Stokes' and Divergence Theorems

Technique: reduce to Green's theorem via the change-of-variables/pullback of the line and surface integrals, then a direct componentwise computation. Write $\mathbf{F} = (P, Q, R)$ and $\mathbf{r}(u,v) = (x(u,v), y(u,v), z(u,v))$, all $C^1$ (we treat the smooth case; piecewise smooth follows by summing over pieces, since both sides of the claimed identity are additive over a decomposition of $S$ into finitely many smooth pieces whose shared internal boundary curves are traversed in opposite directions by adjacent pieces and hence cancel — this is exactly the cancellation used in Green's theorem for regions with holes or unions of simple regions).

Step 1: Reduce the line integral over $\partial S$ to a line integral over $\partial D$. By definition of the line integral of a vector field along a parametrized curve, and since $\partial S = \mathbf{r}(\partial D)$ with the induced orientation matching $\mathbf{r}$ applied to the counterclockwise orientation of $\partial D$,

Call the integrand's coefficient of $du$ as $A(u,v) := P x_u + Q y_u + R z_u$ and the coefficient of $dv$ as $B(u,v) := P x_v + Q y_v + R z_v$, so the right side is $\oint_{\partial D} (A\, du + B\, dv)$.

Step 2: Apply Green's theorem in the $(u,v)$-plane. Since $\partial D$ is a piecewise smooth simple closed curve oriented counterclockwise and $A, B$ are $C^1$ functions of $(u,v)$ (as compositions and products of $C^1$ functions, using that $P,Q,R$ are $C^1$ and $x,y,z$ are $C^1$), Green's Theorem applies directly:

Step 3: Compute $B_u - A_v$ and show it equals $(\nabla \times \mathbf{F}) \cdot (\mathbf{r}_u \times \mathbf{r}_v)$. Differentiate using the chain rule, remembering $P,Q,R$ are functions of $(x,y,z) = \mathbf{r}(u,v)$:

Carrying this out fully for all three terms of $B$ and subtracting the analogous expansion of $A_v$, every second-derivative term ($P x_{uv}$, etc.) is symmetric in $u,v$ (since $x,y,z \in C^2$ is not even needed if we only assume $\mathbf{r}\in C^1$ and work formally — in fact for the standard proof one assumes $\mathbf{r} \in C^2$ so mixed partials are equal by Clairaut's theorem) and cancels between $B_u$ and $A_v$. What survives is exactly:

The right side is precisely the dot product

since $\mathbf{r}_u \times \mathbf{r}_v = (y_u z_v - z_u y_v,\ z_u x_v - x_u z_v,\ x_u y_v - y_u x_v)$ and $\nabla \times \mathbf{F} = (R_y - Q_z, P_z - R_x, Q_x - P_y)$ by definition. This is verified by direct expansion of the dot product, term by term, matching each of the three products above.

Step 4: Recognize the surface integral of the curl. By definition of the flux integral,

Conclusion. Chaining Steps 1–4,

For a piecewise smooth $S$ built from finitely many smooth patches $S_1,\dots,S_k$ glued along curves, apply the above to each $S_i$ and sum; each internal gluing curve is traversed twice with opposite orientation by the induced orientations of adjacent patches (by consistency of the right-hand rule across the gluing), so those line integral contributions cancel, leaving exactly $\oint_{\partial S}$ over the true outer boundary. $\blacksquare$

Technique: reduce to three separate one-variable applications of the Fundamental Theorem of Calculus via Fubini's theorem, exactly as Green's theorem is proved for simple regions, then patch together via additivity for the general (piecewise-simple) case. By linearity of both sides in $\mathbf{F} = (P,Q,R)$, it suffices to prove the three separate identities

and add them. We prove the first; the other two are identical after permuting coordinate names.

Step 1: Use $z$-simplicity. Since $\Omega$ is $z$-simple, write

for some region $D$ in the $xy$-plane and $C^1$ functions $\phi_1 \le \phi_2$. The boundary $\partial\Omega$ then consists of: the top $S_2 = \{z = \phi_2(x,y)\}$ with outward normal pointing upward-ish, the bottom $S_1 = \{z=\phi_1(x,y)\}$ with outward normal pointing downward-ish, and a lateral part $S_0$ (possibly empty) consisting of vertical line segments (where the boundary of $D$ is traced with $z$ ranging over $[\phi_1,\phi_2]$), on which the outward normal is horizontal, so $\mathbf{k}\cdot\mathbf{n} = 0$ there.

Step 2: Compute the right side by Fubini and the Fundamental Theorem of Calculus.

using the Fundamental Theorem of Calculus in the innermost integral (valid since $R$ is $C^1$, so $R_z$ is continuous in $z$ for fixed $(x,y)$), and Fubini's theorem to justify iterating the triple integral as an iterated double-then-single integral (valid since $R_z$ is continuous on the bounded region $\Omega$, hence integrable).

Step 3: Compute the left side directly. Parametrize $S_2$ by $(x,y) \mapsto (x,y,\phi_2(x,y))$ over $D$. Then $\mathbf{r}_x \times \mathbf{r}_y = (-\partial_x\phi_2, -\partial_y\phi_2, 1)$, which has positive $z$-component, matching the outward (upward) orientation on the top. So

Parametrize $S_1$ the same way by $(x,y)\mapsto(x,y,\phi_1(x,y))$; this gives normal $(-\partial_x\phi_1,-\partial_y\phi_1,1)$ with positive $z$-component, but the outward normal on the bottom surface points downward, so the outward-oriented flux is the negative of the one computed from this parametrization:

On the lateral surface $S_0$, the outward unit normal is horizontal (its $z$-component is $0$, since the lateral surface is generated by moving vertically along the boundary of $D$), so $\mathbf{k}\cdot\mathbf{n} = 0$ pointwise and

Summing,

Step 4: Conclude. Comparing Steps 2 and 3, both equal the same double integral over $D$, so

By the symmetric argument using $y$-simplicity and $x$-simplicity of $\Omega$ (relabeling coordinates), the analogous identities hold for $Q\mathbf{j}$ and $P\mathbf{i}$. Adding all three:

General $\Omega$. If $\Omega$ is only piecewise-simple, partition it into finitely many simple pieces $\Omega_1,\dots,\Omega_m$ (each simultaneously $x$-, $y$-, $z$-simple) glued along shared internal boundary surfaces. Apply the identity to each piece and sum. Each internal gluing surface receives opposite outward normals from the two adjacent pieces (since "outward from $\Omega_i$" is "inward to $\Omega_j$" on their shared face), so those flux contributions cancel in pairs, and the sum of the volume integrals is $\iiint_\Omega \nabla\cdot\mathbf{F}\,dV$ by additivity of the integral over the partition. What remains on the left is exactly $\iint_{\partial\Omega}\mathbf{F}\cdot d\mathbf{S}$. $\blacksquare$