Daftar transformasi koordinat berikut memuat transformasi sistem-sistem koordinat yang paling umum digunakan.
Diketahui (x, y) pada sistem koordinat Kartesius baku, serta r dan θ pada sistem koordinat polar baku.
Dari koordinat polar ke koordinat Kartesius
sunting
x
=
r
cos
θ
{\displaystyle x=r\,\cos \theta \quad }
y
=
r
sin
θ
{\displaystyle y=r\,\sin \theta \quad }
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
(
cos
θ
−
r
sin
θ
sin
θ
r
cos
θ
)
{\displaystyle {\frac {\partial (x,y)}{\partial (r,\theta )}}={\begin{pmatrix}\cos \theta &-r\,\sin \theta \\\sin \theta &r\,\cos \theta \end{pmatrix}}}
det
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
r
{\displaystyle \det {\frac {\partial (x,y)}{\partial (r,\theta )}}=r}
Dari koordinat Kartesius ke koordinat polar
sunting
r
=
x
2
+
y
2
{\displaystyle r={\sqrt {x^{2}+y^{2}}}}
θ
′
=
arctan
|
y
x
|
{\displaystyle \theta ^{\prime }=\arctan \left|{\frac {y}{x}}\right|}
Catatan: penghitungan
θ
′
{\displaystyle \theta ^{\prime }}
0
<
θ
<
π
2
{\displaystyle 0<\theta <{\frac {\pi }{2}}}
θ
{\displaystyle \theta }
θ
{\displaystyle \theta }
θ
{\displaystyle \theta }
Untuk
θ
′
{\displaystyle \theta ^{\prime }}
θ
=
θ
′
{\displaystyle \theta =\theta ^{\prime }}
Untuk
θ
′
{\displaystyle \theta ^{\prime }}
θ
=
π
−
θ
′
{\displaystyle \theta =\pi -\theta ^{\prime }}
Untuk
θ
′
{\displaystyle \theta ^{\prime }}
θ
=
π
+
θ
′
{\displaystyle \theta =\pi +\theta ^{\prime }}
Untuk
θ
′
{\displaystyle \theta ^{\prime }}
θ
=
2
π
−
θ
′
{\displaystyle \theta =2\pi -\theta ^{\prime }}
Nilai
θ
{\displaystyle \theta }
θ
{\displaystyle \theta }
tan
θ
{\displaystyle \tan \theta }
−
π
2
<
θ
<
+
π
2
{\displaystyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}
π
{\displaystyle \pi }
fungsi invers hanya menghasilkan nilai dalam domain fungsi itu, tetapi terbatas pada satu periode saja. Jadi, rentang fungsi invers hanyalah setengah lingkaran.
Perhatikan bahwa dapat pula digunakan
r
=
x
2
+
y
2
{\displaystyle r={\sqrt {x^{2}+y^{2}}}}
θ
′
=
2
arctan
y
x
+
r
{\displaystyle \theta ^{\prime }=2\arctan {\frac {y}{x+r}}}
Dari koordinat log-polar ke koordinat Kartesius
sunting
{
x
=
e
ρ
cos
θ
,
y
=
e
ρ
sin
θ
.
{\displaystyle {\begin{cases}x=e^{\rho }\cos \theta ,\\y=e^{\rho }\sin \theta .\end{cases}}}
Dengan menggunakan bilangan kompleks
(
x
,
y
)
=
x
+
i
y
′
{\displaystyle (x,y)=x+iy'}
x
+
i
y
=
e
ρ
+
i
θ
{\displaystyle x+iy=e^{\rho +i\theta }\,}
yaitu diberikan oleh fungsi eksponensial kompleks.
Dari koordinat Kartesius ke koordinat log-polar
sunting
{
ρ
=
log
x
2
+
y
2
,
θ
=
arctan
y
x
.
