Parametric surfaces and boundary theorem derivations

A boundary theorem is not a license to exchange an arbitrary surface with another. Regularity, orientation and the region between surfaces determine whether the theorem applies. By deriving Stokes on a parameterized patch and divergence on a graph-bounded solid, we see exactly where derivatives, area vectors and cancellation enter.

1 Surface area vectors and graph flux

Let $r(u,v)$ be a $C^1$ regular parameterization, so $r_u\times r_v\ne0$. A small parameter rectangle is approximated by a parallelogram with sides $r_u\Delta u$ and $r_v\Delta v$. Its area is $\|r_u\times r_v\|\Delta u\Delta v$ and its oriented area vector is $(r_u\times r_v)\Delta u\Delta v$. Thus scalar area uses the norm while flux uses the vector itself.

For an upward-oriented graph $r(x,y)=(x,y,h(x,y))$,

$$r_x\times r_y=(-h_x,-h_y,1).$$

Therefore for $F=(P,Q,R)$,

$$\iint_SF\cdot n\,dS=\iint_D[-P(x,y,h)h_x-Q(x,y,h)h_y+R(x,y,h)]dx\,dy.$$

The slope factors already encode the area enlargement and the normal direction. Multiplying again by $\sqrt{1+h_x^2+h_y^2}$ would double count area. For downward orientation, negate the whole expression.

Worked cap. For the paraboloid $z=1-x^2-y^2$ over the unit disk and $F=(0,0,z)$, upward flux is

$$\iint_D(1-x^2-y^2)dA=2\pi\int_0^1(1-r^2)r\,dr=\frac\pi2.$$

This is a flux calculation, not the surface's scalar area.

2 Stokes theorem on one regular patch

Assume $D$ is a bounded planar region with positively oriented piecewise smooth boundary to which Green's theorem applies. Let $r$ be a one-to-one $C^2$ regular parameterization on a neighborhood of $\overline D$, and let $F$ be $C^1$ near the image. Orient the surface by $r_u\times r_v$ and its boundary by the image of the positive parameter boundary.

Set

$$A(u,v)=F(r(u,v))\cdot r_u,\qquad B(u,v)=F(r(u,v))\cdot r_v.$$

The boundary chain rule gives $F\cdot dr=A\,du+B\,dv$. Differentiate:

$$B_u-A_v=(DF(r)r_u)\cdot r_v-(DF(r)r_v)\cdot r_u,$$

because $r_{vu}=r_{uv}$ cancels the two $F$ terms. Expanding the three components of $DF$ and collecting antisymmetric pairs gives the identity

$$B_u-A_v=(\nabla\times F)(r)\cdot(r_u\times r_v).$$

For example, the $P_y$ terms have coefficient $r_{u,y}r_{v,x}-r_{v,y}r_{u,x}$, matching the negative $P_y$ contribution to the third curl component; the other pairs match in the same way. Green's theorem now yields

$$\oint_{\partial S}F\cdot dr=\iint_S(\nabla\times F)\cdot n\,dS.$$

This is a proof of Stokes under the stated patch hypotheses. For a compact oriented surface decomposed into finitely many such compatible patches, apply it patchwise. Each internal edge is traversed in opposite directions by its two neighboring patches, so its line integrals cancel. The remaining boundary is the external boundary with the induced orientation. This extension assumes such a finite compatible patch decomposition.

Worked circulation. Let $F=(-y/2,x/2,0)$ and use the upward paraboloid cap above. Its curl is $(0,0,1)$, so the surface curl flux is the projected disk area $\pi$. The boundary is the counterclockwise unit circle at $z=0$; parameterization gives line-integral integrand $1/2$, yielding $\pi$. Reversing the surface normal reverses the boundary direction and both signs.

3 Divergence on a graph slab

Let $D$ be a bounded plane region with a piecewise smooth positively oriented boundary, and let $a,b$ be $C^1$ on a neighborhood with $a<b$. Consider

$$\Omega=\{(x,y,z):(x,y)\in D,\ a(x,y)\le z\le b(x,y)\}.$$

Let $F=(P,Q,R)$ be $C^1$ near its closure. Define $\bar P(x,y)=\int_a^bP(x,y,z)dz$ and similarly $\bar Q$. The variable-limit differentiation rule gives

$$\int_a^bP_xdz=\bar P_x-P(x,y,b)b_x+P(x,y,a)a_x,$$

with an analogous formula for $Q_y$. The remaining derivative integrates to $R(x,y,b)-R(x,y,a)$.

