Regular Surfaces and Surface Integrals: Examples and Counterexamples

Direct differentiation using [L1] and the coordinate formula in [L3] gives ${\phi }_{\varphi }\times {\phi }_{\theta }=R\sin \varphi \phi$, the outward area vector, whose norm is $R^2\sin \varphi$; by [L3] this is the area density $J_{\phi }$. It is nonzero in the parameter interior; the longitude seam and poles lie on the boundary, so [L2] gives a regular patch.

By [L2], the area is ${\int }_0^{2\pi }{\int }_0^{\pi }{R}^2\sin \varphi d\varphi d\theta =4\pi R^2$.

Since $F(\phi )=\phi$, its dot product with the outward area vector is $R^3\sin \varphi$, whose integral is $4\pi R^3$.

On the northern hemisphere $0\le \varphi \le \pi \mathrm{/}2$, the height is $R\cos \varphi$, so its scalar integral is ${\int }_0^{2\pi }{\int }_0^{\pi \mathrm{/}2}{R}^3\sin \varphi \cos \varphi d\varphi d\theta =\pi R^3$.

Steps 2.1, 2.2, and 2.3 establish the area, outward flux, and hemisphere scalar integral with the orientation stated.

Using [L1], ${\sigma }_{\theta }\times {\sigma }_z=(R\cos \theta ,R\sin \theta ,0)$ is the outward side area vector, whose norm is the density $R$ of [L3]. The cap area vectors are vertical with density $\rho$ in polar parameters, directed down at $z=0$ and up at $z=H$.

The patch intersections are boundary circles, whose preimages lie in parameter boundaries and have content zero. Thus [L2] makes these patches a compatible presentation.

By [L3] the patch areas are the integrals of those densities, giving side area $2\pi RH$ and cap area $\pi R^2$ each; [L2] sums them to the total area $2\pi RH+2\pi R^2$.

By [L3] each patch flux is the integral of $F$ dotted with that patch's area vector. The field $F$ has zero dot product with both vertical cap area vectors, while on the side its dot product with the outward area vector is $R^2$; integration and [L2] give total flux $2\pi R^{2}H$.

Summing the compatible patch values proves both formulas.

Differentiating with [L1] and using the coordinate formula in [L3] gives $\mathrm{\parallel }{\phi }_{\theta }\times {\phi }_{\varphi }{\mathrm{\parallel }}_2=b(a+b\cos \varphi )$, which by [L3] is the area density $J_{\phi }$. Since $a>b>0$ it is positive; the only repeated parameter values occur on the rectangle boundary seams, so [L2] gives a regular patch.

By [L2] and step 1.1, the area is ${\int }_0^{2\pi }{\int }_0^{2\pi }b(a+b\cos \varphi )d\varphi d\theta$.

The full-period integral of cosine is zero, so the inner integral is $2\pi ab$ and the outer integral gives $4{\pi }^{2}ab$.

This is the asserted torus area, with positivity of $a-b$ having discharged the possible degeneracy.

On the graph, $F(x,y,xy)=(x^{2}y,x^2{y}^2,xy)$, and its dot product with the downward area vector $(y,x,-1)$ is $x^2{y}^2+x^3{y}^2-xy$.

By [L2], the integral is $1\mathrm{/}9+1\mathrm{/}12-1\mathrm{/}4=-1\mathrm{/}18$.

Reversing the orientation negates flux, so the upward value is $1\mathrm{/}18$.

The profile is positive for $0<s\le H$ and vanishes only at the apex endpoint $s=0$, so [L1] applies.

By [L1] and [L2], the area is $2\pi (R\mathrm{/}H)\sqrt{1+R^2\mathrm{/}{H}^2}{\int }_0^{H}sds=\pi RH\sqrt{1+R^2\mathrm{/}{H}^2}$.

Since $H>0$, this equals $\pi R\sqrt{R^2+H^2}$. The apex is a parameter-boundary point and contributes no separate term.

The parameter derivatives are ${\phi }_u=(1,0,0)$ and ${\phi }_v=(0,0,0)$, so [L2] gives ${\phi }_u\times {\phi }_v=0$ at every point.

Also $\phi (u,v)$ is independent of $v$, so the map is not injective on the interior, and its image is exactly the line segment $\{(u,0,0):0\le u\le 1\}$.

Both regularity requirements in [L1] fail throughout the interior, and the image is one-dimensional. This supplies the claimed collapse.

The derivatives of $\phi$ are $(1,0,0)$ and $(0,1,0)$, with cross product $(0,0,1)$; those of $\psi$ are reversed, with cross product $(0,0,-1)$. Both maps meet [L2], and $\det Dh=-1$.

By [L3] the two area integrands are the constant $\mathrm{\parallel }(0,0,\pm 1){\mathrm{\parallel }}_2=1$ and the two flux integrands are $(0,0,1)\cdot (0,0,1)=1$ and $(0,0,1)\cdot (0,0,-1)=-1$. Integrating over the unit square gives area $1$ for both maps and fluxes $1$ and $-1$.