{\displaystyle {\begin{cases}\rho =\log {\sqrt {x^{2}+y^{2}}},\\\theta =\arctan {\frac {y}{x}}.\end{cases}}}
Dari koordinat bipolar ke koordinat Kartesius
sunting
x
=
a
sinh
τ
cosh
τ
−
cos
σ
{\displaystyle x=a\ {\frac {\sinh \tau }{\cosh \tau -\cos \sigma }}}
y
=
a
sin
σ
cosh
τ
−
cos
σ
{\displaystyle y=a\ {\frac {\sin \sigma }{\cosh \tau -\cos \sigma }}}
Dari koordinat two-center bipolar ke koordinat Kartesius
sunting
x
=
r
1
2
−
r
2
2
4
c
{\displaystyle x={\frac {r_{1}^{2}-r_{2}^{2}}{4c}}}
y
=
±
1
4
c
16
c
2
r
1
2
−
(
r
1
2
−
r
2
2
+
4
c
2
)
2
{\displaystyle y=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-(r_{1}^{2}-r_{2}^{2}+4c^{2})^{2}}}}
[ 1] Dari koordinat two-center bipolar ke koordinat polar
sunting
r
=
r
1
2
+
r
2
2
−
2
c
2
2
{\displaystyle r={\sqrt {\frac {r_{1}^{2}+r_{2}^{2}-2c^{2}}{2}}}}
θ
=
arctan
[
8
c
2
(
r
1
2
+
r
2
2
−
2
c
2
)
r
1
2
−
r
2
2
−
1
]
{\displaystyle \theta =\arctan \left[{\sqrt {{\frac {8c^{2}(r_{1}^{2}+r_{2}^{2}-2c^{2})}{r_{1}^{2}-r_{2}^{2}}}-1}}\right]}
di mana 2c adalah jarak antara kutub-kutub.
sunting
x
=
∫
cos
[
∫
κ
(
s
)
d
s
]
d
s
{\displaystyle x=\int \cos \left[\int \kappa (s)\,ds\right]ds}
y
=
∫
sin
[
∫
κ
(
s
)
d
s
]
d
s
{\displaystyle y=\int \sin \left[\int \kappa (s)\,ds\right]ds}
Dari koordinat Kartesius ke panjang dan kurva Arc
sunting
κ
=
x
′
y
″
−
y
′
x
″
(
x
′
2
+
y
′
2
)
3
/
2
{\displaystyle \kappa ={\frac {x'y''-y'x''}{(x'^{2}+y'^{2})^{3/2}}}}
s
=
∫
a
t
x
′
2
+
y
′
2
d
t
{\displaystyle s=\int _{a}^{t}{\sqrt {x'^{2}+y'^{2}}}\,dt}
Dari koordinat polar ke panjang dan kurva Arc
sunting
κ
=
r
2
+
2
r
′
2
−
r
r
″
(
r
2
+
r
′
2
)
3
/
2
{\displaystyle \kappa ={\frac {r^{2}+2r'^{2}-rr''}{(r^{2}+r'^{2})^{3/2}}}}
s
=
∫
a
ϕ
1
+
y
′
2
d
ϕ
{\displaystyle s=\int _{a}^{\phi }{\sqrt {1+y'^{2}}}\,d\phi }
Diketahui (x, y, z) pada sistem koordinat Kartesius baku, dan (ρ, θ, φ) pada koordinat spherical , dengan sudut θ diukur dari aksis +Z axis. Sebagaimana φ mempunyai rentang 360° pertimbangan yang sama dengan koordinat polar (2 dimensi) diterapkan bilamana diambil suatu arctangen. θ mempunyai rentang 180°, dari 0° ke 180°, dan tidak bermasalah jika dihitung dari suatu arckosinus, tetapi perhatikan untuk suatu arctangen.
Jika, dalam definisi alternatif, θ dipilih untuk rentang dari −90° ke +90°, dengan arah yang berlawanan dibandingkan definisi sebelumnya, maka dapat dihitung secara unik dari suatu arcsinus, tetapi hati-hati dengan arckotangen. Dalam kasus ini semua rumus berikut semua argumen θ harus ditukar sinus dan kosinus-nya, dan sebagai turunan juga harus ditukar tanda plus dan minusnya.
