to the original Cartesian coordinate, determine the quadrant in which ${\displaystyle \theta }$ lies (ex (3,-3) [Cartesian] lies in QIV), then use the following to solve for ${\displaystyle \theta }$:

• For ${\displaystyle \theta ^{\prime }}$ in QI:

  ${\displaystyle \theta =\theta ^{\prime }}$

• For ${\displaystyle \theta ^{\prime }}$ in QII:

  ${\displaystyle \theta =\pi -\theta ^{\prime }}$

• For ${\displaystyle \theta ^{\prime }}$ in QIII:

  ${\displaystyle \theta =\pi +\theta ^{\prime }}$

• For ${\displaystyle \theta ^{\prime }}$ in QIV:

  ${\displaystyle \theta =2\pi -\theta ^{\prime }}$

The value for ${\displaystyle \theta }$ must be solved for in this manner because for all values of ${\displaystyle \theta }$, ${\displaystyle \tan \theta }$ is only defined for ${\displaystyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}$, and is periodic (with period ${\displaystyle \pi }$). This means that the inverse function will only give values in the domain of the function, but restricted to a single period. Hence, the range of the inverse function is only half a full circle.

Note that one can also use

${\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta ^{\prime }&=2\arctan {\frac {y}{x+r}}\end{aligned}}}$

${\displaystyle {\begin{aligned}r&={\sqrt {\frac {r_{1}^{2}+r_{2}^{2}-2c^{2}}{2}}}\\\theta &=\arctan \left[{\sqrt {{\frac {8c^{2}(r_{1}^{2}+r_{2}^{2}-2c^{2})}{r_{1}^{2}-r_{2}^{2}}}-1}}\right]\end{aligned}}}$

Where 2c is the distance between the poles.

${\displaystyle {\begin{aligned}\rho &=\log {\sqrt {x^{2}+y^{2}}},\\\theta &=\arctan {\frac {y}{x}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\kappa &={\frac {x'y''-y'x''}{({x'}^{2}+{y'}^{2})^{\frac {3}{2}}}}\\s&=\int _{a}^{t}{\sqrt {{x'}^{2}+{y'}^{2}}}\,dt\end{aligned}}}$

${\displaystyle {\begin{aligned}\kappa &={\frac {r^{2}+2{r'}^{2}-rr''}{(r^{2}+{r'}^{2})^{\frac {3}{2}}}}\\s&=\int _{a}^{\phi }{\sqrt {r^{2}+{r'}^{2}}}\,d\phi \end{aligned}}}$

${\displaystyle {\begin{aligned}x&=\rho \,\sin \theta \,\cos \phi \\y&=\rho \,\sin \theta \,\sin \phi \\z&=\rho \,\cos \theta \\{\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\phi )}}&={\begin{pmatrix}\sin \theta \cos \phi &\rho \cos \theta \cos \phi &-\rho \sin \theta \sin \phi \\\sin \theta \sin \phi &\rho \cos \theta \sin \phi &\rho \sin \theta \cos \phi \\\cos \theta &-\rho \sin \theta &0\end{pmatrix}}\end{aligned}}}$

So for the volume element:

${\displaystyle dx\;dy\;dz=\det {\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\phi )}}d\rho \;d\theta \;d\phi =\rho ^{2}\sin \theta \;d\rho \;d\theta \;d\phi }$

${\displaystyle {\begin{aligned}x&=r\,\cos \theta \\y&=r\,\sin \theta \\z&=z\,\\{\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}&={\begin{pmatrix}\cos \theta &-r\sin \theta &0\\\sin \theta &r\cos \theta &0\\0&0&1\end{pmatrix}}\end{aligned}}}$

So for the volume element:

${\displaystyle dV=dx\;dy\;dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}dr\;d\theta \;dz=r\;dr\;d\theta \;dz}$

${\displaystyle {\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arctan \left({\frac {\sqrt {x^{2}+y^{2}}}{z}}\right)=\arccos \left({\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\right)\\\phi &=\arctan \left({\frac {y}{x}}\right)=\arccos \left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)=\arcsin \left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)\\{\frac {\partial \left(\rho ,\theta ,\phi \right)}{\partial \left(x,y,z\right)}}&={\begin{pmatrix}{\frac {x}{\rho }}&{\frac {y}{\rho }}&{\frac {z}{\rho }}\\{\frac {xz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&{\frac {yz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&-{\frac {\sqrt {x^{2}+y^{2}}}{\rho ^{2}}}\\{\frac {-y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}&0\\\end{pmatrix}}\end{aligned}}}$

See also the article on atan2 for how to elegantly handle some edge cases.

So for the element:

${\displaystyle d\rho \ d\theta \ d\phi =\det {\frac {\partial (\rho ,\theta ,\phi )}{\partial (x,y,z)}}dx\ dy\ dz={\frac {1}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}dx\ dy\ dz}$

${\displaystyle {\begin{aligned}\rho &={\sqrt {r^{2}+h^{2}}}\\\theta &=\arctan {\frac {r}{h}}\\\phi &=\phi \\{\frac {\partial (\rho ,\theta ,\phi )}{\partial (r,h,\phi )}}&={\begin{pmatrix}{\frac {r}{\sqrt {r^{2}+h^{2}}}}&{\frac {h}{\sqrt {r^{2}+h^{2}}}}&0\\{\frac {h}{r^{2}+h^{2}}}&{\frac {-r}{r^{2}+h^{2}}}&0\\0&0&1\\\end{pmatrix}}\\\det {\frac {\partial (\rho ,\theta ,\phi )}{\partial (r,h,\phi )}}&={\frac {1}{\sqrt {r^{2}+h^{2}}}}\end{aligned}}}$

${\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta &=\arctan {\left({\frac {y}{x}}\right)}\\z&=z\quad \end{aligned}}}$

${\displaystyle {\frac {\partial (r,\theta ,h)}{\partial (x,y,z)}}={\begin{pmatrix}{\frac {x}{\sqrt {x^{2}+y^{2}}}}&{\frac {y}{\sqrt {x^{2}+y^{2}}}}&0\\{\frac {-y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}&0\\0&0&1\end{pmatrix}}}$

${\displaystyle {\begin{aligned}r&=\rho \sin \phi \\h&=\rho \cos \phi \\\theta &=\theta \\{\frac {\partial (r,h,\theta )}{\partial (\rho ,\phi ,\theta )}}&={\begin{pmatrix}\sin \phi &\rho \cos \phi &0\\\cos \phi &-\rho \sin \phi &0\\0&0&1\\\end{pmatrix}}\\\det {\frac {\partial (r,h,\theta )}{\partial (\rho ,\phi ,\theta )}}&=-\rho \end{aligned}}}$

${\displaystyle {\begin{aligned}s&=\int _{0}^{t}{\sqrt {{x'}^{2}+{y'}^{2}+{z'}^{2}}}\,dt\\[3pt]\kappa &={\frac {\sqrt {(z''y'-y''z')^{2}+(x''z'-z''x')^{2}+(y''x'-x''y')^{2}}}{({x'}^{2}+{y'}^{2}+{z'}^{2})^{\frac {3}{2}}}}\\[3pt]\tau &={\frac {x'''(y'z''-y''z')+y'''(x''z'-x'z'')+z'''(x'y''-x''y')}{{(x'y''-x''y')}^{2}+{(x''z'-x'z'')}^{2}+{(y'z''-y''z')}^{2}}}\end{aligned}}}$