The calculation agrees with [L1]: the coordinate swap preserves area and reverses the flux sign.

By [L1], $A(T)=2\pi {\int }_1^T{x}^{-1}\sqrt{1+x^{-4}}dx$.

Since $\sqrt{1+x^{-4}}\ge 1$ for $x\ge 1$, [L2] gives $A(T)\ge 2\pi {\int }_1^{T}dx\mathrm{/}x=2\pi L(T)$.

The lower bound is unbounded by [L2], so the family of compact-truncation areas is unbounded. The noncompact horn itself was not treated as one compact patch.

Fix $T>1$. The profile $r(x)=x^{-1}$ is continuous and positive on $[1,T]$, so [L1] applies with $a=1$, $b=T$ and gives the truncation volume $V(T)=\pi {\int }_1^T{x}^{-2}dx$. By [L2] at $p=2$, these volumes tend to the finite value $\pi$.

On a neighbourhood of $[1,T]$ the same $r$ is $C^1$ with $r^{\mathrm{\prime }}(x)=-x^{-2}$ by [L4], and it is positive throughout, so the hypotheses of [L3] hold and $A(T)=2\pi {\int }_1^T{x}^{-1}\sqrt{1+x^{-4}}dx$. Since $\sqrt{1+x^{-4}}\ge 1$ on $[1,T]$, [L5] gives $A(T)\ge 2\pi {\int }_1^T{x}^{-1}dx$. By [L2] at $p=1$, the right side is unbounded as $T\to \mathrm{\infty }$.

Thus the horn has finite improper volume but unbounded compact-truncation lateral area, refuting the implication. The truncation $T=1$ is excluded because [L1] and [L3] need $a\le b$ and $a<b$ respectively; the refutation concerns the unbounded endpoint and uses only $T>1$.

Every vertex lies on the cylinder. Parametrize each triangular face affinely over the standard parameter triangle. Its constant parameter tangents are two edge vectors, so [L1] makes its area one half of their cross-product norm. Each of the $m$ bands contains $2n$ congruent triangles; expanding those edge-vector cross products gives total area $A_{m,n}=2rn\sin (\pi \mathrm{/}n)\sqrt{H^2+m^2{r}^2(1-\cos (\pi \mathrm{/}n){)}^2}\mathrm{.}$

Take $m=n^3$. The maximum edge length is bounded by the sum of the vertical step $H\mathrm{/}{n}^3$ and a circular chord of angle at most $2\pi \mathrm{/}n$, so the mesh tends to zero by [L2].

By [L1], $n^2(1-\cos (\pi \mathrm{/}n))=2(n\sin (\pi \mathrm{/}(2n)){)}^2\to {\pi }^2\mathrm{/}2$ using [L2]. Hence $n^3(1-\cos (\pi \mathrm{/}n))$ grows like $({\pi }^2\mathrm{/}2)n$ and diverges by [L3].

Also $n\sin (\pi \mathrm{/}n)\to \pi$ by [L2]. Substitution in step 1.1 and step 2.1 shows $A_{n^3,n}\to +\mathrm{\infty }$.

Thus these inscribed lanterns have mesh tending to zero while their areas diverge, proving the stated counterexample.

By [L1], the set of areas of inscribed triangulated polyhedral surfaces for this fixed cylinder is unbounded above.

Its supremum is therefore not a finite real number. The lateral cylinder is the surface of revolution of the constant profile $r$ on $[0,H]$, which is smooth on all of $\mathbb{R}$ and positive, so [L2] gives it the finite surface-integral area $2\pi {\int }_0^{H}r\sqrt{1+0}ds=2\pi rH$.

Consequently the proposed supremum does not equal surface area, even when arbitrarily small mesh is imposed.

By [L1], $\phi$ is continuous, and its first two coordinates make it injective on $D$.

Along $v=0$, the third component is $\mathrm{\mid }u\mathrm{\mid }$. Its right difference quotient at $0$ is $1$ and its left difference quotient is $-1$, so the $u$-partial derivative of $\phi$ does not exist at the interior point $(0,0)$ by [L2].

Thus the cross-product integrand required by [L2] is unavailable at an interior point despite continuity and injectivity, refuting the statement. A polar cone parametrization moves the apex failure to a parameter-boundary point and is a different parametrization.

By [L1], ${\phi }_u\times {\phi }_v=(0,0,1)$ and ${\psi }_s\times {\psi }_t=(0,0,-1)$. Both maps are injective, so these nonzero interior cross products make them regular patches.

The two flux integrands are therefore $1$ and $-1$ on the unit square, so the fluxes are $1$ and $-1$.

The coordinate swap has determinant $-1$, and [L2] explains the sign change. Since the two values differ, the orientation-free statement is false.