Del in cylindrical and spherical coordinates

Del in cylindrical and spherical coordinates

This is a list of some vector calculus formulae for working with common curvilinear coordinate systems.

Coordinate conversions

Del in cylindrical and spherical coordinates

<table><thead><tr><th></th><th>From</th></tr></thead><tbody><tr><td>Cartesian</td><td>Cylindrical</td><td>Spherical</td></tr><tr><td>To</td><td>Cartesian</td><td>{\displaystyle {\begin{aligned}x&=x\\y&=y\\z&=z\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}x&=\rho \cos \varphi \\y&=\rho \sin \varphi \\z&=z\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}x&=r\sin \theta \cos \varphi \\y&=r\sin \theta \sin \varphi \\z&=r\cos \theta \\\end{aligned}}}</td></tr><tr><td>Cylindrical</td><td>{\displaystyle {\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}}}\\\varphi &=\arctan \left({\frac {y}{x}}\right)\\z&=z\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}\rho &=\rho \\\varphi &=\varphi \\z&=z\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}\rho &=r\sin \theta \\\varphi &=\varphi \\z&=r\cos \theta \end{aligned}}}</td></tr><tr><td>Spherical</td><td>{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arccos \left({\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\right)\\\varphi &=\arctan \left({\frac {y}{x}}\right)\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}r&={\sqrt {\rho ^{2}+z^{2}}}\\\theta &=\arctan {\left({\frac {\rho }{z}}\right)}\\\varphi &=\varphi \end{aligned}}}</td><td>{\displaystyle {\begin{aligned}r&=r\\\theta &=\theta \\\varphi &=\varphi \end{aligned}}}</td></tr></tbody></table>

Note that the operation ${\displaystyle \arctan \left({\frac {A}{B}}\right)}$ must be interpreted as the two-argument inverse tangent, atan2.

Unit vector conversions

Del in cylindrical and spherical coordinates

<table><thead><tr><th></th><th>Cartesian</th><th>Cylindrical</th><th>Spherical</th></tr></thead><tbody><tr><td>Cartesian</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&={\hat {\mathbf {x} }}\\{\hat {\mathbf {y} }}&={\hat {\mathbf {y} }}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&=\cos \varphi {\hat {\boldsymbol {\rho }}}-\sin \varphi {\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {y} }}&=\sin \varphi {\hat {\boldsymbol {\rho }}}+\cos \varphi {\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&=\sin \theta \cos \varphi {\hat {\mathbf {r} }}+\cos \theta \cos \varphi {\hat {\boldsymbol {\theta }}}-\sin \varphi {\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {y} }}&=\sin \theta \sin \varphi {\hat {\mathbf {r} }}+\cos \theta \sin \varphi {\hat {\boldsymbol {\theta }}}+\cos \varphi {\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&=\cos \theta {\hat {\mathbf {r} }}-\sin \theta {\hat {\boldsymbol {\theta }}}\end{aligned}}}</td></tr><tr><td>Cylindrical</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&={\frac {x{\hat {\mathbf {x} }}+y{\hat {\mathbf {y} }}}{\sqrt {x^{2}+y^{2}}}}\\{\hat {\boldsymbol {\varphi }}}&={\frac {-y{\hat {\mathbf {x} }}+x{\hat {\mathbf {y} }}}{\sqrt {x^{2}+y^{2}}}}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&={\hat {\boldsymbol {\rho }}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&=\sin \theta {\hat {\mathbf {r} }}+\cos \theta {\hat {\boldsymbol {\theta }}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&=\cos \theta {\hat {\mathbf {r} }}-\sin \theta {\hat {\boldsymbol {\theta }}}\end{aligned}}}</td></tr><tr><td>Spherical</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&={\frac {x{\hat {\mathbf {x} }}+y{\hat {\mathbf {y} }}+z{\hat {\mathbf {z} }}}{\sqrt {x^{2}+y^{2}+z^{2}}}}\\{\hat {\boldsymbol {\theta }}}&={\frac {\left(x{\hat {\mathbf {x} }}+y{\hat {\mathbf {y} }}\right)z-\left(x^{2}+y^{2}\right){\hat {\mathbf {z} }}}{{\sqrt {x^{2}+y^{2}+z^{2}}}{\sqrt {x^{2}+y^{2}}}}}\\{\hat {\boldsymbol {\varphi }}}&={\frac {-y{\hat {\mathbf {x} }}+x{\hat {\mathbf {y} }}}{\sqrt {x^{2}+y^{2}}}}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&={\frac {\rho {\hat {\boldsymbol {\rho }}}+z{\hat {\mathbf {z} }}}{\sqrt {\rho ^{2}+z^{2}}}}\\{\hat {\boldsymbol {\theta }}}&={\frac {z{\hat {\boldsymbol {\rho }}}-\rho {\hat {\mathbf {z} }}}{\sqrt {\rho ^{2}+z^{2}}}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&={\hat {\mathbf {r} }}\\{\hat {\boldsymbol {\theta }}}&={\hat {\boldsymbol {\theta }}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\\end{aligned}}}</td></tr></tbody></table>