Integrating over $D$, the terms involving the top values are exactly the upward graph flux through $z=b$; the bottom terms are exactly the downward graph flux through $z=a$. The remaining planar divergence integral is

$$\iint_D(\bar P_x+\bar Q_y)dA =\oint_{\partial D}(\bar P\nu_x+\bar Q\nu_y)ds,$$

by Green's flux form, where $\nu$ is the outward planar unit normal. On a counterclockwise unit-speed boundary parameterization $(x(s),y(s))$, $\nu=(y'(s),-x'(s))$. The side surface parameterized by $(x(s),y(s),z)$ has outward area vector $(y',-x',0)ds\,dz$, so this last boundary integral is precisely its flux. Adding top, bottom and side proves

$$\iiint_\Omega\nabla\cdot F\,dV=\iint_{\partial\Omega}F\cdot n\,dS.$$

For finite decompositions into graph slabs, internal surface fluxes cancel because the outward normals oppose. If top and bottom meet on the boundary, an inward-domain limit can sometimes extend the result, but the extra integrals and shrinking side flux must be checked. Singular fields, nonregular boundaries or domains requiring additional limiting arguments are not covered just by writing the formula.

4 Comparing surfaces safely

If $F$ is $C^1$ throughout a solid between two surfaces and has zero divergence there, their consistently oriented fluxes agree after including any side boundary. The side contribution cannot be omitted unless it vanishes or is absent. For curl flux, Stokes shows surfaces with the same oriented boundary give the same curl integral when each satisfies its hypotheses and the field is regular on their neighborhoods.

A singularity invalidates a casual replacement. The radial field $F=x/\|x\|^3$ has divergence zero away from the origin but outward sphere flux $4\pi$ around it. The divergence theorem cannot be applied to the full enclosed ball because the field is undefined at the origin. On a shell excluding the origin, the outer flux is canceled by the inward-oriented inner boundary flux, exactly as the theorem requires.

Proper three-dimensional slicing on the graph slab

Let $D\subset\mathbb R^2$ be compact Jordan, let $a,b$ be continuous on $D$ with $a\le b$, and put $\Omega=\{(p,z):p\in D,\ a(p)\le z\le b(p)\}$. Choose a compact height interval $J$ of positive length containing every fiber. The solid is compact, and its boundary is contained in the two graphs and $\partial D\times J$. Each graph is Jordan null: subdivide a containing base box into squares of side $h$, and over each square meeting $D$ enclose the graph in a vertical box of height at most $2\omega(\sqrt2h)+h$, where $\omega$ is a common continuity modulus for $a,b$ on $D$. There are $O(h^{-2})$ such squares, so total cover volume tends to zero. The side set is null because a finite base cover of $\partial D$ with area below $\eta$ becomes a prism cover of volume at most $|J|\eta$. Thus $\Omega$ is Jordan measurable by the proved boundary criterion.

For continuous $h:\Omega\to\mathbb R$, define $g(p)=\int_{a(p)}^{b(p)}h(p,z)dz$. No ambient extension of $h$ is needed. On the overlap of two nearby fibers, uniform continuity of $h$ controls the integrand difference; on their unmatched ends, a bound $M$ for $|h|$ controls the error. In particular

$$|g(p)-g(q)|\le |J|\omega_h(\|p-q\|) +M\bigl(|a(p)-a(q)|+|b(p)-b(q)|\bigr)\longrightarrow0.$$

Hence $g$ is continuous on $D$. The zero extensions $h_0$ and $g_0$ to containing three- and two-dimensional boxes are Riemann integrable, by compact continuity and the Jordan boundary criterion. Every vertical section of $h_0$ is continuous on its fiber, zero outside it, and has at most the two endpoint jumps, so is integrable with integral $g_0(p)$. On any finite product grid, integrate the constant lower and upper cell bounds first in height and then over the base. They bound $\int g_0$ between exactly the three-dimensional lower and upper sums of $h_0$. Since those sums can have arbitrarily small gap, squeezing proves

$$\iiint_\Omega h\,dV=\iint_D\left(\int_{a(p)}^{b(p)}h(p,z)dz\right)dA.$$

This proves the actual slab slicing used in the divergence derivation. Allowing a zero-length fiber in this integration lemma does not remove the strict separation and flux-limit conditions in the graph-slab divergence proof.

Area definitions and changes of surface coordinates

For an injective regular $C^1$ patch on a compact Jordan parameter region, define its scalar area integral with density $f$ by $\int_D f(r)\|r_u\times r_v\|$, and its oriented flux by the area-vector integral above. These definitions agree across $C^1$ diffeomorphic parameter changes satisfying the proper two-dimensional substitution hypotheses. Indeed, for $(u,v)=\psi(s,t)$, bilinearity and antisymmetry of cross product give

$$(r\circ\psi)_s\times(r\circ\psi)_t =\det D\psi\,(r_u\times r_v)\circ\psi.$$

Taking norms supplies the absolute Jacobian for scalar area; keeping the sign supplies the induced orientation for flux. Apply the proved two-dimensional proper substitution. Finite compatible patches add along edges of parameter area zero. This is the precise surface-integral model being used, and does not invoke an unrestricted measure-theoretic surface-area theorem.

