Surface Integrals & the Divergence Theorem

A parameterized surface in $\mathbb{R}^3$ is a $C^1$ function $\mathbf{r}: D^* \to \mathbb{R}^3$, where $D^* \subseteq \mathbb{R}^2$ is a bounded, closed region (the parameter domain). We write $\mathbf{r}(u, v) = (x(u,v),\; y(u,v),\; z(u,v))$.

The surface is regular (or smooth) if the cross product $\mathbf{r}_u \times \mathbf{r}_v \neq \mathbf{0}$ for all $(u, v)$ in the interior of $D^*$. This ensures the tangent plane is well-defined at every point — the two tangent vectors are linearly independent, and the surface has no cusps or self-intersections locally.

For vectors $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{b} = (b_1, b_2, b_3)$ in $\mathbb{R}^3$, the cross product is:

$\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1).$

Equivalently, using the determinant mnemonic:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.$

Key properties:

1. Orthogonality: $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0$ and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0$ — the cross product is perpendicular to both factors.
2. Area: $\|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\|\,\|\mathbf{b}\|\sin\theta$ — the magnitude equals the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$.
3. Anti-commutativity: $\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b})$ — swapping the factors reverses the direction.

For a regular parameterized surface $\mathbf{r}: D^* \to \mathbb{R}^3$, the surface area element is:

$dS = \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv.$

The surface area of $S$ is:

$\text{Area}(S) = \iint_{D^*} \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv.$

The factor $\|\mathbf{r}_u \times \mathbf{r}_v\|$ is the area of the infinitesimal parallelogram spanned by the tangent vectors — it measures how much the parameterization stretches area at each point.

The Jacobian matrix of the parameterization is the $3 \times 2$ matrix $J_{\mathbf{r}} = [\mathbf{r}_u \mid \mathbf{r}_v]$, with columns $\mathbf{r}_u$ and $\mathbf{r}_v$. The Gram matrix is the $2 \times 2$ matrix:

$J_{\mathbf{r}}^T J_{\mathbf{r}} = \begin{pmatrix} \mathbf{r}_u \cdot \mathbf{r}_u & \mathbf{r}_u \cdot \mathbf{r}_v \\ \mathbf{r}_v \cdot \mathbf{r}_u & \mathbf{r}_v \cdot \mathbf{r}_v \end{pmatrix} = \begin{pmatrix} \|\mathbf{r}_u\|^2 & \mathbf{r}_u \cdot \mathbf{r}_v \\ \mathbf{r}_u \cdot \mathbf{r}_v & \|\mathbf{r}_v\|^2 \end{pmatrix}.$

By the Lagrange identity:

$\|\mathbf{r}_u \times \mathbf{r}_v\|^2 = \|\mathbf{r}_u\|^2 \|\mathbf{r}_v\|^2 - (\mathbf{r}_u \cdot \mathbf{r}_v)^2 = \det(J_{\mathbf{r}}^T J_{\mathbf{r}}).$

So $dS = \sqrt{\det(J_{\mathbf{r}}^T J_{\mathbf{r}})}\,du\,dv$. This is the natural generalization of the 1D Jacobian $|\det J_\phi|$ from Topic 14: when the map goes from $\mathbb{R}^k$ to $\mathbb{R}^n$ with $k \le n$, the area scaling factor is $\sqrt{\det(J^T J)}$.

Parameterize the sphere of radius $R$ using spherical coordinates $(\theta, \phi)$ with $\theta \in [0, 2\pi]$ (azimuthal) and $\phi \in [0, \pi]$ (polar):

$\mathbf{r}(\theta, \phi) = (R\sin\phi\cos\theta,\; R\sin\phi\sin\theta,\; R\cos\phi).$

The tangent vectors are:

$\mathbf{r}_\theta = (-R\sin\phi\sin\theta,\; R\sin\phi\cos\theta,\; 0),$

$\mathbf{r}_\phi = (R\cos\phi\cos\theta,\; R\cos\phi\sin\theta,\; -R\sin\phi).$

We compute the cross product component by component:

$\mathbf{r}_\theta \times \mathbf{r}_\phi = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -R\sin\phi\sin\theta & R\sin\phi\cos\theta & 0 \\ R\cos\phi\cos\theta & R\cos\phi\sin\theta & -R\sin\phi \end{vmatrix}.$

The $\hat{\mathbf{i}}$-component: $(R\sin\phi\cos\theta)(-R\sin\phi) - (0)(R\cos\phi\sin\theta) = -R^2\sin^2\phi\cos\theta$.

The $\hat{\mathbf{j}}$-component: $(0)(R\cos\phi\cos\theta) - (-R\sin\phi\sin\theta)(-R\sin\phi) = -R^2\sin^2\phi\sin\theta$.

