From polar coordinates

$${\displaystyle {\begin{aligned}x&=r\cos \theta \\y&=r\sin \theta \\[5pt]{\frac {\partial (x,y)}{\partial (r,\theta )}}&={\begin{bmatrix}\cos \theta &-r\sin \theta \\\sin \theta &{\phantom {-}}r\cos \theta \end{bmatrix}}\\[5pt]{\text{Jacobian}}=\det {\frac {\partial (x,y)}{\partial (r,\theta )}}&=r\end{aligned}}}$$

From log-polar coordinates

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

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

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

That is, it is given by the complex exponential function.

From bipolar coordinates

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

From 2-center bipolar coordinates

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

From Cesàro equation

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

From Cartesian coordinates

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

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

$${\displaystyle \theta ={\begin{cases}{\hphantom {2}}\theta '&({\text{for }}\theta '{\text{ in QI: }}{\phantom {IIV}}0<\theta '<{\frac {\pi }{2}})\\{\hphantom {2}}\pi -\theta '&({\text{for }}\theta '{\text{ in QII: }}{\phantom {IV}}{\frac {\pi }{2}}<\theta '<\pi )\\{\hphantom {2}}\pi +\theta '&({\text{for }}\theta '{\text{ in QIII: }}{\phantom {II}}\pi <\theta '<{\frac {3\pi }{2}})\\2\pi -\theta '&({\text{for }}\theta '{\text{ in QIV: }}{\phantom {I}}{\frac {3\pi }{2}}<\theta '<2\pi )\end{cases}}}$$

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 ${\textstyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}$, 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

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

From 2-center bipolar coordinates

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

Where 2c is the distance between the poles.

In Cartesian coordinates

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

In polar coordinates

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

From spherical coordinates

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

So for the volume element:

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

From cylindrical coordinates

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

So for the volume element:

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

From Cartesian coordinates

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

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

So for the element:

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

From cylindrical coordinates

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

From Cartesian coordinates

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

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

From spherical coordinates

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