Let
(
x
,
y
)
{\displaystyle (x,y)}
Cartesian coordinates , and
(
r
,
θ
)
{\displaystyle (r,\theta )}
polar coordinates .
To Cartesian coordinates
edit
From polar coordinates
edit
x
=
r
cos
θ
y
=
r
sin
θ
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
[
cos
θ
−
r
sin
θ
sin
θ
−
r
cos
θ
]
Jacobian
=
det
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
r
{\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
edit
x
=
e
ρ
cos
θ
,
y
=
e
ρ
sin
θ
.
{\displaystyle {\begin{aligned}x&=e^{\rho }\cos \theta ,\\y&=e^{\rho }\sin \theta .\end{aligned}}}
By using complex numbers
(
x
,
y
)
=
x
+
i
y
′
{\displaystyle (x,y)=x+iy'}
x
+
i
y
=
e
ρ
+
i
θ
{\displaystyle x+iy=e^{\rho +i\theta }}
That is, it is given by the complex exponential function.
From bipolar coordinates
edit
x
=
a
sinh
τ
cosh
τ
−
cos
σ
y
=
a
sin
σ
cosh
τ
−
cos
σ
{\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
edit
x
=
1
4
c
(
r
1
2
−
r
2
2
)
y
=
±
1
4
c
16
c
2
r
1
2
−
(
r
1
2
−
r
2
2
+
4
c
2
)
2
{\displaystyle {\begin{aligned}x&={\frac {1}{4c}}\left(r_{1}^{2}-r_{2}^{2}\right)\\[1ex]y&=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-\left(r_{1}^{2}-r_{2}^{2}+4c^{2}\right)^{2}}}\end{aligned}}}
edit
x
=
∫
cos
[
∫
κ
(
s
)
d
s
]
d
s
y
=
∫
sin
[
∫
κ
(
s
)
d
s
]
d
s
{\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}}}
To polar coordinates
edit
From Cartesian coordinates
edit
r
=
x
2
+
y
2
θ
′
=
arctan
|
y
x
|
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=\arctan \left|{\frac {y}{x}}\right|\end{aligned}}}
θ
′
{\displaystyle \theta '}
0
<
θ
<
π
2
{\textstyle 0<\theta <{\frac {\pi }{2}}}
θ
,
{\displaystyle \theta ,}
θ
{\displaystyle \theta }
θ
:
{\displaystyle \theta :}
θ
=
{
θ
′
for
θ
′
in QI:
0
<
θ
′
<
π
2
π
−
θ
′
for
θ
′
in QII:
π
2
<
θ
′
<
π
π
+
θ
′
for
θ
′
in QIII:
π
<
θ
′
<
3
π
2
2
π
−
θ
′
for
θ
′
in QIV:
3
π
2
<
θ
′
<
2
π
{\displaystyle \theta ={\begin{cases}\theta '&{\text{for }}\theta '{\text{ in QI: }}&0<\theta '<{\frac {\pi }{2}}\\[1.2ex]\pi -\theta '&{\text{for }}\theta '{\text{ in QII: }}&{\frac {\pi }{2}}<\theta '<\pi \\[1.2ex]\pi +\theta '&{\text{for }}\theta '{\text{ in QIII: }}&\pi <\theta '<{\frac {3\pi }{2}}\\[1.2ex]2\pi -\theta '&{\text{for }}\theta '{\text{ in QIV: }}&{\frac {3\pi }{2}}<\theta '<2\pi \end{cases}}}
The value for
θ
{\displaystyle \theta }
θ
{\displaystyle \theta }
tan
θ
{\displaystyle \tan \theta }
−
π
2
<
θ
<
+
π
2
{\textstyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}
π
{\displaystyle \pi }
Note that one can also use
r
=
x
2
+
y
2
θ
′
=
2
arctan
y
x
+
r
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=2\arctan {\frac {y}{x+r}}\end{aligned}}}
From 2-center bipolar coordinates
edit
r
=
r
1
2
+
r
2
2
−
2
c
2
2
θ
=
arctan
[
8
c
2
(
r
1
2
+
r
2
2
−
2
c
2
)
r
1
2
−
r
2
2
−
1
]
{\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.
To log-polar coordinates from Cartesian coordinates
edit
ρ
=
log
x
2
+
y
2
,
θ
=
arctan
y
x
.
