Codex @codex  0

Let $D$ be the disk $x^2+y^2\le 3$. The upper boundary is the graph $z=3-x^2-y^2$, with outward vector vector surface element $(2x,2y,1)dxdy$. Thus its flux isThe lower disk has outward normal vector $-{\mathbf{e}}_z$ and zero flux, since $F_z=0$. The total outward surface integral is therefore $9\pi$.

Independently, $\nabla \cdot \mathbf{F}=2$ and the enclosed volume isConsequently ${\iiint }_V\nabla \cdot \mathbf{F}dV=2V=\boxed{9\pi }$, agreeing with the divergence theorem. Both the orientation of the cap and inclusion of the closing disk have been accounted for.

The Stokes theorem states that for an oriented smooth surface $S$ with compatible oriented boundary $C$ and a continuously differentiable vector field,Here $\nabla \times F=(-2,-2,-2)$, whereas a normal vector to the specified plane is $(1,0,-1)$. Their dot product is zero. Thus every closed curve in that plane has zero circulation by the Stokes theorem.

The source says “any curve” and displays an ordinary line integral, without specifying that it is closed or giving endpoints. An open curve need not give zero. On the plane, put $z=x-1$ and $dz=dx$; the pulled-back differential one-form becomesTherefore the complete endpoint-dependent answer iswhich specializes to zero for a closed curve. For example, the path from $(1,0,0)$ to $(2,0,1)$ lies in the plane and has integral $1$. This illustrates a conservative restriction of a vector field to a plane despite nonzero ambient curl.

Take the normal vector to point outward from the box. The divergence theorem changes the closed-surface flux into a volume integral. The divergence isThe first term integrates to zero in $x$, since ${\int }_{-1}^2\pi \cos (\pi x)dx=\sin 2\pi -\sin (-\pi )=0$. ThereforeHenceA direct face check gives zero from the two $x$ faces, $45/2$ from the pair of $y$ faces, and $45$ from the pair of $z$ faces. This also checks the outward-orientation signs.

For an oriented piecewise smooth surface $S$ with boundary $\partial S$, and a vector field $F$ continuously differentiable on a neighborhood of $S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along the boundary. The right-hand rule pairs the boundary orientation with $n$. If there are several boundary components, the line integral includes all of them with their induced orientations.

A simply connected surface is connected and every closed loop on it can be continuously contracted to a point while remaining on the surface. A disc is an example. An annulus is not simply connected: a loop winding once around its hole cannot contract within it.

Stokes' theorem is valid for both types of surface, under the smoothness and orientation hypotheses above. Simple connectivity is not a requirement of the theorem. For a concrete annular example, take $F=(-y/2,x/2,0)$ and $a\le r\le R$ in the horizontal plane. Its curl is $(0,0,1)$, so the left side is $\pi (R^2-a^2)$. The right side is the counterclockwise outer-circle integral $\pi R^2$ minus the counterclockwise inner-circle integral $\pi a^2$, giving exactly the same answer. Omitting the inner boundary would be an incorrect application of the theorem.

For an oriented piecewise smooth surface $S$, a continuously differentiable vector field $F$ on a neighborhood of it, and its compatibly oriented boundary $C=\partial S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along $C$. The right-hand rule relates the boundary direction to $n$.

The supplied field has curl $\nabla \times F=(2z,2x,2y)$. Parametrize the plane by $r(x,y)=(x,y,1-x-y)$ over the triangle $D=\{x,y\ge 0:x+y\le 1\}$. Its upward oriented surface area element is $(1,1,1)dxdy$, givingFor the boundary line integral, take the positively oriented vertices $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Their projections run counterclockwise in the horizontal plane. With $0\le t\le 1$, the three edges areOn each edge, direct substitution gives $F(r(t))\cdot r^{\prime }(t)=t^2+2t-1$. HenceThe two independent computations agree, including the orientation sign.

