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For a ring torus of major radius $A>0$, circular torus coordinates use distance $r$ and angle $\alpha$ in the circular cross-section, and azimuth $\beta$ around the central axis. The coordinate tangent vectors are the three partial derivatives of the displayed position. They are mutually orthogonal, with scale factors of orthogonal coordinates $1,r,A+r\sin \alpha$. For $0<r<A$ their ordered Jacobian determinant is $r(A+r\sin \alpha )>0$, so $dV=r(A+r\sin \alpha )drd\alpha d\beta$. At $r=0$ the cross-section angle degenerates. These are tubular coordinates for a circular torus, rather than the confocal coordinate convention also called toroidal coordinates.

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.

For $a>0$, $u\ge 0$, $0\le v<2\pi$ and real $z$, these orthogonal coordinates have coordinate tangents $r_u=(a\sinh u\cos v,a\cosh u\sin v,0)$, $r_v=(-a\cosh u\sin v,a\sinh u\cos v,0)$ and $r_z=(0,0,1)$. Their scale factors of orthogonal coordinates are $h_u=h_v=a\sqrt{{\sinh }^{2}u+{\sin }^{2}v}$ and $h_z=1$; the ordered Jacobian determinant is positive away from the two focal lines. Positive constant $u$ gives confocal elliptical cylinders, while generic constant $v$ gives a hyperbolic-cylinder half-branch selected by the signs of $x$ and $y$. The limiting $u=0$ surface is the strip $y=0$, $|x|\le a$, with arbitrary $z$. Constant $z$ gives horizontal planes.

Set $x=a\cosh \rho \cos u\cos v$, $y=a\cosh \rho \cos u\sin v$, $z=a\sinh \rho \sin u$, with $a>0$, $\rho \ge 0$ and $|u|\le \pi /2$. Differentiating the position vector gives pairwise orthogonal coordinate tangents and the displayed scale factors of orthogonal coordinates. Thus the positive volume density is $a^3\cosh \rho \cos u({\sinh }^2\rho +{\sin }^{2}u)$. The ordered Jacobian determinant is negative for the order $(\rho ,u,v)$; volume uses its absolute value. Positive constant $\rho$ surfaces are oblate spheroids. For $0<|u|<\pi /2$, constant $u$ gives the half of the one-sheet hyperboloid $(x^2+y^2)/(a^2{\cos }^{2}u)-z^2/(a^2{\sin }^{2}u)=1$ with the sign of $z$ fixed by $u$. At $\rho =0$ the coordinates cover the equatorial disk, while $u=0$ covers the exterior equatorial plane.

Orthogonal coordinates have pairwise orthogonal coordinate tangent directions. Their metric is diagonal, with scale factors of orthogonal coordinates $h_i=|\partial \mathbf{x}/\partial q_i|$. Spherical coordinates and cylindrical coordinates are examples.

The displayed three-dimensional parabolic coordinates use $u,v\ge 0$ and an azimuthal angle $\varphi$. Their coordinate tangent vectors have pairwise zero dot products, with scale factors of orthogonal coordinates $h_u=h_v=\sqrt{u^2+v^2}$ and $h_{\varphi }=uv$. The Jacobian determinant is $uv(u^2+v^2)$. Constant $u$ or $v$ gives a family of confocal circular elliptic paraboloids; the coordinate system degenerates on the axis where $uv=0$. For example, $0\le u\le v\le 1$ fills the cap $(x^2+y^2-1)/2\le z\le 0$, whose rim is the unit circle and whose vertex is $(0,0,-1/2)$.

These orthogonal coordinates extend two confocal families of parabolas parallel to an axis. In the displayed scaling their scale factors of orthogonal coordinates are $h_u=h_v=2\sqrt{u^2+v^2}$ and $h_w=1$, so the Laplacian is $\Delta \varphi =({\varphi }_{uu}+{\varphi }_{vv})/[4(u^2+v^2)]+{\varphi }_{ww}$ away from $u=v=0$. With $u\ge 0$, signed $v$ distinguishes the upper and lower half-planes; the two signs at $u=0$ describe the same negative $x$ half-axis. Fixed positive $u^2$ or $v^2$ describes a parabolic cylinder. This differs from the rotational three-dimensional parabolic coordinates, whose coordinate surfaces are elliptic paraboloids.

Differentiate the position vector with respect to each of the spherical polar coordinates. The derivatives areTheir lengths give the scale factors of orthogonal coordinates:Dividing the derivatives by these lengths yieldsEach has unit length. For example, ${\mathbf{e}}_r\cdot {\mathbf{e}}_{\theta }=\sin \theta \cos \theta ({\cos }^2\varphi +{\sin }^2\varphi )-\cos \theta \sin \theta =0$; the other two dot products vanish because their first two terms cancel. Thus these are mutually orthogonal vectors, and the total differential has the required form. The coordinates degenerate at $r=0$ and on the polar axis, but this does not affect an area integral.

On the parametrized surface $\theta =\alpha$, use $r,\varphi$ as parameters. The surface integral area element isFor $0\le \alpha \le \pi$, its lateral area, with no closing disk, is thereforeAt $\alpha =0,\pi$ the surface collapses to a line and its area is zero.

Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

Vector identity. Work in Cartesian coordinates and use the Einstein summation convention. Contraction of the two Levi-Civita symbols givesThus the convective acceleration identity isEach term is a geometrically defined vector field: the directional derivative of a vector along itself, the gradient of a scalar field, and a cross product with the curl. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate basis, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.

Spherical metric. LetDifferentiating gives the coordinate tangent vectorsTheir pairwise dot products vanish, proving that spherical coordinates are orthogonal coordinates. Their lengths are the scale factors of orthogonal coordinates:These formulas use the usual chart $r>0,0<\theta <\pi$; the origin and polar axis are coordinate singularities.

Azimuthal field. Write the speed as $U(r)$ to distinguish it from the velocity field $\mathbf{u}=U(r)e_{\varphi }$. Using the curl in spherical coordinates,Also $\frac{1}{2}\nabla |\mathbf{u}{|}^2=U{U}^{\prime }{e}_r$. Since $e_{\varphi }\times e_r=e_{\theta }$ and $e_{\varphi }\times e_{\theta }=-e_r$,The $U{U}^{\prime }$ terms cancel in the convective acceleration identity, givingThis centripetal acceleration of an azimuthal flow points toward the polar axis. A streamline is a circle of cylindrical radius $r\sin \theta$, so its acceleration is $-U^2{e}_{\rho }/(r\sin \theta )$, with $e_{\rho }=\sin \theta e_r+\cos \theta e_{\theta }$. This independent geometric argument recovers both nonzero components and explains why $U^{\prime }$ cannot appear: the particle travels at fixed $r$.

Coordinate surfaces and sketch. Adding and subtracting the defining relations gives $s=u^2+v^2$ and $x=u^2-v^2$. Since $y^2=s^2-x^2$, we have $y^2=4u^2{v}^2$. For $u^2=c>0$, eliminate $v^2$ to obtainThis is a parabolic cylinder parallel to the $z$ axis, opening towards negative $x$, with cross-sectional vertex $(c,0)$ and focus at the origin. Similarly, $v^2=d>0$ givesa parabolic cylinder opening towards positive $x$, with vertex $(-d,0)$ and the same focus. The degenerate $u^2=0$ surface is the half-plane $y=0,x\le 0$; $v^2=0$ gives $y=0,x\ge 0$. Constant $w$ gives the horizontal planes $z=w$.

Thus the requested cross-sections consist of $x=1-y^2/4$, $x=2-y^2/8$, $x=y^2/4-1$, $x=y^2/8-2$ and the two horizontal half-axes. The parabolas of the two families meet orthogonally.

Signed coordinates. Take $u\ge 0$ and choose the sign of $v$ to agree with $y$ away from $y=0$. Then the Cartesian coordinates areEvery coordinate triple gives exactly one point. For $u>0$, $v$ and $y$ have the same sign. A fixed signed $v\ne 0$ selects the upper or lower half of its parabolic cylinder; the entire cylinder corresponds to fixed $v^2$. On the cut $u=0$, the two values $v=\pm \sqrt{-x}$ represent the same point of the negative $x$ half-axis. This boundary identification is unavoidable: there $y=0$ even when $v\ne 0$, so the sign prescription is understood by approaching from either side of the cut. The interior $u>0$ gives a unique coordinate chart off that half-axis.

Orthogonality and metric. Differentiate the position vector:The pairwise dot products vanish. Their lengths give the scale factors of orthogonal coordinates and hence the Riemannian metric:These are parabolic cylindrical coordinates. The differential calculation is regular for $u^2+v^2>0$; the $z$ axis is the coordinate singularity where the two planar tangent vectors vanish.

Laplacian. Put $h=2\sqrt{u^2+v^2}$. The physical components of the gradient of a $w$-independent scalar field are ${\varphi }_u/h,{\varphi }_v/h,0$. Applying the divergence formula cancels the two equal scale factors of orthogonal coordinates inside the derivatives:Therefore, away from the coordinate singularity,

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

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.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

Differentiate the position vector, using the angular factor from the scale factors of orthogonal coordinates:The basis is an orthonormal basis, so all cross terms in its squared length vanish. The Euclidean line element is thereforeIn particular, an angular change $d\varphi$ at fixed radius traverses a physical distance $\rho |d\varphi |$.

Dividing the coordinate tangents by their scale factors of orthogonal coordinates gives an orthonormal basisEvaluate the Cartesian cross products componentwise:ConsequentlyThe ordered cylindrical basis is right-handed: its scalar triple product ${\mathbf{e}}_{\rho }\cdot ({\mathbf{e}}_{\varphi }\times {\mathbf{e}}_z)$ equals $+1$.

Write the position vector as $\mathbf{r}(\rho ,\varphi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$. Its coordinate tangent vectors areTheir pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, areThus the cylindrical coordinate system is orthogonal wherever it is a nonsingular coordinate chart. At $\rho =0$, the angular tangent vector vanishes and the azimuth is undefined; this is the usual coordinate singularity on the axis, not a failure of the Euclidean geometry.

The radial unit vector in the spherical coordinate system isThe coordinate tangent in the polar direction has length $r$, so its corresponding unit vector isDifferentiating ${\mathbf{e}}_r$ at fixed $\varphi$ therefore givesThe factor is not $r$: the corresponding scale factors of orthogonal coordinates occur in the derivative of the position vector, whereas the question differentiates its unit direction. The formula is understood away from the coordinate singularities where a complete spherical basis is not uniquely specified.