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A steradian is the solid angle subtended at a sphere's center by a patch of surface area $R^2$. The whole sphere subtends $4\pi \mathrm{sr}$. In the SI, the steradian has unit one; retaining its symbol identifies a solid angle, just as radians identify an angle. A dimensionless quantity still needs its physical meaning specified.^[ref]

For the radial profile $(\rho (t),z(t))$, with $\rho >0$, use the parametrized surface $\mathbf{R}(t,\theta )=(\rho \cos \theta ,\rho \sin \theta ,z)$, $0\le \theta <2\pi$. The two tangent vectors are orthogonal and have Euclidean norms $\sqrt{{\rho }^{\prime 2}+z^{\prime 2}}$ and $\rho$. Their cross product gives the displayed surface area. Geometric surface area assumes a single covering of the radial profile. A general three-dimensional generating curve must first be replaced by its actual radius $\rho (t)=\sqrt{x(t{)}^2+y(t{)}^2}$; its spatial arc length cannot simply replace the radial profile's arc length.

For three radius-$a$ circular cylinders with mutually perpendicular axes through one point, the boundary of their common intersection has three congruent cylindrical surface pieces. On $x^2+y^2=a^2$, the height restriction is $|z|\le a\min (|\sin \theta |,|\cos \theta |)$. Eight congruent sectors give surface area $8a^2(2-\sqrt{2})$ per piece. Pairwise intersections are curves of zero surface area, so the three pieces add.

Let a simple closed regular planar curve be invariant under point reflection through a center and remain strictly on one side of a coplanar axis of rotation, at distance $d$ from its center. If $L$ is its arc length, its surface of revolution has surface area $2\pi dL$. To prove this, choose coordinates with the center at the origin and the axis of rotation at $y=-d$. The surface area formula is $2\pi \oint (d+y)ds$. Pair each point with its reflected point: the arc length element $ds$ is unchanged and $y$ changes sign. Thus $\oint yds=0$, proving the result. The one-side condition prevents the surface from passing through the axis of rotation or requiring an absolute value with changing sign.

For finite-range interactions and bounded wall couplings, only sites within an interaction range of the wall have altered local environments. In a regular container of linear size $L$ and dimension $d$, the surface area scales as $A\sim L^{d-1}$ while volume scales as $V\sim L^d$. A boundary contribution of order $A$ therefore gives an order-$1/L$ correction to free-energy density. Away from a critical phenomenon, a regular surface contribution often has the form $F=Vf+A\gamma +\cdots$, with $\gamma$ a wall Helmholtz free energy per area. Critical finite-size rounding, edges and long-range interactions require separate care; vanishing relative boundary size does not guarantee a simple surface expansion in every regime.

An exact check uses the zero-field nearest-neighbour Ising model on a chain of $N\ge 3$ sites with fixed coupling $J>0$ and $K=J/(k_{B}T)$. The open chain has $N-1$ bonds, while the periodic ring has $N$. Their finite partition functions and dimensionless Helmholtz free energies per spin $g_N=F_N/(N{k}_{B}T)$ areFor the open chain, sum the last Ising spin successively: each removed bond contributes $2\cosh K$, leaving a factor of two. For the ring, the transfer matrix for the one-dimensional Ising model has eigenvalues $2\cosh K$ and $2\sinh K$, and its $N$th power's trace gives the displayed sum. The two ends are a constant-size boundary in one dimension, so the open-chain total correction is $k_{B}T\log \cosh K$, and its per-spin correction is exactly proportional to $1/N$.

Both sets of finite results approach the same $g_{\infty }$ at the fixed plotted $K=0.8$. Multiplying the correction by $N$ exposes the open-chain boundary term, while the ring correction vanishes exponentially at fixed finite $K$. This model's thermodynamic limit remains a real analytic function at every positive temperature: having a thermodynamic limit does not require a phase transition.

For a liquid interface with constant surface tension $\gamma$, the mechanical surface energy is the energy associated with its surface area $A$:The additive constant $C$ does not affect the work needed to change surface area. For several interfaces with the same surface tension, $A$ means their total surface area; the two faces of a thin soap film must both be counted. ^[ref]

With assumed $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, creating $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ of one interface requires $7.2\mu \mathrm{J}$ of reversible work. In the sliding-rim example illustrated under surface tension, moving a $20\mathrm{mm}$ rim outwards by $1\mathrm{mm}$ creates $4.0\times 10^{-5}{\mathrm{m}}^2$ across the two faces and requires $2.88\mu \mathrm{J}$. Independently, multiplying its $2.88\mathrm{mN}$ holding force by the displacement gives the same work.