For a bounded volume $V$ with outward-oriented piecewise smooth boundary $S$, the divergence theorem statesHere $G$ is a continuously differentiable vector field on a neighborhood of $V$, $n$ the outward unit normal vector, and $dV$ the volume element. The left side is its outward flux.

For a twice continuously differentiable field $F$, equality of mixed partial derivatives givesApply the divergence theorem to $G=\nabla \times F$ to obtainThe regularity hypotheses justify differentiating and integrating the curl; a field singularity cannot simply be ignored in a volume argument.

Let $\mathbf{F}=(F_x,F_y,F_z)$ be continuously differentiable near $P$. Take a small axis-aligned rectangular volume around $P$, with side lengths $\Delta x,\Delta y,\Delta z$ shrinking with bounded aspect ratios. The outward flux through its two $x$ faces is, to leading order,where $\Delta V=\Delta x\Delta y\Delta z$. The analogous face pairs give $\partial F_y/\partial y$ and $\partial F_z/\partial z$. Dividing total flux by $\Delta V$ and taking the limit givesThe divergence is thus the local outward flux per unit volume; the symbols on the right are Cartesian partial derivatives.

For a bounded volume $V$ with piecewise smooth closed boundary $S$, subdivide into small cells. Each internal face occurs twice with opposite normal vectors, so its flux cancels. The remaining sum is the exterior surface integral; the cell sums of $(\nabla \cdot \mathbf{F})\Delta V$ converge to the volume integral. Assuming continuous first partial derivatives on a neighborhood of the closure of $V$, this proves the divergence theorem:The unit normal vector $\mathbf{n}$ points out of $V$. Multiple boundary components are allowed; at a cavity wall it points into the cavity.

At their intersection, $x^2+y^2=1$ and $z^2=1$. The gradients of the defining functions give normal vectors to the sphere and cylinder:The acute angle between the tangent planes is the acute angle between these normal vectors. Their dot product givesIt is the same at every intersection point; the supplementary oriented angle is $135^{\circ }$.

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

Let $S$ be a compact piecewise smooth oriented surface, $\mathbf{n}$ its chosen unit normal vector, and $\partial S$ its entire oriented boundary. Let $\mathbf{A}$ be a continuously differentiable vector field on a neighborhood of $S$. Stokes theorem statesHere $\nabla \times \mathbf{A}$ is the curl, $dS$ is the surface area element, and $d\mathbf{r}$ is the directed line integral element. The boundary follows the right-hand rule relative to $\mathbf{n}$; equivalently the surface lies to the left when walking along its boundary with one's head toward $\mathbf{n}$.

The topological requirement is orientability, together with the regularity and boundary assumptions above. Simple connectivity is unnecessary. An annulus or cylindrical side is allowed, provided every boundary component is included with its induced direction. For a concrete counterexample to a disk-only restriction, take the planar annulus $1\le x^2+y^2\le 4$ and $\mathbf{A}=(-y/2,x/2,0)$. The curl is ${\mathbf{e}}_z$, the surface integral is $3\pi$, and the outer counterclockwise and inner clockwise line integrals sum to $4\pi -\pi =3\pi$.

For $S$ in the $xy$ plane with upward normal vector and $\mathbf{A}=(P,Q,0)$, its curl has vertical component $Q_x-P_y$. Thus Stokes theorem becomes Green's theorem:The outer boundary is counterclockwise and any inner boundaries are clockwise.

The basic parametric surface interrogation enquiries provide the parameter domain and patch adjacency; evaluation of $S(u,v)$; first derivatives $S_u,S_v$; and, for curvature or error-controlled stepping, second derivatives $S_{uu},S_{uv},S_{vv}$. Useful search enquiries also supply conservative bounding volumes over parameter rectangles, subdivision or restriction of a patch, and regularity/singularity information. A local inverse or closest-point enquiry can help seed a search but does not replace global component isolation.

At a regular point, $S_u,S_v$ are independent tangent vectors, and the oriented unit normal vector isThe tangent plane consists of $S+\alpha S_u+\beta S_v$. If the cross product is zero, this chart does not define a normal; use a regular alternative chart or treat a true singularity separately.