<table><thead><tr><th></th><th>Cartesian</th><th>Cylindrical</th><th>Spherical</th></tr></thead><tbody><tr><td>Cartesian</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&={\hat {\mathbf {x} }}\\{\hat {\mathbf {y} }}&={\hat {\mathbf {y} }}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&={\frac {x{\hat {\boldsymbol {\rho }}}-y{\hat {\boldsymbol {\varphi }}}}{\sqrt {x^{2}+y^{2}}}}\\{\hat {\mathbf {y} }}&={\frac {y{\hat {\boldsymbol {\rho }}}+x{\hat {\boldsymbol {\varphi }}}}{\sqrt {x^{2}+y^{2}}}}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {x} }}&={\frac {x\left({\sqrt {x^{2}+y^{2}}}{\hat {\mathbf {r} }}+z{\hat {\boldsymbol {\theta }}}\right)-y{\sqrt {x^{2}+y^{2}+z^{2}}}{\hat {\boldsymbol {\varphi }}}}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}\\{\hat {\mathbf {y} }}&={\frac {y\left({\sqrt {x^{2}+y^{2}}}{\hat {\mathbf {r} }}+z{\hat {\boldsymbol {\theta }}}\right)+x{\sqrt {x^{2}+y^{2}+z^{2}}}{\hat {\boldsymbol {\varphi }}}}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}\\{\hat {\mathbf {z} }}&={\frac {z{\hat {\mathbf {r} }}-{\sqrt {x^{2}+y^{2}}}{\hat {\boldsymbol {\theta }}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}\end{aligned}}}</td></tr><tr><td>Cylindrical</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&=\cos \varphi {\hat {\mathbf {x} }}+\sin \varphi {\hat {\mathbf {y} }}\\{\hat {\boldsymbol {\varphi }}}&=-\sin \varphi {\hat {\mathbf {x} }}+\cos \varphi {\hat {\mathbf {y} }}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&={\hat {\boldsymbol {\rho }}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&={\hat {\mathbf {z} }}\\\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\boldsymbol {\rho }}}&={\frac {\rho {\hat {\mathbf {r} }}+z{\hat {\boldsymbol {\theta }}}}{\sqrt {\rho ^{2}+z^{2}}}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\{\hat {\mathbf {z} }}&={\frac {z{\hat {\mathbf {r} }}-\rho {\hat {\boldsymbol {\theta }}}}{\sqrt {\rho ^{2}+z^{2}}}}\end{aligned}}}</td></tr><tr><td>Spherical</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&=\sin \theta \left(\cos \varphi {\hat {\mathbf {x} }}+\sin \varphi {\hat {\mathbf {y} }}\right)+\cos \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\theta }}}&=\cos \theta \left(\cos \varphi {\hat {\mathbf {x} }}+\sin \varphi {\hat {\mathbf {y} }}\right)-\sin \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\varphi }}}&=-\sin \varphi {\hat {\mathbf {x} }}+\cos \varphi {\hat {\mathbf {y} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&=\sin \theta {\hat {\boldsymbol {\rho }}}+\cos \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\theta }}}&=\cos \theta {\hat {\boldsymbol {\rho }}}-\sin \theta {\hat {\mathbf {z} }}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}{\hat {\mathbf {r} }}&={\hat {\mathbf {r} }}\\{\hat {\boldsymbol {\theta }}}&={\hat {\boldsymbol {\theta }}}\\{\hat {\boldsymbol {\varphi }}}&={\hat {\boldsymbol {\varphi }}}\\\end{aligned}}}</td></tr></tbody></table>