This makes the preference for a round isolated droplet precise. At fixed volume $V$, the three-dimensional isoperimetric inequality giveswith equality for a sphere. Thus a sphere minimizes surface energy among sufficiently regular closed shapes enclosing the same volume, when surface tension is uniform and other energy contributions are negligible. This is an application of the isoperimetric inequality, rather than a proof of it. Walls, wetting, and gravity can change the shape by adding constraints or other energy contributions. ^[ref]

The differential rule $d{E}_{\mathrm{s}}=\gamma dA$ integrates directly to $E_{\mathrm{s}}=\gamma A+C$ because surface tension is constant. At fixed liquid volume, decreasing surface area can drive viscous fluid flow; the released surface energy can supply viscous dissipation. This mechanical bookkeeping is useful for predicting which deformations can release energy. If surface tension depends on temperature or composition, distinguish this mechanical area-work relation from the complete interfacial thermodynamic free energy and internal energy.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

A toroid is a ring-shaped surface of revolution obtained by rotating a closed planar profile about an axis in its plane that stays outside the profile. A circular profile gives a torus; a rectangular profile gives a ring with rectangular cross-section. The enclosed solid supplies the core geometry in a toroidal winding, while the mathematical surface is its boundary. The distinction matters when computing a surface area versus a volume.

The common intersection of three equal-radius circular cylinders with mutually perpendicular axes. The displayed inequalities describe a centered example of radius $a$. It is a Steinmetz solid; its surface area is given by the surface area of a tricylinder formula.

Turgor pressure is the internal hydrostatic pressure above the external pressure in a biological cell. When water enters a walled biological cell by osmosis, expansion loads the cell wall. Rising pressure can oppose further net entry while the cell membrane retains its selective permeability. Mechanical support therefore does not require the cell wall to block water.

As an original mechanical check, idealize a small flat patch of cell wall of surface area $A$ under uniform turgor pressure $\Delta P=P_{\mathrm{in}}-P_{\mathrm{out}}$. Its outward force is $F=\Delta PA$. Doubling $A$ at fixed pressure doubles the load; equal internal and external pressures give zero net pressure load. This is a force balance, not a measured wall strength or a complete model of a curved, growing biological cell. It says nothing about which solutes cross the cell membrane.

For the faces of a closed polyhedron, take outward unit normal vectors. Applying the divergence theorem to an arbitrary constant vector field $c$ givesSince $c$ is arbitrary, the vector areas sum to zero. This is an oriented statement; the positive scalar surface areas do not cancel. The same proof works for every piecewise smooth closed boundary to which the divergence theorem applies.

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.

For $-\pi \le t\le \pi$, this parametric curve closes with a corner on a unit circular cylinder. Its speed is $\cosh t$, so its arc length is $2\sinh \pi$. The cylindrical surface between the curve and $z=0$ has the same surface area, because unrolling gives the integral ${\int }_{-\pi }^{\pi }\cosh tdt$.

The Young–Laplace equation relates the pressure jump across a stationary interface to its curvature and surface tension. Choose a unit normal vector $\mathbf{n}$ pointing from the inside region to the outside region, and define $\kappa ={\nabla }_s\cdot \mathbf{n}$ to be positive for a convex sphere. In mechanical equilibrium, with constant surface tension and no additional normal viscous stresses,Here $k_1,k_2$ are the signed principal curvatures in this convention. Their sum is twice the corresponding signed mean curvature; stating the convention prevents a factor-of-two ambiguity. The Young–Laplace equation is a local normal stress balance for one interface. ^[ref]

A spherical liquid droplet of radius $R$ in a gas has one interface, with $k_1=k_2=1/R$. An ideal thin soap bubble has two liquid–gas interfaces. Adding their pressure jumps givesFor the soap bubble, $\Delta p$ now means the inner gas pressure minus the outer gas pressure. The second formula assumes equal surface tensions on both faces and film thickness negligible relative to $R$, so their radii can be identified. Retaining distinct radii gives $2\gamma /R_{\mathrm{inner}}+2\gamma /R_{\mathrm{outer}}$. A gas bubble inside bulk liquid has only one interface and therefore uses $2\gamma /R$. ^[ref]

With the assumed constant toy surface tension $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, a droplet with $R=1\mathrm{mm}$ has excess pressure $144\mathrm{Pa}$; at $R=0.5\mathrm{mm}$ it has $288\mathrm{Pa}$. A thin soap bubble assigned the same $\gamma$ and $R=1\mathrm{mm}$ has excess pressure $288\mathrm{Pa}$. These are differences from the surrounding pressure, rather than absolute pressures. Halving the radius doubles the capillary pressure; doubling the interface count also doubles it at fixed radius and surface tension.

The spherical case has a short work argument. A virtual increase in radius gives $dV=4\pi R^{2}dR$ and $dA=8\pi RdR$. At mechanical equilibrium, the net pressure work balances the increase in surface energy:For a thin soap bubble, both interfaces gain approximately the same surface area, doubling the right-hand side and giving $4\gamma /R$. This is a virtual work balance between neighboring spherical configurations, rather than an assumption that a fixed-volume droplet actually expands. The general Young–Laplace equation replaces the spherical area-to-volume ratio by local curvature, making geometry directly measurable through pressure.