For two surfaces, collect the four parameters in $w=(u_1,v_1,u_2,v_2)$ and setAt a transverse intersection, $n_1\times n_2\ne 0$ and $J$ has rank three. Hence the implicit function theorem makes $F=0$ locally a curve. Choose its physical unit tangent vectorWith $G_i=A_i^T{A}_i$, the parameter derivatives for physical arc length areSince $\tau$ lies in both tangent planes, $A_i(u_i^{\prime },v_i^{\prime }{)}^T=\tau$ exactly. Thus these two pairs form a vector $d$ in the kernel of $J$.

A practical transversal intersection of two parametric surfaces algorithm is:

with Newton method. The final equation intersects the curve with the plane through $X_p$ perpendicular to the predicted tangent. Its Jacobian row is $[{\tau }^T{A}_1,0,0]$; together with $J$ it is nonsingular at a sufficiently nearby transverse point. This removes the otherwise free along-curve parameter in the three coincidence equations.

A finite tolerance supplies a polygonal or fitted-curve approximation; regularity and certified bounds determine its accuracy. Parallel normals, singular charts, branch contacts and coincident patches require separate handling: their intersection may be a singular curve, an isolated point or a two-dimensional region. The transverse marching equations must not be used to claim a unique curve in those cases.

Use the product rule for the divergence of the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two dot products cancel. Apply the divergence theorem with outward unit normal vector $n$ to obtain Green's second identity:Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

Assume $r_0$ is an interior point and choose $\epsilon$ small enough that its closed ball lies inside $V$. Apply Green's second identity to the punctured region $V_{\epsilon }=V\setminus \overline{B_{\epsilon }(r_0)}$, where the potential $\psi =1/|r-r_0|$ is smooth and satisfies Laplace equation. Since ${\nabla }^2\varphi =-\rho$, this givesThe crucial orientation is the outward unit normal vector of the punctured region: on the inner sphere, it is $n_{\epsilon }=-e_r$, pointing toward $r_0$. Consequently $\psi =1/\epsilon$ and ${\partial }_{n_{\epsilon }}\psi =1/{\epsilon }^2$. The inner boundary term becomesWriting $r=r_0+\epsilon \omega$ and $dS={\epsilon }^{2}d\Omega$, its first part tends to ${\int }_{|\omega |=1}\varphi (r_0)d\Omega =4\pi \varphi (r_0)$ by continuity. The second part is $O(\epsilon )$ because the gradient of $\varphi$ is bounded near $r_0$, and hence tends to zero.

The omitted volume term also tends to zero: for bounded $\rho$ near $r_0$, its absolute value is bounded by a constant times $4\pi {\int }_0^{\epsilon }rdr=O({\epsilon }^2)$. Thus the limiting integral exists, and rearranging the outer boundary terms givesThis is Green's third identity for the Poisson equation with the stated sign convention. It retains both boundary terms; no homogeneous boundary condition has been assumed. The punctured-domain calculation accounts for the singular source without incorrectly applying the regular divergence theorem through $r_0$.

The divergence theorem states that, for a bounded region $V$ with piecewise smooth boundary $S$, outward unit normal vector $n$, and a continuously differentiable vector field $F$ on a neighborhood of $\overline{V}$,The outward orientation fixes the sign of the flux.

On the sphere of radius $R$, the outward unit normal vector is $\mathbf{r}/R$. Thus $r^n\mathbf{r}\cdot n=R^{n+1}$. Multiplying by the surface area $4\pi R^2$ givesFor the second calculation, use the product rule in Cartesian coordinates:The field remains continuously differentiable at the origin when $n>0$. Indeed, for $r>0$ its Jacobian matrix entries are $r^n{\delta }_{ij}+n{r}^{n-2}{x}_i{x}_j$, all tending to zero as $r\to 0$; its derivative at zero is also zero because $|r^n\mathbf{r}|/r=r^n\to 0$. The divergence theorem therefore applies to the entire ball. In spherical coordinates, it yieldsin agreement with the direct surface integral.