Semua pembagian oleh nol menghasilkan kasus-kasus khusus dengan arah di sepanjang aksis-aksis utama dan dalam praktik dapat dipecahkan dengan mudah melalui observasi.
x
=
ρ
sin
θ
cos
ϕ
{\displaystyle {x}=\rho \,\sin \theta \,\cos \phi \quad }
y
=
ρ
sin
θ
sin
ϕ
{\displaystyle {y}=\rho \,\sin \theta \,\sin \phi \quad }
z
=
ρ
cos
θ
{\displaystyle {z}=\rho \,\cos \theta \quad }
∂
(
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 {\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}}}
sehingga untuk volume elemen:
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 ,\phi )}}d\rho \;d\theta \;d\phi =\rho ^{2}\sin \theta \;d\rho \;d\theta \;d\phi \;}
x
=
r
cos
θ
{\displaystyle {x}={r}\,\cos \theta }
y
=
r
sin
θ
{\displaystyle {y}={r}\,\sin \theta }
z
=
h
{\displaystyle {z}={h}\,}
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
h
)
=
(
cos
θ
−
r
sin
θ
0
sin
θ
r
cos
θ
0
0
0
1
)
{\displaystyle {\frac {\partial (x,y,z)}{\partial (r,\theta ,h)}}={\begin{pmatrix}\cos \theta &-r\sin \theta &0\\\sin \theta &r\cos \theta &0\\0&0&1\end{pmatrix}}}
sehingga untuk volume elemen:
d
x
d
y
d
z
=
det
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
h
)
d
r
d
θ
d
h
=
r
d
r
d
θ
d
h
{\displaystyle dx\;dy\;dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,h)}}dr\;d\theta \;dh={r}\;dr\;d\theta \;dh\;}
ρ
=
x
2
+
y
2
+
z
2
{\displaystyle {\rho }={\sqrt {x^{2}+y^{2}+z^{2}}}}
ϕ
=
arctan
(
y
x
)
=
arccos
(
x
x
2
+
y
2
)
=
arcsin
(
y
x
2
+
y
2
)
{\displaystyle {\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)}
θ
=
arctan
(
x
2
+
y
2
z
)
=
arccos
(
z
x
2
+
y
2
+
z
2
)
{\displaystyle {\theta }=\arctan \left({\frac {\sqrt {x^{2}+y^{2}}}{z}}\right)=\arccos \left({\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\right)}
∂
(
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 {\frac {\partial (\rho ,\theta ,\phi )}{\partial (x,y,z)}}={\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}}}
sehingga untuk volume elemen:
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\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}
ρ
=
r
2
+
h
2
{\displaystyle {\rho }={\sqrt {r^{2}+h^{2}}}}
ϕ
=
ϕ
{\displaystyle {\phi }=\phi \quad }
θ
=
arctan
r
h
{\displaystyle {\theta }=\arctan {\frac {r}{h}}}
∂
(
ρ
,
θ
,
ϕ
)
∂
(
r
,
ϕ
,
h
)
=
(
r
r
2
+
h
2
0
h
r
2
+
h
2
−
r
r
2
+
h
2
0
h
r
2
+
h
2
0
1
0
)
{\displaystyle {\frac {\partial (\rho ,\theta ,\phi )}{\partial (r,\phi ,h)}}={\begin{pmatrix}{\frac {r}{\sqrt {r^{2}+h^{2}}}}&0&{\frac {h}{\sqrt {r^{2}+h^{2}}}}\\{\frac {-r}{r^{2}+h^{2}}}&0&{\frac {h}{r^{2}+h^{2}}}\\0&1&0\end{pmatrix}}}