The $\hat{\mathbf{k}}$-component: $(-R\sin\phi\sin\theta)(R\cos\phi\sin\theta) - (R\sin\phi\cos\theta)(R\cos\phi\cos\theta) = -R^2\sin\phi\cos\phi$.

So $\mathbf{r}_\theta \times \mathbf{r}_\phi = -R^2\sin\phi\,(R\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi)$. Wait — let us be more careful. Factoring:

$\mathbf{r}_\theta \times \mathbf{r}_\phi = (-R^2\sin^2\phi\cos\theta,\; -R^2\sin^2\phi\sin\theta,\; -R^2\sin\phi\cos\phi).$

$= -R^2\sin\phi\,(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi).$

The vector $(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi)$ is the outward unit normal $\hat{\mathbf{n}}$ on the sphere, so $\mathbf{r}_\theta \times \mathbf{r}_\phi = -R^2\sin\phi\,\hat{\mathbf{n}}$. (The negative sign means this particular parameterization gives the inward normal; we can reverse it by swapping the order of the cross product.) The magnitude is:

$\|\mathbf{r}_\theta \times \mathbf{r}_\phi\| = R^2\sin\phi \quad (\text{since } \sin\phi \ge 0 \text{ for } \phi \in [0, \pi]).$

The surface area is:

$\text{Area}(S^2_R) = \int_0^{2\pi}\int_0^\pi R^2\sin\phi\,d\phi\,d\theta = R^2 \cdot 2\pi \cdot [-\cos\phi]_0^\pi = R^2 \cdot 2\pi \cdot 2 = 4\pi R^2.$

The cylinder of radius $R$ and height $h$ is parameterized by:

$\mathbf{r}(\theta, z) = (R\cos\theta,\; R\sin\theta,\; z), \quad \theta \in [0, 2\pi],\; z \in [0, h].$

The tangent vectors are $\mathbf{r}_\theta = (-R\sin\theta,\; R\cos\theta,\; 0)$ and $\mathbf{r}_z = (0, 0, 1)$. The cross product is:

$\mathbf{r}_\theta \times \mathbf{r}_z = (R\cos\theta,\; R\sin\theta,\; 0),$

with magnitude $\|\mathbf{r}_\theta \times \mathbf{r}_z\| = R$. The lateral surface area is:

$\text{Area} = \int_0^{2\pi}\int_0^h R\,dz\,d\theta = 2\pi R h.$

A surface defined as the graph of a function $z = g(x, y)$ over a region $D \subseteq \mathbb{R}^2$ has the natural parameterization $\mathbf{r}(x, y) = (x, y, g(x,y))$. The tangent vectors are:

$\mathbf{r}_x = (1, 0, g_x), \quad \mathbf{r}_y = (0, 1, g_y).$

The cross product is:

$\mathbf{r}_x \times \mathbf{r}_y = (-g_x, -g_y, 1),$

with magnitude $\|\mathbf{r}_x \times \mathbf{r}_y\| = \sqrt{1 + g_x^2 + g_y^2}$. So the surface area element for a graph is:

$dS = \sqrt{1 + g_x^2 + g_y^2}\,dA,$

where $dA = dx\,dy$. This is a formula worth memorizing: for a graph surface, the area element is the Euclidean area $dA$ scaled by the factor $\sqrt{1 + \|\nabla g\|^2}$, which measures how much the surface tilts away from horizontal.

The surface area $\text{Area}(S) = \iint_{D^*} \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv$ is independent of the parameterization. If $\tilde{\mathbf{r}} = \mathbf{r} \circ \phi$ is a reparameterization via a $C^1$ diffeomorphism $\phi: \tilde{D}^* \to D^*$, then $\text{Area}$ computed via $\tilde{\mathbf{r}}$ equals $\text{Area}$ computed via $\mathbf{r}$.

By the chain rule, the Jacobian of $\tilde{\mathbf{r}} = \mathbf{r} \circ \phi$ is $J_{\tilde{\mathbf{r}}} = J_{\mathbf{r}} \cdot J_\phi$, where $J_\phi$ is the $2 \times 2$ Jacobian of the reparameterization. The tangent vectors transform as:

$\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t = (\mathbf{r}_u \times \mathbf{r}_v) \cdot \det J_\phi.$

Taking magnitudes and applying the change of variables theorem (Topic 14):

$\iint_{\tilde{D}^*} \|\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t\|\,ds\,dt = \iint_{\tilde{D}^*} \|\mathbf{r}_u \times \mathbf{r}_v\| \cdot |\det J_\phi|\,ds\,dt = \iint_{D^*} \|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv.$

The $|\det J_\phi|$ from the cross product formula is absorbed by the $|\det J_\phi|^{-1}$ from the change-of-variables substitution, leaving the original integral. $\square$

Let $S$ be a regular parameterized surface $\mathbf{r}: D^* \to \mathbb{R}^3$, and let $f: S \to \mathbb{R}$ be continuous. The scalar surface integral of $f$ over $S$ is:

$\iint_S f\,dS = \iint_{D^*} f(\mathbf{r}(u,v))\,\|\mathbf{r}_u \times \mathbf{r}_v\|\,du\,dv.$

This is a double Riemann integral (Topic 13) of the composite function $(u,v) \mapsto f(\mathbf{r}(u,v)) \cdot \|\mathbf{r}_u \times \mathbf{r}_v\|$ over the parameter domain $D^*$.