The full component cancellation in Stokes

Write $p=r_u$, $q=r_v$ and $F=(P,Q,R)$. In $(DFp)\cdot q-(DFq)\cdot p$, the diagonal derivatives cancel. The other terms collect as

$$(R_y-Q_z)(p_yq_z-p_zq_y) +(P_z-R_x)(p_zq_x-p_xq_z) +(Q_x-P_y)(p_xq_y-p_yq_x),$$

which is exactly $(\nabla\times F)\cdot(p\times q)$. Together with $C^2$ equality of $r_{uv},r_{vu}$ and the already proved Green identity, this closes the patch derivation component by component. The finite extension requires an actual finite decomposition with no interior overlap except zero-area edges, and every internal edge paired with the opposite induced direction.

Why the variable-limit derivative is valid

Locally fix a constant height $z_*$ and put $H(x,y,z)=\int_{z_*}^zP(x,y,w)dw$. Continuity of $P_x$ on a compact neighborhood makes the difference quotients for $H_x$ converge uniformly in $w$ to $P_x$, so $H_x=\int_{z_*}^zP_xdw$; the error is bounded by interval length times the uniform error. The fundamental theorem gives $H_z=P$. Apply the ordinary chain rule to $H(x,y,b(x,y))-H(x,y,a(x,y))$ to obtain the variable-limit formula in Section 3. The same argument applies to $Q$. Thus the graph-slab proof requires $a,b\in C^1$ on a neighborhood of the compact base, strictly $a<b$ there, $F\in C^1$ near the slab closure, and a base in the proved Green class. Replacing $a<b$ by boundary contact or using a new class of solids requires its own limiting estimate. The paraboloid exercise supplies such an estimate for that particular contact; it is not a general extension theorem.

Proof of the sphere area used above

For the sphere of radius $R>0$, put

$$s(\phi,\theta)=(R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi).$$

Direct differentiation gives $\|s_\phi\times s_\theta\|=R^2\sin\phi$, positive when $0<\phi<\pi$. On a compact truncated rectangle $\delta\le\phi\le\pi-\delta$, $-\pi+\eta\le\theta\le\pi-\eta$, where $0<\delta<\pi/3$ and $0<\eta<\pi/2$, this is a regular once-covered patch. Its area is

$$A_{\delta,\eta} =R^2(2\pi-2\eta)\int_\delta^{\pi-\delta}\sin\phi\,d\phi =2R^2(2\pi-2\eta)\cos\delta.$$

It remains to control the omitted surface area, rather than merely volume. Near either pole the sphere is the regular graph $z=\pm\sqrt{R^2-x^2-y^2}$. On the disk of radius $R\sin\delta$, its graph area factor is $R/\sqrt{R^2-x^2-y^2}\le1/\cos\delta\le2$. Each omitted cap therefore has area at most $2\pi R^2\sin^2\delta$, tending to zero. This bound uses the graph formula and planar disk area, not the sphere area being proved.

For the remaining angular strip, use the alternative nonsingular azimuth interval $[\pi-\eta,\pi+\eta]$, still with $\delta\le\phi\le\pi-\delta$. It is a regular patch whose area is at most $2\eta R^2\int_0^\pi\sin\phi\,d\phi=4\eta R^2$. Thus the total omitted area is bounded by $4\pi R^2\sin^2\delta+4\eta R^2\to0$. The finitely many shared patch edges have zero parameter area and do not alter the sum. Passing to the limit proves $\operatorname{Area}(S_R)=4\pi R^2$. Multiplying by the constant outward normal component $R$ proves the earlier position-field flux $4\pi R^3$ as well.

Why the particular spherical shell admits the divergence check

The shell in Exercise 4 can be obtained from the proved graph slabs with explicit limits. Write $\rho=\sqrt{x^2+y^2}$ and choose $0<\varepsilon<1/4$. Over $\rho\le1-\varepsilon$, take the two slabs

$$\sqrt{1-\rho^2}\le z\le\sqrt{4-\rho^2},\qquad -\sqrt{4-\rho^2}\le z\le-\sqrt{1-\rho^2}.$$

Over $1+\varepsilon\le\rho\le2-\varepsilon$, take the slab $-\sqrt{4-\rho^2}\le z\le\sqrt{4-\rho^2}$. Every top and bottom is $C^1$ on a neighborhood of its compact base and they are strictly separated. The disk bases satisfy the proved Green hypotheses; divide the annular base into four quadrants, each with continuous upper/lower and left/right graph bounds, and cancel the radial edges. Thus each displayed truncated solid is in the proved finite class.