det
∂
(
ρ
,
θ
,
ϕ
)
∂
(
r
,
ϕ
,
h
)
=
1
r
2
+
h
2
{\displaystyle \det {\frac {\partial (\rho ,\theta ,\phi )}{\partial (r,\phi ,h)}}={\frac {1}{\sqrt {r^{2}+h^{2}}}}}
r
=
x
2
+
y
2
{\displaystyle r={\sqrt {x^{2}+y^{2}}}}
θ
=
{
0
if
x
=
0
and
y
=
0
arcsin
(
y
r
)
if
x
≥
0
−
arcsin
(
y
r
)
+
π
if
x
<
0
{\displaystyle \theta ={\begin{cases}0&{\mbox{if }}x=0{\mbox{ and }}y=0\\\arcsin({\frac {y}{r}})&{\mbox{if }}x\geq 0\\-\arcsin({\frac {y}{r}})+\pi &{\mbox{if }}x<0\\\end{cases}}}
h
=
z
{\displaystyle h=z\quad }
Note that many computer systems may offer a more concise function for computing
θ
{\displaystyle \theta }
atan2 (y,x)
∂
(
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}}}
r
=
ρ
sin
ϕ
{\displaystyle r=\rho \sin \phi \,}
θ
=
θ
{\displaystyle \theta =\theta \,}
h
=
ρ
cos
ϕ
{\displaystyle h=\rho \cos \phi \,}
∂
(
r
,
θ
,
h
)
∂
(
ρ
,
θ
,
ϕ
)
=
(
sin
ϕ
0
ρ
cos
ϕ
0
1
0
cos
ϕ
0
−
ρ
sin
ϕ
)
{\displaystyle {\frac {\partial (r,\theta ,h)}{\partial (\rho ,\theta ,\phi )}}={\begin{pmatrix}\sin \phi &0&\rho \cos \phi \\0&1&0\\\cos \phi &0&-\rho \sin \phi \end{pmatrix}}}
det
∂
(
r
,
θ
,
h
)
∂
(
ρ
,
θ
,
ϕ
)
=
−
ρ
{\displaystyle \det {\frac {\partial (r,\theta ,h)}{\partial (\rho ,\theta ,\phi )}}=-\rho }
sunting
s
=
∫
0
t
x
′
2
+
y
′
2
+
z
′
2
d
t
{\displaystyle s=\int _{0}^{t}{\sqrt {x'^{2}+y'^{2}+z'^{2}}}\,dt}
κ
=
(
z
″
y
′
−
y
″
z
′
)
2
+
(
x
″
z
′
−
z
″
x
′
)
2
+
(
y
″
x
′
−
x
″
y
′
)
2
(
x
′
2
+
y
′
2
+
z
′
2
)
3
/
2
{\displaystyle \kappa ={\frac {\sqrt {(z''y'-y''z')^{2}+(x''z'-z''x')^{2}+(y''x'-x''y')^{2}}}{(x'^{2}+y'^{2}+z'^{2})^{3/2}}}}
τ
=
z
‴
(
x
′
y
″
−
y
′
x
″
)
+
z
″
(
x
‴
y
′
−
x
′
y
‴
)
+
z
′
(
x
″
y
‴
−
x
‴
y
″
)
(
x
′
2
+
y
′
2
+
z
′
2
)
(
x
″
2
+
y
″
2
+
z
″
2
)
{\displaystyle \tau ={\frac {z'''(x'y''-y'x'')+z''(x'''y'-x'y''')+z'(x''y'''-x'''y'')}{(x'^{2}+y'^{2}+z'^{2})(x''^{2}+y''^{2}+z''^{2})}}}
^ Weisstein, Eric W.. "Bipolar Coordinates." Treasure Troves . 26 May 1999. Sociology and Anthropology China. 14 February 2007 [1] Diarsipkan 2007-12-12 di Wayback Machine .
![{\displaystyle \theta =\arctan \left[{\sqrt {{\frac {8c^{2}(r_{1}^{2}+r_{2}^{2}-2c^{2})}{r_{1}^{2}-r_{2}^{2}}}-1}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/010a0c267d31acc3d2a743ad7a94ee3e2efaacf9)
![{\displaystyle x=\int \cos \left[\int \kappa (s)\,ds\right]ds}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02cff907588352a0c15c8e8ecb7e972c745f378f)
![{\displaystyle y=\int \sin \left[\int \kappa (s)\,ds\right]ds}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71982132e04a316224eeb7859edb3ae157a2f402)
