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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.

Departure of an intrinsic nuclear shape from spherical symmetry. Here intrinsic means referred to axes attached to a model nuclear charge distribution. A nuclear quantum state with definite spin need not have the same laboratory charge density as that intrinsic distribution. In particular, the ground state of a deformed even-even nucleus can have $I=0$, a spherical laboratory expectation value, and a nonzero intrinsic nuclear quadrupole moment.

For an illustrative uniform charge density inside a spheroid, let the semiaxes be $(a,a,c)$ and keep its volume equal to that of a sphere of radius $R$:Thus $\eta >1$ gives a prolate spheroid, $\eta =1$ a sphere, and $\eta <1$ an oblate spheroid. These are three-dimensional equal-volume shapes: equal areas of their drawn cross-sections would be a different constraint. Real nuclear charge distributions have diffuse surfaces and need not be uniform, axially symmetric or confined to one shape.

Use the stellar position second-moment tensor $I_{ij}=\int \rho x_i{x}_j{d}^{3}x$. This is the $I$ entering the tensor virial theorem, rather than the engineering moment-of-inertia tensor $\mathrm{tr}(I)1-I$. For a density pattern rotating steadily about its third principal axis, choose $t=0$ when its in-plane axes are principal axes. Then $I(0)=\mathrm{diag}(I_1,I_2,I_3)$ andIn particular, $I_{11}=I_1{\cos }^2\Omega t+I_2{\sin }^2\Omega t$, $I_{22}=I_1{\sin }^2\Omega t+I_2{\cos }^2\Omega t$, and $I_{12}=(I_1-I_2)\sin \Omega t\cos \Omega t$, while $I_{33}=I_3$ and $I_{13}=I_{23}=0$. Two differentiations at $t=0$ prove the inertia-tensor acceleration of a steadily rotating stellar pattern:Both the off-diagonal acceleration and the in-plane trace acceleration vanish at this instant.

Write $\mathbf{u}=\overline{\mathbf{v}}$ for the local mean streaming. The stellar kinetic-energy tensor, stellar pressure-energy tensor, and stellar potential-energy tensor have the conventionsFor an isolated self-gravitating system, $\frac{1}{2}{\ddot{I}}_{ij}=2T_{ij}+{\Pi }_{ij}+W_{ij}$. Subtracting its second diagonal component from the first givessoThis quotient requires $I_{11}\ne I_{22}$. In an axisymmetric density distribution it becomes $0/0$: rotating the pattern about its symmetry axis does not change the density, and this equation cannot determine a pattern speed. Stellar streaming need not be rigid rotation with angular frequency $\Omega$.

Next add the first two diagonal virial equations and use the third. Their acceleration terms vanish, soFor $T_{33}=0$, ${\Pi }_{33}=-W_{33}$. With the planar virial trace ${\Pi }_{\perp }={\Pi }_{11}+{\Pi }_{22}$ and $W_{\perp }=W_{11}+W_{22}$, the prescribed mass-weighted quantities satisfy ${\Pi }_{\perp }=2M{\sigma }_0^2$, $2(T_{11}+T_{22})=M{V}_0^2$, and $1-\delta =2{\Pi }_{33}/{\Pi }_{\perp }$. Therefore $W_{33}=-(1-\delta )M{\sigma }_0^2$ and the planar equation becomesEliminating $M{\sigma }_0^2$ givesThe ${\nu }_0$ in the numerator of the PDF's displayed relation is a notation slip: the velocity subsequently defined is $V_0$. The relation follows from the traces, not from replacing a galaxy by a solid rotating body.

For the axisymmetric oblate spheroid, rotational symmetry gives ${\Pi }_{11}={\Pi }_{22}=M{\sigma }_0^2$, ${\Pi }_{33}=M{\sigma }_0^2(1-\delta )$ and zero integrated off-diagonal random stresses. A line-of-sight unit vector can be chosen as $\mathbf{n}=(\sin i,0,\cos i)$, with $i=0$ face-on. Contracting the random stress tensor with this vector gives the inclination projection of an axisymmetric stellar velocity tensor:This is a mass-weighted random velocity dispersion after subtracting streaming, with consistent aperture weights. Along the projected major axis, azimuthal streaming lies along the direction whose line-of-sight projection is $\sin i$; a rotation amplitude $V_0$ consequently projects asThere is a necessary observational convention here. The globally defined $V_0^2=2(T_{11}+T_{22})/M$ is a streaming mean square. The displayed amplitude relation holds if the rotation curve is constant, or if the measured rotation-amplitude statistic is weighted consistently to represent this intrinsic quantity. Axisymmetry alone does not equate an arbitrary observed peak rotation to the global $V_0$. If $V_r$ instead means the full-aperture line-of-sight streaming root-mean-square, azimuthal averaging yieldsand the corresponding root-mean-square is $V_0\sin i/\sqrt{2}$. Keeping amplitude and root-mean-square distinct prevents a spurious factor of two in the kinetic energy.

Finally, let the intrinsic equatorial and polar semiaxes be $a$ and $c$, and put $q_t=c/a=1-{ϵ}_t$. An intrinsic boundary has $(x^2+y^2)/a^2+z^2/c^2=1$. Its sky minor-axis coordinate is $Y=y\cos i-z\sin i$. Eliminating the line-of-sight coordinate, or maximizing $Y$ on this ellipse, gives the projected minor semiaxis $\sqrt{a^2{\cos }^{2}i+c^2{\sin }^{2}i}$; the major semiaxis remains $a$. Thus the projected axis ratio of an oblate spheroid isSubtracting from unity provesThe face-on limit is round with zero projected rotation amplitude; the edge-on limit reveals the intrinsic flattening and in-plane random velocity dispersion. These limits check both the geometry and the convention for inclination.

Choose intrinsic Cartesian coordinates in which the oblate spheroid isLet $\ell$ be distance along the line of sight and let $y$ be the sky coordinate in the plane containing the line of sight and the symmetry axis. A rotation through the inclination $i$ givesSubstitution into the ellipsoid equation and minimization over $\ell$, equivalently requiring the quadratic in $\ell$ to have zero discriminant on the projected boundary, givesTherefore the projected axis ratio of an oblate spheroid isIt correctly gives $Q=1$ face-on and $Q=q$ edge-on.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.

Put $s^2=x^2+y^2$. For $\rho >0$, elimination of $u$ givesThese are oblate spheroids with equatorial radius $\cosh \rho$ and polar radius $\sinh \rho$, flattened along the $z$ axis. At $\rho =0$ the limiting surface is the disk $z=0$, $s\le 1$.

For $0<|u|<\pi /2$, elimination of $\rho$ instead givesThese are one-sheet hyperboloids of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho \ge 0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\ge 1$; at $u=\pm \pi /2$ it degenerates to the corresponding half of the $z$ axis.

Finally, fixed $v$ specifies a vertical half-plane through the $z$ axis, with horizontal direction $(\cos v,\sin v)$. The two values $v$ and $v+\pi$ together give a full plane. These surfaces form the mutually orthogonal families of the hyperbolic parametrization of oblate spheroidal coordinates.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

If the symmetry axis of an oblate spheroid makes angle $i$ with the line of sight, its projected axis ratio satisfiesThus a face-on spheroid appears round and an edge-on spheroid reveals its intrinsic axis ratio.

A spheroid is an ellipsoid of revolution obtained by rotating an ellipse about a principal axis. Rotation about the shorter axis gives an oblate spheroid; rotation about the longer axis gives a prolate spheroid. Thus the oblate form is a member of the general family.