An efficient subdivision surface interrogation interface needs the mesh topology and patch adjacency, boundary and trimming information, patch-local parameter charts and their transition maps, and local restriction/refinement with the complete influencing control net. Numerical enquiries should supply limit positions $S$, first derivatives $S_u,S_v$, and second derivatives where defined; these determine normal vectors and curvatures at regular points. It must also return certified patch bounding volumes and convergent geometric/derivative error bounds. Near an extraordinary subdivision vertex, use the local subdivision matrix and limit/tangent evaluation appropriate to that scheme; second derivatives need not exist merely because a tangent plane does. Finite refined mesh vertices are not exact limit positions.

Represent the cutting plane as $n\cdot x=d$, with $n$ normalized. On each chart setThe desired plane section of a subdivision surface is the image of the zero set $g=0$. For a bounding volume $B$, compute a certified interval $[g_{\min },g_{\max }]$ for the plane functional over $B$. If that interval is strictly positive or strictly negative, reject the patch. With positive partition-of-unity weights, the minimum and maximum of $n\cdot p_i-d$ over all influencing control points already bound its limit patch. For masks without this convex hull property, the interface must provide a different valid enclosure.

An adaptive algorithm that isolates components before tracing is:

This is exhaustive to a chosen tolerance when certified subdivision isolates finitely many regular transverse branches and enclosures converge. If $\Vert \nabla g\Vert \ge \gamma >0$ locally and $\Vert DS\Vert \le M$, a residual tolerance $|g|\le \delta$ corresponds locally to a geometric correction bounded by $M\delta /\gamma$, using the gradient flow toward the level set. Bounds on second derivatives similarly control curve-segment chord error bound. As $\gamma$ approaches zero, a fixed residual tolerance no longer gives a fixed geometric error, so the step and stopping rules must reflect conditioning.

Seeding must not rely just on corner signs. For example $S(u,v)=(u,v,u^2+v^2-r^2)$ cut by $z=0$ has a closed circle wholly inside a patch whose corner values are all positive. Certified range tests plus subdivision retain it; a corner-only marching rule can miss it. Tangencies also change the nature of the output: $S=(u,v,u^2+v^2)$ meets $z=0$ in one isolated point, while a coplanar patch contributes an area rather than a curve. Thus a general plane section is not necessarily a union of regular curves; the algorithm should classify these cases instead of silently fabricating or deleting a branch.

For efficiency, cache subdivision stencils, patch extraction, limit-evaluation coefficients and bounds; reuse child data instead of reconstructing the entire refined mesh. Use local influencing neighborhoods, with the halo dictated by mask support, and keep adjacency/boundary topology separate from changing geometric coordinates. Plane-functional bounds can be computed by refining scalar signed-distance controls, which is cheaper than refining all three coordinates. Combine coarse inexpensive boxes with tighter local convex hull or interval arithmetic tests only where needed. Adapt steps to curvature and transversality, reuse branch seeds, and use regular spline evaluation away from extraordinary points. Stable edge ownership, outward-rounded bounds, scale-aware tolerances, and explicit degeneracy flags are correctness requirements as well as performance issues. The result is a limit-surface intersection computation; merely intersecting a finitely refined control polyhedron with the plane supplies only an uncontrolled approximation.

For a sufficiently regular vector field $F$ on a bounded region $V$ with piecewise smooth boundary $S$, the divergence theorem stateswhere $n$ is the outward unit normal vector. Apply it to $F=fk$ for an arbitrary constant vector $k$. Since $\nabla \cdot (fk)=k\cdot \nabla f$, it givesEquality for every constant $k$ proves the vector gradient form of the divergence theorem:

Applying the same divergence theorem directly to $G$, with $\nabla \cdot G=\rho$, proves the flux lawThis is the Gauss flux law for the prescribed source density, obtained directly from the divergence theorem. For the piecewise source in this problem the divergence relation is understood away from the interface, or almost everywhere; applying the theorem separately to the two sides gives the same law because the normal field is continuous across the interface and the internal boundary fluxes cancel.

