Vector calculus and conservation laws

A flux field describes transport through oriented surfaces. Conservation converts that boundary transport into a local differential equation. Line integrals, surface integrals and the divergence and Stokes theorems express different aspects of this relationship, with orientation and domain assumptions determining the signs and validity.

1 Line integrals and potentials

For a piecewise $C^1$ curve $r:[a,b]\to\mathbb R^3$ and continuous vector field $F$, its oriented line integral is

$$\int_CF\cdot dr=\int_a^bF(r(t))\cdot r'(t)dt.$$

Reversing orientation reverses the sign. Orientation-preserving regular reparameterizations leave it unchanged by the one-variable substitution rule. A field is conservative if $F=\nabla\phi$ for a single-valued differentiable potential on the domain.

Gradient theorem. If $\phi$ is $C^1$ near the curve, then $\int_C\nabla\phi\cdot dr=\phi(r(b))-\phi(r(a))$.

Proof. The chain rule makes the integrand $d\phi(r(t))/dt$. Apply the fundamental theorem on each smooth piece and telescope endpoint values. $\square$

Thus closed-loop integrals of conservative fields vanish. If $\phi\in C^2$, its gradient has zero curl by equality of mixed partials. The converse needs a domain condition: a $C^1$ curl-free field on an open domain star shaped about one point in $\mathbb R^3$ has a potential; a full proof for this explicit sufficient domain condition is given below. Curl-free alone is not enough on a domain with holes.

For $F=(-y/(x^2+y^2),x/(x^2+y^2),0)$ off the $z$ axis, curl is zero, yet its integral around the unit circle is $2\pi$. A global single-valued angular potential does not exist on that domain.

2 Surface integrals divergence and curl

A parameterized surface $r(u,v)$ has oriented area vector $(r_u\times r_v)du\,dv$ where the cross product is nonzero. Its flux is

$$\iint_SF\cdot n\,dS=\iint_DF(r(u,v))\cdot(r_u\times r_v)du\,dv.$$

Changing orientation negates the flux. Scalar area integrals instead use $\|r_u\times r_v\|$.

For $F=(P,Q,R)$,

$$\nabla\cdot F=P_x+Q_y+R_z,$$

and

$$\nabla\times F=(R_y-Q_z,\ P_z-R_x,\ Q_x-P_y).$$

Divergence measures net outward flux density; curl records local circulation density with the right-hand orientation convention. They are different operations and have different output types.

3 Boundary theorems with hypotheses

Divergence theorem. For a rectangular box, or a bounded solid $\Omega$ with an actual finite compatible graph-slab decomposition satisfying the conditions proved in the parameterized-boundary chapter, and a $C^1$ field on an open neighborhood of its closure,

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

with outward normal. On a rectangular box, the proof is direct: opposite $x$ faces contribute $\int\int[P(b,y,z)-P(a,y,z)]dy\,dz=\iiint P_x$ by the fundamental theorem and Fubini. Add the analogous $y$ and $z$ pairs. The stated finite extension follows by adding those proved pieces and canceling internal faces; arbitrary piecewise smooth regions are not included merely by that phrase. Singularities inside the solid cannot be ignored.

Stokes theorem. For a compact oriented surface $S$ given by an actual finite compatible decomposition into the injective $C^2$ regular patches over the Green regions specified below, with compatible oriented boundary and $F\in C^1$ on an open neighborhood of $S$,

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

where the boundary orientation follows the right-hand rule from the surface normal. A planar version is Green's theorem: for a plane region $D$ in the explicit elementary or finite-decomposition class proved below, with its positive boundary orientation and $C^1$ fields on a neighborhood of its closure,

$$\oint_{\partial D}P\,dx+Q\,dy=\iint_D(Q_x-P_y)dA.$$

For an $x$-simple region, integrating $-P_y$ gives the difference between top and bottom boundary contributions; the analogous $y$-simple argument handles $Q_x$. Decomposition extends the identity when internal boundary pieces cancel.

Worked check. For $F=(x,y,z)$, divergence is 3. The outward flux through the sphere of radius $R>0$ is $3(4\pi R^3/3)=4\pi R^3$. Directly, $F\cdot n=R$ and area is $4\pi R^2$, giving the same value.

4 From integral conservation to a local equation

Let $c(x,t)$ be concentration, $j(x,t)$ outward flux per unit area, and $s(x,t)$ local production rate. On any fixed control volume,

$$\frac d{dt}\iiint_\Omega c\,dV=-\iint_{\partial\Omega}j\cdot n\,dS+\iiint_\Omega s\,dV.$$

If the fields are sufficiently smooth to differentiate under the integral and apply divergence, this becomes $\iiint_\Omega(c_t+\nabla\cdot j-s)dV=0$. If the integrand is continuous and the identity holds for every small control volume, then

$$c_t+\nabla\cdot j=s.$$

Indeed, a nonzero continuous value would preserve one sign on a small neighborhood, giving a nonzero integral there. With $j=-K\nabla c$, the equation is $c_t=\nabla\cdot(K\nabla c)+s$. When $K=kI$ with constant scalar $k$, it simplifies to $c_t=k\Delta c+s$. Spatially variable coefficients and anisotropic tensors must remain inside the divergence.