{\displaystyle {\begin{aligned}\rho &=\log {\sqrt {x^{2}+y^{2}}},\\\theta &=\arctan {\frac {y}{x}}.\end{aligned}}}
Arc-length and curvature
edit
In Cartesian coordinates
edit
κ
=
x
′
y
″
−
y
′
x
″
(
x
′
2
+
y
′
2
)
3
2
s
=
∫
a
t
x
′
2
+
y
′
2
d
t
{\displaystyle {\begin{aligned}\kappa &={\frac {x'y''-y'x''}{\left({x'}^{2}+{y'}^{2}\right)^{\frac {3}{2}}}}\\s&=\int _{a}^{t}{\sqrt {{x'}^{2}+{y'}^{2}}}\,dt\end{aligned}}}
In polar coordinates
edit
κ
=
r
2
+
2
r
′
2
−
r
r
″
(
r
2
+
r
′
2
)
3
2
s
=
∫
a
φ
r
2
+
r
′
2
d
φ
{\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}}}
Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates , with θ the angle measured away from the +Z axis (as [1] , see conventions in spherical coordinates ). As φ has a range of 360° the same considerations as in polar (2 dimensional) coordinates apply whenever an arctangent of it is taken. θ has a range of 180°, running from 0° to 180°, and does not pose any problem when calculated from an arccosine, but beware for an arctangent.
If, in the alternative definition, θ is chosen to run from −90° to +90°, in opposite direction of the earlier definition, it can be found uniquely from an arcsine, but beware of an arccotangent. In this case in all formulas below all arguments in θ should have sine and cosine exchanged, and as derivative also a plus and minus exchanged.
All divisions by zero result in special cases of being directions along one of the main axes and are in practice most easily solved by observation.
To Cartesian coordinates
edit
From spherical coordinates
edit
x
=
ρ
sin
θ
cos
φ
y
=
ρ
sin
θ
sin
φ
z
=
ρ
cos
θ
∂
(
x
,
y
,
z
)
∂
(
ρ
,
θ
,
φ
)
=
(
sin
θ
cos
φ
ρ
cos
θ
cos
φ
−
ρ
sin
θ
sin
φ
sin
θ
sin
φ
ρ
cos
θ
sin
φ
ρ
sin
θ
cos
φ
cos
θ
−
ρ
sin
θ
0
)
{\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:
d
x
d
y
d
z
=
det
∂
(
x
,
y
,
z
)
∂
(
ρ
,
θ
,
φ
)
d
ρ
d
θ
d
φ
=
ρ
2
sin
θ
d
ρ
d
θ
d
φ
{\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
edit
x
=
r
cos
θ
y
=
r
sin
θ
z
=
z
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
z
)
=
(
cos
θ
−
r
sin
θ
0
sin
θ
r
cos
θ
0
0
0
1
)
{\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:
d
V
=
d
x
d
y
d
z
=
det
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
z
)
d
r
d
θ
d
z
=
r
d
r
d
θ
d
z
{\displaystyle dV=dx\,dy\,dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}\,dr\,d\theta \,dz=r\,dr\,d\theta \,dz}
To spherical coordinates
edit
From Cartesian coordinates
edit
ρ
=
x
2
+
y
2
+
z
2
θ
=
arctan
(
x
2
+
y
2
z
)
=
arccos
(
z
x
2
+
y
2
+
z
2
)
φ
=
arctan
(
y
x
)
=
arccos
(
x
x
2
+
y
2
)
=
arcsin
(
y
x
2
+
y
2
)
∂
(
ρ
,
θ
,
φ
)
∂
(
x
,
y
,
z
)
=
(
x
ρ
y
ρ
z
ρ
x
z
ρ
2
x
2
+
y
2
y
z
ρ
2
x
2
+
y
2
−
x
2
+
y
2
ρ
2
−
y
x
2
+
y
2
x
x
2
+
y
2
0
)
{\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:
d
ρ
d
θ
d
φ
=
det
∂
(
ρ
,
θ
,
φ
)
∂
(
x
,
y
,
z
)
d
x
d
y
d
z
=
1
x
2
+
y
2
x
2
+
y
2
+
z
2
d
x
d
y
d
z
{\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
edit
ρ
=
r
2
+
h
2
θ
=
arctan
r
h
φ
=
φ
∂
(
ρ
,
θ
,
φ
)
∂
(
r
,
h
,
φ
)
=
(
r
r
2
+
h
2
h
r
2
+
h
2
0
h
r
2
+
h
2
−
r
r
2
+
h
2
0
0
0
1
)
det
∂
(
ρ