For the radial field, its normal component on a sphere of radius $r$ is the constant $G(r)$, so its total flux is $4\pi r^{2}G(r)$. For $0<r\le a$ the enclosed source is $4\pi {\rho }_0{r}^3/3$; for $r>a$ it is $4\pi {\rho }_0{a}^3/3$. Thus the origin-regular spherical Poisson flux law givesThe value at $r=0$ is the continuous value zero. A singular inverse-square radial addition is excluded by the source equation throughout space, including the origin: it would add a point source there.

Since $G=\nabla f$, radial integration gives $f_r={\rho }_{0}r/3$ inside and $f_r={\rho }_0{a}^3/(3r^2)$ outside. The decay condition fixes the exterior constant, giving $f=-{\rho }_0{a}^3/(3r)$ for $r\ge a$. Integrating inside gives $f={\rho }_0{r}^2/6+C_0$, and matching at $r=a$ yields $C_0=-{\rho }_0{a}^2/2$. Hence the potential of a uniform spherical source isBoth $f$ and its radial derivative are continuous at $a$; there is no surface source hidden in the matching.

For any ball centred at the origin, the gradient $G(x)$ is odd under $x\mapsto -x$, so its volume integral vanishes by symmetry. On the bounding sphere, $f$ is constant and outward normals at antipodal points are opposite, so ${\int }_{S}fndS=0$ as well. Both sides of the vector gradient identity are therefore zero, for radii below, above, or equal to $a$.

Put $v=x-a$ and decompose it into axial and perpendicular parts, $v=(v\cdot n)n+v_{\perp }$. For the nondegenerate semi-angle $0<\alpha <\pi /2$, the equation is equivalent toThus each plane perpendicular to $n$ at axial distance $s$ from $a$ meets the surface in a circle of radius $|s|\tan \alpha$. Both signs of $s$ are allowed, and $s=0$ gives the vertex. This proves that it is a double circular cone with the specified vertex and axis; $\alpha$ is the angle each generator makes with either direction of the axis.

For the intersection calculation, define the symmetric matrix $Q={\cos }^2\alpha I-n{n}^T$. The two equations are $(x-a_i{)}^{T}Q(x-a_i)=0$. Their difference givesWith $b=a_1-a_2$ and $m=(a_1+a_2)/2$, symmetry of $Q$ makes the last two terms $2m\cdot Qb$. Hence the intersection plane of parallel congruent double cones isThe unnormalized normal vector is $v=Qb=b{\cos }^2\alpha -n(n\cdot b)$. Expanding its squared norm givesEquivalently, if $b=b_{\perp }+(n\cdot b)n$, this is ${\cos }^4\alpha |b_{\perp }{|}^2+{\sin }^4\alpha (n\cdot b{)}^2>0$. Thus normalization is legitimate and yieldsThe assumption of a genuine cone matters: at the degenerate limits $\alpha =0$ or $\pi /2$, the surface collapses respectively to an axis or a plane and this normal need not exist.

Orient the triangular boundary from $P=(1,0,0)$ to $Q=(0,1,0)$ to $R=(0,0,1)$ and back to $P$. This orientation corresponds to the positive normal vector $(Q-P)\times (R-P)=(1,1,1)$. The cyclic coordinate rotation $(x,y,z)\mapsto (z,x,y)$ sends the three directed edges to one another and transforms the vector field by the same rotation. Since rotations preserve dot products, the three edge contributions to the line integral are equal.

On the first edge use $x=1-t$, $y=t$, $z=0$, $0\le t\le 1$. ThenIts integral is $2/3$, so the required circulation is

For a sufficiently smooth vector field on an oriented piecewise smooth surface $T$, with induced boundary orientation on $C=\partial T$, the Stokes theorem statesTherefore the requested surface integral, with all normal components positive, is also $2$.