Del formula

Del in cylindrical and spherical coordinates

$${\displaystyle A{x}{\hat {\mathbf {x} }}+A{y}{\hat {\mathbf {y} }}+A{z}{\hat {\mathbf {z} }}}$$

Differential elements

<table><thead><tr><th>Operation</th><th>Cartesian coordinates (x, y, z)</th><th>Cylindrical coordinates (ρ, φ, z)</th><th>Spherical coordinatesα (r, θ, φ)</th></tr></thead><tbody><tr><td>Differential displacement dℓ</td><td>{\displaystyle dx\,{\hat {\mathbf {x} }}+dy\,{\hat {\mathbf {y} }}+dz\,{\hat {\mathbf {z} }}}</td><td>{\displaystyle d\rho \,{\hat {\boldsymbol {\rho }}}+\rho \,d\varphi \,{\hat {\boldsymbol {\varphi }}}+dz\,{\hat {\mathbf {z} }}}</td><td>{\displaystyle dr\,{\hat {\mathbf {r} }}+r\,d\theta \,{\hat {\boldsymbol {\theta }}}+r\,\sin \theta \,d\varphi \,{\hat {\boldsymbol {\varphi }}}}</td></tr><tr><td>Differential normal area dS</td><td>{\displaystyle {\begin{aligned}dy\,dz&\,{\hat {\mathbf {x} }}\\{}+dx\,dz&\,{\hat {\mathbf {y} }}\\{}+dx\,dy&\,{\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}\rho \,d\varphi \,dz&\,{\hat {\boldsymbol {\rho }}}\\{}+d\rho \,dz&\,{\hat {\boldsymbol {\varphi }}}\\{}+\rho \,d\rho \,d\varphi &\,{\hat {\mathbf {z} }}\end{aligned}}}</td><td>{\displaystyle {\begin{aligned}r^{2}\sin \theta \,d\theta \,d\varphi &\,{\hat {\mathbf {r} }}\\{}+r\sin \theta \,dr\,d\varphi &\,{\hat {\boldsymbol {\theta }}}\\{}+r\,dr\,d\theta &\,{\hat {\boldsymbol {\varphi }}}\end{aligned}}}</td></tr><tr><td>Differential volume dV</td><td>{\displaystyle dx\,dy\,dz}</td><td>{\displaystyle \rho \,d\rho \,d\varphi \,dz}</td><td>{\displaystyle r^{2}\sin \theta \,dr\,d\theta \,d\varphi }</td></tr></tbody></table>

Calculation rules

• ${\displaystyle \operatorname {div} \,\operatorname {grad} f\equiv \nabla \cdot \nabla f\equiv \nabla ^{2}f}$ • ${\displaystyle \operatorname {curl} \,\operatorname {grad} f\equiv \nabla \times \nabla f=\mathbf {0} }$ • ${\displaystyle \operatorname {div} \,\operatorname {curl} \mathbf {A} \equiv \nabla \cdot (\nabla \times \mathbf {A} )=0}$ • ${\displaystyle \operatorname {curl} \,\operatorname {curl} \mathbf {A} \equiv \nabla \times (\nabla \times \mathbf {A} )=\nabla (\nabla \cdot \mathbf {A} )-\nabla ^{2}\mathbf {A} }$ (Lagrange's formula for del) • ${\displaystyle \nabla ^{2}(fg)=f\nabla ^{2}g+2\nabla f\cdot \nabla g+g\nabla ^{2}f}$ • ${\displaystyle \nabla ^{2}\left(\mathbf {P} \cdot \mathbf {Q} \right)=\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} -\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} +2\nabla \cdot \left[\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} \right]\quad }$ (From )