For a $C^1$ field on a neighborhood of the closed shell, the field and divergence are bounded. The omitted cylindrical bands near $\rho=1,2$ have volume tending to zero, bounded by height 4 times their planar annulus areas. Omitted spherical boundary belts also have area tending to zero. To see this independently of a whole-sphere area formula, parameterize an equatorial belt on radius $a$ by $(\sqrt{a^2-z^2}\cos\theta,\sqrt{a^2-z^2}\sin\theta,z)$; its cross-product norm is $a$. A belt $z_1\le z\le z_2$ has area $2\pi a(z_2-z_1)$; in particular $|z|\le h$ has area $4\pi a h$. The inner- and outer-equatorial half-heights omitted near $\rho=1,2$ are at most $\sqrt{1-(1-\varepsilon)^2}$ and $\sqrt{4-(2-\varepsilon)^2}$, both tending to zero. In addition, omitting the band $1-\varepsilon<\rho<1+\varepsilon$ removes two outer-sphere latitude belts. Put $\Delta z=\sqrt{4-(1-\varepsilon)^2}-\sqrt{4-(1+\varepsilon)^2}\to0$. Their combined area is $8\pi\Delta z\to0$. These account for all the spherical boundary omissions.

The artificial outer cylinder at $\rho=2-\varepsilon$ has vanishing height and area. The cylinders at $\rho=1-\varepsilon$ and $\rho=1+\varepsilon$ have opposing normals. Their parameterized flux integrands converge uniformly on fixed bounded height ranges to the same integrand at $\rho=1$; the small missing central height and endpoint height differences tend to zero. Boundedness and uniform continuity therefore make their combined flux tend to zero. Taking limits in the finite graph-slab identities proves divergence for this particular shell and every such regular field. Its radial example is regular on a neighborhood of the shell, although it is singular in the excluded central ball. This supplies the needed special-domain proof without asserting an unproved extension to all piecewise smooth solids.

5 Graded practice

E1 Foundation. Find upward flux of $F=(0,0,1)$ through $z=x^2+y^2$ over a disk of radius $R>0$. Is it the scalar surface area?

E2 Stokes. For $F=(-y,x,0)$, compute circulation around the counterclockwise circle $x^2+y^2=4$, $z=3$, using any admissible spanning surface.

E3 Slab divergence. For $F=(x,y,z)$ and the solid below $z=1-x^2-y^2$ and above $z=0$, find total outward flux and check the bottom contribution.

E4 Singularity. Apply the divergence theorem to $F=x/\|x\|^3$ on the shell $1\le\|x\|\le2$, keeping both boundary orientations.

E5 Orientation proof. Explain why the common edge of two consistently oriented adjacent patches cancels in Stokes but the scalar surface areas add rather than cancel.

6 Full solutions

S1. The graph flux formula gives integrand 1 over the projected disk, so flux is $\pi R^2$. Scalar area has integrand $\sqrt{1+4x^2+4y^2}$ and is larger for $R>0$. Flux measures a projected oriented quantity, not total area.

S2. Curl is $(0,0,2)$. The flat upward disk at $z=3$ gives $2\cdot\pi\cdot2^2=8\pi$. The boundary direction is counterclockwise when viewed from above, consistent with that normal.

S3. For the graph-slab derivation, first use disks of radius $1-\varepsilon$, where top and bottom are strictly separated, then let $\varepsilon\downarrow0$. The bounded smooth integrands converge in the volume and cap integrals, and the shrinking side has height $1-(1-\varepsilon)^2\to0$, so its flux tends to zero. Divergence is 3. The volume is $2\pi\int_0^1(1-r^2)rdr=\pi/2$, so total flux is $3\pi/2$. On the bottom, outward normal is $(0,0,-1)$ and $z=0$, so flux is zero. On the cap, the area-vector formula gives $2x^2+2y^2+(1-x^2-y^2)=1+r^2$, integrating to $3\pi/2$.

S4. Divergence is zero throughout the shell. On radius 2, field magnitude is $1/4$ and outward area is $16\pi$, giving $4\pi$. At the inner sphere the shell's outward normal points toward the origin, opposite the radial field, giving $-4\pi$. Total is zero, agreeing with the volume integral.

S5. Boundary orientation is induced by the surface orientation, and the two patches place their common edge on opposite sides of their interiors. Therefore their edge tangent directions oppose, and oriented line integrals negate. Scalar area uses the norm of the cross product, which is nonnegative and does not change when orientation reverses, so areas add.

Summary

Area vectors prevent double counting and fix flux signs. Stokes on a regular patch is Green's theorem plus a chain-rule identity; divergence on a graph slab is one-variable integration plus planar flux. Finite compatible decompositions extend both by internal-boundary cancellation. Singularities and missing side boundaries are concrete obstructions, not technical details to ignore.