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Let $\rho =\sqrt{x^2+y^2}=uv$. The condition $u\le v$ gives $z\le 0$. On the outer coordinate surface $v=1$, elimination of $u$ gives $z=({\rho }^2-1)/2$. On $u=v$, the surface is the plane $z=0$, with $\rho =u^2\le 1$. Thus the closure of the body isIt is a solid circular cap bounded below by an elliptic paraboloid of revolution and above by the unit disk in $z=0$. Its vertex is $\boxed{(0,0,-1/2)}$, its height is $\boxed{1/2}$, and its rim is the circle $\boxed{x^2+y^2=1,z=0}$. The symmetry axis is the $z$ axis. The coordinate surface $u=0$ maps to that axis rather than to an additional two-dimensional boundary. These are parabolic coordinates.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.

The solid $0\le z\le h$, $x^2+y^2\le az$, with $a,h>0$, has horizontal slice area $\pi az$. Thus its volume is ${\int }_0^h\pi azdz=\pi a{h}^2/2$. For a positive constant mass density, rotational symmetry puts its centre of mass on the vertical axis, andThe curved surface area, excluding the horizontal opening, is $\pi a^2[(1+4h/a{)}^{3/2}-1]/6$. Parametrizing the elliptic paraboloid by $(r\cos \theta ,r\sin \theta ,r^2/a)$ gives the area element of a surface $r\sqrt{1+4r^2/a^2}drd\theta$, with $0\le r\le \sqrt{ah}$, proving this last formula. It is also a translated and reflected specialization of paraboloid sector surface measures.