Cartesian derivation

Del in cylindrical and spherical coordinates

${\displaystyle {\begin{aligned}\operatorname {div} \mathbf {A} &=\lim {V\to 0}{\frac {\iint {\partial V}\mathbf {A} \cdot d\mathbf {S} }{\iiint {V}dV}}\\[1ex]&={\frac {\left[A{x}(x{+}dx)-A{x}(x)\right]dy\,dz+\left[A{y}(y{+}dy)-A{y}(y)\right]dx\,dz+\left[A{z}(z{+}dz)-A{z}(z)\right]dx\,dy}{dx\,dy\,dz}}\\&={\frac {\partial A{x}}{\partial x}}+{\frac {\partial A{y}}{\partial y}}+{\frac {\partial A{z}}{\partial z}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){x}&=\lim {S^{\perp \mathbf {\hat {x}} }\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d\mathbf {\ell } }{\iint {S}dS}}\\[1ex]&={\frac {\left[A{z}(y{+}dy)-A{z}(y)\right]dz-\left[A{y}(z{+}dz)-A{y}(z)\right]dy}{dy\,dz}}\\&={\frac {\partial A{z}}{\partial y}}-{\frac {\partial A{y}}{\partial z}}\end{aligned}}}$

The expressions for ${\displaystyle (\operatorname {curl} \mathbf {A} ){y}}$ and ${\displaystyle (\operatorname {curl} \mathbf {A} ){z}}$ are found in the same way.

Cylindrical derivation

Del in cylindrical and spherical coordinates

${\displaystyle {\begin{aligned}\operatorname {div} \mathbf {A} &=\lim {V\to 0}{\frac {\iint {\partial V}\mathbf {A} \cdot d\mathbf {S} }{\iiint {V}dV}}\\&={\frac {\left[A{\rho }(\rho {+}d\rho )(\rho {+}d\rho )-A{\rho }(\rho )\rho \right]d\phi \,dz+\left[A{\phi }(\phi {+}d\phi )-A{\phi }(\phi )\right]d\rho \,dz+\left[A{z}(z{+}dz)-A{z}(z)\right]d\rho (\rho {+}d\rho /2)\,d\phi }{\rho \,d\phi \,d\rho \,dz}}\\&={\frac {1}{\rho }}{\frac {\partial (\rho A{\rho })}{\partial \rho }}+{\frac {1}{\rho }}{\frac {\partial A{\phi }}{\partial \phi }}+{\frac {\partial A{z}}{\partial z}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){\rho }&=\lim {S^{\perp {\hat {\boldsymbol {\rho }}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d{\boldsymbol {\ell }}}{\iint {S}dS}}\\[1ex]&={\frac {A{\phi }(z)\left(\rho +d\rho \right)\,d\phi -A{\phi }(z+dz)\left(\rho +d\rho \right)\,d\phi +A{z}(\phi +d\phi )\,dz-A{z}(\phi )\,dz}{\left(\rho +d\rho \right)\,d\phi \,dz}}\\[1ex]&=-{\frac {\partial A{\phi }}{\partial z}}+{\frac {1}{\rho }}{\frac {\partial A{z}}{\partial \phi }}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){\phi }&=\lim {S^{\perp {\boldsymbol {\hat {\phi }}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d{\boldsymbol {\ell }}}{\iint {S}dS}}\\&={\frac {A{z}(\rho )\,dz-A{z}(\rho +d\rho )\,dz+A{\rho }(z+dz)\,d\rho -A{\rho }(z)\,d\rho }{d\rho \,dz}}\\&=-{\frac {\partial A{z}}{\partial \rho }}+{\frac {\partial A{\rho }}{\partial z}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){z}&=\lim {S^{\perp {\hat {\boldsymbol {z}}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d\mathbf {\ell } }{\iint {S}dS}}\\[1ex]&={\frac {A{\rho }(\phi )\,d\rho -A{\rho }(\phi +d\phi )\,d\rho +A{\phi }(\rho +d\rho )(\rho +d\rho )\,d\phi -A{\phi }(\rho )\rho \,d\phi }{\rho \,d\rho \,d\phi }}\\[1ex]&=-{\frac {1}{\rho }}{\frac {\partial A{\rho }}{\partial \phi }}+{\frac {1}{\rho }}{\frac {\partial (\rho A{\phi })}{\partial \rho }}\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {curl} \mathbf {A} &=(\operatorname {curl} \mathbf {A} ){\rho }{\hat {\boldsymbol {\rho }}}+(\operatorname {curl} \mathbf {A} ){\phi }{\hat {\boldsymbol {\phi }}}+(\operatorname {curl} \mathbf {A} ){z}{\hat {\boldsymbol {z}}}\\[1ex]&=\left({\frac {1}{\rho }}{\frac {\partial A{z}}{\partial \phi }}-{\frac {\partial A{\phi }}{\partial z}}\right){\hat {\boldsymbol {\rho }}}+\left({\frac {\partial A{\rho }}{\partial z}}-{\frac {\partial A{z}}{\partial \rho }}\right){\hat {\boldsymbol {\phi }}}+{\frac {1}{\rho }}\left({\frac {\partial (\rho A{\phi })}{\partial \rho }}-{\frac {\partial A{\rho }}{\partial \phi }}\right){\hat {\boldsymbol {z}}}\end{aligned}}}$

