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For a right-handed regular system of orthogonal coordinates, the gradient in orthogonal coordinates gives $\nabla u_i={\mathbf{e}}_i/h_i$. Thus $\varphi {\mathbf{e}}_1=(\varphi h_1)\nabla u_1$, and the curl product rule givesAlso $\nabla u_2\times \nabla u_3={\mathbf{e}}_1/(h_2{h}_3)$ and the divergence of this cross product is zero, by the divergence-of-a-cross-product identity and the vanishing curl of each gradient. Applying the divergence product rule to $\varphi {\mathbf{e}}_1=(\varphi h_2{h}_3)(\nabla u_2\times \nabla u_3)$ givesCyclic permutation and addition give the full component formulas for any vector field. This derivation accounts for the varying coordinate basis without differentiating its unit vectors explicitly. The factors $h_i>0$ are scale factors of orthogonal coordinates; the chart must avoid coordinate singularities.

Compare the total differential $d\varphi ={\sum }_i{\partial }_{u_i}\varphi d{u}_i$ with $d\varphi =\nabla \varphi \cdot d\mathbf{r}$ and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{u}_i$. The orthonormality of the coordinate basis gives the gradient in orthogonal coordinates:To obtain the curl without neglecting the variation of the basis vectors, write $\varphi {\mathbf{e}}_1=(\varphi h_1)\nabla u_1$. The curl of a gradient vanishes, and the product rule givesLikewise $\nabla u_2\times \nabla u_3={\mathbf{e}}_1/(h_2{h}_3)$ because the system is right-handed. This cross product has zero divergence: $\nabla \cdot (\nabla u_2\times \nabla u_3)=\nabla u_3\cdot (\nabla \times \nabla u_2)-\nabla u_2\cdot (\nabla \times \nabla u_3)=0$. The divergence product rule applied to $(\varphi h_2{h}_3)(\nabla u_2\times \nabla u_3)$ now givesCyclic permutation and linearity give the general component formulas described by divergence and curl from orthogonal coordinate gradients:andIn cylindrical coordinates, $(h_r,h_{\theta },h_z)=(1,r,1)$. The specified field has no $z$ dependence or $z$ component, soIt is an irrotational vector field for every $r>0$ exactly when $\boxed{p=3,\alpha =4}$. This statement is local to an angular coordinate chart: the component proportional to $\theta$ is not single-valued under $\theta \mapsto \theta +2\pi$.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\varphi /dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in Cartesian coordinatesThe origin is excluded because the scalar field is singular there.

The Stokes theorem states ${\int }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{\ell }$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf{n}$.

For the convective acceleration identity, use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$. It givesRearranging proves the required vector identity.

For orthogonal coordinates, define the scale factors of orthogonal coordinates $h_i=|\partial \mathbf{r}/\partial q_i|$ and tangent unit vectors ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$. They form an orthonormal basis, and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$. Equating $d\psi ={\sum }_i{\partial }_{q_i}\psi d{q}_i$ with $\nabla \psi \cdot d\mathbf{r}$ identifies each component. Thus the gradient in orthogonal coordinates isThis holds on a regular coordinate chart where all scale factors are positive.

In Cartesian coordinates, $\nabla \Phi =({\Phi }_x,{\Phi }_y,{\Phi }_z)$. For orthogonal coordinates, let $\mathbf{r}(q_1,q_2,q_3)$ be the position, $h_i=|\partial \mathbf{r}/\partial q_i|$ the scale factors of orthogonal coordinates, and ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$ the corresponding unit vectors. Since $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ and $d\Phi =\nabla \Phi \cdot d\mathbf{r}={\sum }_i{\Phi }_{q_i}d{q}_i$, comparison of independent increments gives the gradient in orthogonal coordinates:For the hyperbolic parametrization of oblate spheroidal coordinates, the three coordinate tangent vectors areTheir pairwise dot products are zero. Their squared Euclidean norms are ${\sinh }^{2}R+{\sin }^2\theta$, ${\sinh }^{2}R+{\sin }^2\theta$, and ${\cosh }^{2}R{\cos }^2\theta$, respectively. Thus the coordinate surfaces meet orthogonally wherever this is a regular coordinate chart, andThe last expression uses the usual latitude range $|\theta |<\pi /2$; without that range a scale factor is its absolute value. Axial or ring degeneracies are excluded from a regular chart.