,
θ
,
φ
)
∂
(
r
,
h
,
φ
)
=
1
r
2
+
h
2
{\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}}}
To cylindrical coordinates
edit
From Cartesian coordinates
edit
r
=
x
2
+
y
2
θ
=
arctan
(
y
x
)
z
=
z
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta &=\arctan {\left({\frac {y}{x}}\right)}\\z&=z\quad \end{aligned}}}
∂
(
r
,
θ
,
h
)
∂
(
x
,
y
,
z
)
=
(
x
x
2
+
y
2
y
x
2
+
y
2
0
−
y
x
2
+
y
2
x
x
2
+
y
2
0
0
0
1
)
{\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
edit
r
=
ρ
sin
φ
h
=
ρ
cos
φ
θ
=
θ
∂
(
r
,
h
,
θ
)
∂
(
ρ
,
φ
,
θ
)
=
(
sin
φ
ρ
cos
φ
0
cos
φ
−
ρ
sin
φ
0
0
0
1
)
det
∂
(
r
,
h
,
θ
)
∂
(
ρ
,
φ
,
θ
)
=
−
ρ
{\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}}}
edit
s
=
∫
0
t
x
′
2
+
y
′
2
+
z
′
2
d
t
κ
=
(
z
″
y
′
−
y
″
z
′
)
2
+
(
x
″
z
′
−
z
″
x
′
)
2
+
(
y
″
x
′
−
x
″
y
′
)
2
(
x
′
2
+
y
′
2
+
z
′
2
)
3
2
τ
=
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
{\displaystyle {\begin{aligned}s&=\int _{0}^{t}{\sqrt {{x'}^{2}+{y'}^{2}+{z'}^{2}}}\,dt\\[3pt]\kappa &={\frac {\sqrt {\left(z''y'-y''z'\right)^{2}+\left(x''z'-z''x'\right)^{2}+\left(y''x'-x''y'\right)^{2}}}{\left({x'}^{2}+{y'}^{2}+{z'}^{2}\right)^{\frac {3}{2}}}}\\[3pt]\tau &={\frac {x'''\left(y'z''-y''z'\right)+y'''\left(x''z'-x'z''\right)+z'''\left(x'y''-x''y'\right)}{{\left(x'y''-x''y'\right)}^{2}+{\left(x''z'-x'z''\right)}^{2}+{\left(y'z''-y''z'\right)}^{2}}}\end{aligned}}}
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9005464a89553d80be52a7b3c648717e7efc509)
![{\displaystyle {\begin{aligned}x&={\frac {1}{4c}}\left(r_{1}^{2}-r_{2}^{2}\right)\\[1ex]y&=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-\left(r_{1}^{2}-r_{2}^{2}+4c^{2}\right)^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/972f7959ed9a099263bb1c1dcdb6e77f925c441a)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cf415cca15fe7faf409efdfa8323225993978f8)
![{\displaystyle \theta ={\begin{cases}\theta '&{\text{for }}\theta '{\text{ in QI: }}&0<\theta '<{\frac {\pi }{2}}\\[1.2ex]\pi -\theta '&{\text{for }}\theta '{\text{ in QII: }}&{\frac {\pi }{2}}<\theta '<\pi \\[1.2ex]\pi +\theta '&{\text{for }}\theta '{\text{ in QIII: }}&\pi <\theta '<{\frac {3\pi }{2}}\\[1.2ex]2\pi -\theta '&{\text{for }}\theta '{\text{ in QIV: }}&{\frac {3\pi }{2}}<\theta '<2\pi \end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c128ed42f71cb67f7323cff563f379284d5e9e4f)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40fd7b0ef6f1fea0e4467981685d16feb513186d)
![{\displaystyle {\begin{aligned}s&=\int _{0}^{t}{\sqrt {{x'}^{2}+{y'}^{2}+{z'}^{2}}}\,dt\\[3pt]\kappa &={\frac {\sqrt {\left(z''y'-y''z'\right)^{2}+\left(x''z'-z''x'\right)^{2}+\left(y''x'-x''y'\right)^{2}}}{\left({x'}^{2}+{y'}^{2}+{z'}^{2}\right)^{\frac {3}{2}}}}\\[3pt]\tau &={\frac {x'''\left(y'z''-y''z'\right)+y'''\left(x''z'-x'z''\right)+z'''\left(x'y''-x''y'\right)}{{\left(x'y''-x''y'\right)}^{2}+{\left(x''z'-x'z''\right)}^{2}+{\left(y'z''-y''z'\right)}^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4956dccc982282eafc49bd83da2ec73d77546d7)
![[1]](https://commons.wikimedia.org/wiki/File:3D_Spherical.svg){kind=link}