Spherical derivation

Del in cylindrical and spherical coordinates

${\displaystyle {\begin{aligned}\operatorname {div} \mathbf {A} &=\lim {V\to 0}{\frac {\iint {\partial V}\mathbf {A} \cdot d\mathbf {S} }{\iiint {V}dV}}\\&={\frac {\left[A{r}(r{+}dr)(r{+}dr)^{2}-A{r}(r)r^{2}\right]\sin \theta \,d\theta \,d\phi +\left[A{\theta }(\theta {+}d\theta )\sin(\theta {+}d\theta )-A{\theta }(\theta )\sin \theta \right]r\,dr\,d\phi +\left[A{\phi }(\phi {+}d\phi )-A{\phi }(\phi )\right]r\,dr\,d\theta }{dr\,r\,d\theta \,r\sin \theta \,d\phi }}\\&={\frac {1}{r^{2}}}{\frac {\partial (r^{2}A{r})}{\partial r}}+{\frac {1}{r\sin \theta }}{\frac {\partial (A{\theta }\sin \theta )}{\partial \theta }}+{\frac {1}{r\sin \theta }}{\frac {\partial A{\phi }}{\partial \phi }}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){r}&=\lim {S^{\perp {\boldsymbol {\hat {r}}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d\mathbf {\ell } }{\iint {S}dS}}\\[1ex]&={\frac {A{\theta }(\phi )r\,d\theta +A{\phi }(\theta +d\theta )r\sin(\theta +d\theta )\,d\phi -A{\theta }(\phi +d\phi )r\,d\theta -A{\phi }(\theta )r\sin(\theta )\,d\phi }{r\,d\theta \,r\sin \theta \,d\phi }}\\&={\frac {1}{r\sin \theta }}{\frac {\partial (A{\phi }\sin \theta )}{\partial \theta }}-{\frac {1}{r\sin \theta }}{\frac {\partial A{\theta }}{\partial \phi }}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){\theta }&=\lim {S^{\perp {\boldsymbol {\hat {\theta }}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d\mathbf {\ell } }{\iint {S}dS}}\\[1ex]&={\frac {A{\phi }(r)r\sin \theta \,d\phi +A{r}(\phi +d\phi )\,dr-A{\phi }(r+dr)(r+dr)\sin \theta \,d\phi -A{r}(\phi )\,dr}{dr\,r\sin \theta \,d\phi }}\\&={\frac {1}{r\sin \theta }}{\frac {\partial A{r}}{\partial \phi }}-{\frac {1}{r}}{\frac {\partial (rA{\phi })}{\partial r}}\end{aligned}}}$