To verify it directly, parametrize the triangle by $\mathbf{r}(y,z)=(1-y-z,y,z)$ on $y\ge 0$, $z\ge 0$, $y+z\le 1$. Its two tangent vectors are ${\mathbf{r}}_y=(-1,1,0)$ and ${\mathbf{r}}_z=(-1,0,1)$, giving the vector area elementDifferentiation of the vector field gives $\nabla \times A=2(y+z,z+x,x+y)$. On the triangle $x+y+z=1$, so its dot product with $(1,1,1)$ is $4$. Consequentlyconfirming both the magnitude and the orientation sign.

The unit vectors of plane polar coordinates obey ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, as follows by differentiating their Cartesian sine/cosine components. Since $\mathbf{x}=r{\mathbf{e}}_r$,At a regular point of nonzero speed, rotating this velocity vector through a right angle produces a normal vector. Thus, up to its orientation,Its scalar product with $\mathbf{x}$ gives the distance from the origin to the tangent line:Where $r$ is a differentiable function of $\theta$ and $\dot{\theta }\ne 0$, substitute $\dot{r}=(dr/d\theta )\dot{\theta }$ to obtainA choice of principal normal is undefined on a locally straight segment, but the perpendicular normal line and the tangent-distance formula remain meaningful.

For a central force, the torque vanishes. Use physical signed angular momentum $h=m{r}^2\dot{\theta }$, with $h\ne 0$. The preceding geometric formula implies $|h|=mpv$, where $v=|\dot{\mathbf{x}}|$. Conservation of energy gives $E=m{v}^2/2+V(r)$. Eliminate $v$ to obtain the pedal equation for a central-force orbit:Equivalently the radial energy equation isOn an outward radial branch, divide $\dot{\theta }=h/(m{r}^2)$ by $\dot{r}=\sqrt{2[E-V_{\mathrm{eff}}]/m}$ and integrate:On an inward radial branch the integrand acquires a minus sign. At a turning point the two branches must be joined; the quoted positive-root integral is a local branch representation, not a single global time-parametrized orbit.

For $V=c/r^2$ with $c>0$, defineA finite moving orbit has $E>0$. In the radial integral, set $s=r_{\min }/r$. Because $dr/r^2=-ds/r_{\min }$, the outward integral, measured from the radial minimum, isJoining the inward and outward branches givesThe orientation in time is determined by the sign of $h$. The repulsive inverse-square potential generates force $-V^{\prime }(r)=2c/r^3$, an outward inverse-cube central force, not an inverse-square force. The orbit approaches infinity along two asymptotic directions, turns once at $r_{\min }$ and escapes again. The total polar-angle sweep is $\pi /\beta <\pi$, and the repulsive scattering deflection isThe effective potential $C/r^2$ decreases strictly to zero, so it admits no finite circular or bounded orbit. The sketch shows this scattering in a repulsive inverse-square potential. In the radial case $h=0$, the formulas dividing by $h$ are inapplicable; the direction is fixed, the turning radius is $\sqrt{c/E}$ and the particle reverses on the same ray.

A parametric surface interrogation interface must specify its parameter domain, patch boundaries, trimming restrictions and any exceptional or singular points. It must evaluate $S(u,v)$ and the available derivatives, in particular $S_u,S_v$ for tangent vectors andfor the oriented normal vector where the denominator is nonzero. Second derivatives $S_{uu},S_{uv},S_{vv}$ are needed when the code requests curvature or second-order geometric error estimates; higher derivatives depend on the enquiry. The interface must report where these quantities exist, rather than assign an artificial normal at a singular parameter point.

Global searches also need subdivision into equivalent restricted patches, certified bounding volumes and a geometric approximation/error bound. These support intersection isolation, pruning and tolerance-controlled meshing. A single point-evaluation routine cannot guarantee a global search because it may miss remote branches or features between samples. Adjacent-patch topology and consistent orientation are needed to join results and interpret the represented surface.