A proved domain condition for a potential

In this chapter use the following explicit sufficient domain condition. An open set $U$ is star shaped about $a\in U$ if every segment $a+t(x-a)$, $0\le t\le1$, stays in $U$ for $x\in U$. For a $C^1$ curl-free field on such a set define

$$\phi(x)=\int_0^1F(a+t(x-a))\cdot(x-a)\,dt.$$

For a fixed $x$, its compact segment has a small neighborhood inside $U$. Continuity of the first derivatives there justifies differentiating this fixed-interval integral: difference quotients converge uniformly in $t$ by the coordinate mean value theorem and uniform continuity. The $j$th partial derivative of the integrand is $F_j+t\sum_i(x_i-a_i)\partial_jF_i$. Zero curl gives $\partial_jF_i=\partial_iF_j$, so this is $d[tF_j(a+t(x-a))]/dt$. Integration gives $\partial_j\phi=F_j(x)$. Hence $F=\nabla\phi$. This proves the converse on star-shaped domains. No proof for every simply connected domain is asserted here; that broader topological extension would require an additional path/homotopy argument. The punctured-axis example fails even the existence of any global potential, by its nonzero loop integral.

Green's theorem on an explicit finite class

For the boundary results below, an elementary plane region has a compact Jordan closure, a positively oriented piecewise $C^1$ boundary, and both a vertical-strip description $\alpha\le x\le\beta$, $g_-(x)\le y\le g_+(x)$ and a horizontal-strip description $\gamma\le y\le\delta$, $h_-(y)\le x\le h_+(y)$ with continuous boundary graphs. Endpoint vertical or horizontal tangents are allowed when the boundary arcs themselves are piecewise $C^1$ under regular parameterizations. Finite regions assembled from such pieces are admitted when interiors are disjoint and every shared boundary arc is paired with the opposite orientation. Rectangles and disks are examples. Assume $P,Q$ are $C^1$ on an open neighborhood of the closure.

The fundamental theorem on every vertical interval gives

$$-\iint_D P_y\,dA=\int_\alpha^\beta[P(x,g_-(x))-P(x,g_+(x))]dx=\oint_{\partial D}P\,dx.$$

The bottom boundary runs to the right and the top to the left; any vertical end pieces have $dx=0$. Similarly, horizontal intervals give

$$\iint_D Q_x\,dA=\int_\gamma^\delta[Q(h_+(y),y)-Q(h_-(y),y)]dy=\oint_{\partial D}Q\,dy.$$

These iterated integrals equal the Jordan integrals: extend the continuous integrand by zero to a containing rectangle, bound the iterated value between its cellwise lower and upper sums, and use uniform continuity on interior cells plus the vanishing total area of cells meeting the Jordan boundary. If an endpoint tangent is vertical in a graph parameter, first truncate that arc away from its endpoints; boundedness of $P,Q$ times the omitted arc lengths makes the endpoint contribution vanish. Adding the identities proves Green. On the stated finite decompositions, interior line integrals cancel and area integrals add, proving the extension. The flux form follows by replacing $(P,Q)$ in Green by $(-V,U)$: $U\nu_x+V\nu_y=U y'-V x'$ along a positively oriented unit-speed boundary.

Conservation with explicit differentiability

For a fixed compact Jordan control volume and a compact time interval, assume $c,c_t$ are continuous on a neighborhood of their product, and $j$ and its first spatial derivatives are continuous there. The time difference quotients of $c$ converge uniformly in space to $c_t$, by the fundamental theorem in time and uniform continuity of $c_t$. The integral error is bounded by the volume times that uniform error, so $d\int c/dt=\int c_t$. Apply the proved divergence theorem on boxes, or on one of the explicitly admissible solids from the parameterized-boundary chapter. If $c_t+\nabla\cdot j-s$ is continuous and its integral is zero on every sufficiently small box, a positive value would remain positive and bounded away from zero on some such box, contradicting its zero integral; the negative case is identical. This supplies the local conservation equation under the stated hypotheses.

5 Graded practice

E1 Foundation. Evaluate the line integral of $F=(2x,2y,2z)$ from $(0,0,0)$ to $(1,2,3)$ along any piecewise smooth path.

E2 Green. Compute $\oint_C(-y\,dx+x\,dy)$ around the counterclockwise circle of radius $a>0$.

E3 Divergence. Find outward flux of $F=(x^2,y,z)$ through the unit cube $[0,1]^3$.

E4 Conservation. If $j=-k\nabla c$ for constant $k>0$ and no source, derive the equation and show that zero boundary flux conserves total mass.

6 Full solutions

S1. Potential is $\phi=x^2+y^2+z^2$, so the integral is $14-0=14$. The path is irrelevant because a potential is globally defined on all of space.

S2. $Q_x-P_y=1-(-1)=2$. Green gives $2\pi a^2$. Direct parameterization $x=a\cos t$, $y=a\sin t$ produces integrand $a^2dt$, confirming the sign.

S3. Divergence is $2x+2$. Its volume integral is $\int_0^1(2x+2)dx=3$. Opposite-face contributions are 1 in each coordinate direction, also totaling 3.

S4. Substitution gives $c_t=k\Delta c$. Integrating and applying divergence gives $dM/dt=-\iint_{\partial\Omega}j\cdot n=0$ under the stated boundary condition. A zero gradient everywhere is not required; internal redistribution can occur while boundary flux vanishes.

Summary

Line and surface integrals carry orientation. Potentials make line integrals depend only on endpoints; divergence and curl connect boundary data to interior derivatives under explicit regularity assumptions. Conservation plus a constitutive flux law yields a local evolution equation, whose coefficients must retain their spatial and tensor structure.