${\displaystyle {\begin{aligned}(\operatorname {curl} \mathbf {A} ){\phi }&=\lim {S^{\perp {\boldsymbol {\hat {\phi }}}}\to 0}{\frac {\int {\partial S}\mathbf {A} \cdot d\mathbf {\ell } }{\iint {S}dS}}\\[1ex]&={\frac {A{r}(\theta )\,dr+A{\theta }(r+dr)(r+dr)\,d\theta -A{r}(\theta +d\theta )\,dr-A{\theta }(r)r\,d\theta }{r\,dr\,d\theta }}\\&={\frac {1}{r}}{\frac {\partial (rA{\theta })}{\partial r}}-{\frac {1}{r}}{\frac {\partial A{r}}{\partial \theta }}\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {curl} \mathbf {A} &=(\operatorname {curl} \mathbf {A} ){r}\,{\hat {\boldsymbol {r}}}+(\operatorname {curl} \mathbf {A} ){\theta }\,{\hat {\boldsymbol {\theta }}}+(\operatorname {curl} \mathbf {A} ){\phi }\,{\hat {\boldsymbol {\phi }}}\\[1ex]&={\frac {1}{r\sin \theta }}\left({\frac {\partial (A{\phi }\sin \theta )}{\partial \theta }}-{\frac {\partial A{\theta }}{\partial \phi }}\right){\hat {\boldsymbol {r}}}+{\frac {1}{r}}\left({\frac {1}{\sin \theta }}{\frac {\partial A{r}}{\partial \phi }}-{\frac {\partial (rA{\phi })}{\partial r}}\right){\hat {\boldsymbol {\theta }}}+{\frac {1}{r}}\left({\frac {\partial (rA{\theta })}{\partial r}}-{\frac {\partial A{r}}{\partial \theta }}\right){\hat {\boldsymbol {\phi }}}\end{aligned}}}$

Unit vector conversion formula

Del in cylindrical and spherical coordinates

The unit vector of a coordinate parameter u is defined in such a way that a small positive change in u causes the position vector ${\displaystyle \mathbf {r} }$ to change in ${\displaystyle \mathbf {u} }$ direction.

Therefore, ${\displaystyle {\frac {\partial {\mathbf {r} }}{\partial u}}={\frac {\partial {s}}{\partial u}}\mathbf {u} }$ where s is the arc length parameter.

For two sets of coordinate systems ${\displaystyle u{i}}$ and ${\displaystyle v{j}}$, according to chain rule, ${\displaystyle {\begin{aligned}d\mathbf {r} &=\sum {i}{\frac {\partial \mathbf {r} }{\partial u{i}}}\,du{i}=\sum {i}{\frac {\partial s}{\partial u{i}}}{\hat {\mathbf {u} }}{i}du{i}=\sum {j}{\frac {\partial s}{\partial v{j}}}{\hat {\mathbf {v} }}{j}\,dv{j}\\&=\sum {j}{\frac {\partial s}{\partial v{j}}}{\hat {\mathbf {v} }}{j}\sum {i}{\frac {\partial v{j}}{\partial u{i}}}\,du{i}\\&=\sum {i}\sum {j}{\frac {\partial s}{\partial v{j}}}{\frac {\partial v{j}}{\partial u{i}}}{\hat {\mathbf {v} }}{j}\,du{i}.\end{aligned}}}$

Now, we isolate the ${\displaystyle i}$th component. For ${\displaystyle i{\neq }k}$, let ${\displaystyle \mathrm {d} u{k}=0}$. Then divide on both sides by ${\displaystyle \mathrm {d} u{i}}$ to get: ${\displaystyle {\frac {\partial s}{\partial u{i}}}{\hat {\mathbf {u} }}{i}=\sum {j}{\frac {\partial s}{\partial v{j}}}{\frac {\partial v{j}}{\partial u{i}}}{\hat {\mathbf {v} }}{j}.}$