The scalar surface integral $\iint_S f\,dS$ is independent of the parameterization, including orientation. The proof is identical to Proposition 1: the $|\det J_\phi|$ from the area element cancels with the $|\det J_\phi|^{-1}$ from the change of variables in the double integral. The absolute value ensures the result holds regardless of whether the reparameterization preserves or reverses orientation.

Let $S$ be the upper hemisphere of radius $R$ (i.e., $x^2 + y^2 + z^2 = R^2$ with $z \ge 0$) and let $f(x,y,z) = z$ be the density. From Example 1, we use the spherical parameterization with $\phi \in [0, \pi/2]$ (upper hemisphere only) and $\|\mathbf{r}_\theta \times \mathbf{r}_\phi\| = R^2\sin\phi$. On the sphere, $z = R\cos\phi$, so:

$\iint_S z\,dS = \int_0^{2\pi}\int_0^{\pi/2} (R\cos\phi)(R^2\sin\phi)\,d\phi\,d\theta = R^3 \int_0^{2\pi} d\theta \int_0^{\pi/2} \sin\phi\cos\phi\,d\phi.$

The inner integral: $\int_0^{\pi/2} \sin\phi\cos\phi\,d\phi = \frac{1}{2}\int_0^{\pi/2}\sin(2\phi)\,d\phi = \frac{1}{2}\left[-\frac{\cos(2\phi)}{2}\right]_0^{\pi/2} = \frac{1}{2}\cdot\frac{1+1}{2} = \frac{1}{2}$.

So $\iint_S z\,dS = R^3 \cdot 2\pi \cdot \frac{1}{2} = \pi R^3$.

The average value of $f$ over $S$ is:

$\bar{f} = \frac{1}{\text{Area}(S)}\iint_S f\,dS,$

directly analogous to $\bar{f} = \frac{1}{b-a}\int_a^b f(x)\,dx$ from single-variable calculus (Topic 7) and $\bar{f} = \frac{1}{L(C)}\int_C f\,ds$ for curves (Topic 15, Example 5). For the hemispherical shell with $f = z$, the average height is $\bar{z} = \frac{\pi R^3}{2\pi R^2} = \frac{R}{2}$ — the average height on the hemisphere is half the radius, which matches geometric intuition (the hemisphere is “top-heavy” in the $z$-direction but the area element weights the equatorial region more heavily).

A surface $S$ is orientable if it admits a continuous unit normal vector field $\hat{\mathbf{n}}: S \to \mathbb{R}^3$ with $\|\hat{\mathbf{n}}\| = 1$ and $\hat{\mathbf{n}}$ perpendicular to the tangent plane at every point. An oriented surface is an orientable surface together with a specific choice of $\hat{\mathbf{n}}$.

For a parameterized surface, the two orientations correspond to $\hat{\mathbf{n}} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\|\mathbf{r}_u \times \mathbf{r}_v\|}$ and $\hat{\mathbf{n}} = -\frac{\mathbf{r}_u \times \mathbf{r}_v}{\|\mathbf{r}_u \times \mathbf{r}_v\|}$.

Not every surface is orientable. The Mobius strip is the classic counterexample: if you start with a normal vector and slide it continuously around the strip, it returns pointing the opposite way. There is no globally consistent choice of “inside” and “outside.”

Two standard conventions govern orientation:

1. Closed surfaces (surfaces that enclose a volume, like a sphere or cube): the outward-pointing normal is the positive orientation. Flux with the outward normal measures net outflow.

2. Surfaces with boundary (for Stokes’ theorem): the right-hand rule determines the orientation. If you curl the fingers of your right hand in the direction of traversal along the boundary curve $C$, your thumb points in the direction of $\hat{\mathbf{n}}$. Equivalently: walking along $C$ with $\hat{\mathbf{n}}$ pointing up from your head, the surface is on your left.

Let $S$ be an oriented surface parameterized by $\mathbf{r}: D^* \to \mathbb{R}^3$ (with the orientation given by $\mathbf{r}_u \times \mathbf{r}_v$), and let $\mathbf{F}: S \to \mathbb{R}^3$ be a continuous vector field. The flux integral (or surface integral of a vector field) is:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_{D^*} \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,du\,dv.$

Equivalently, writing $d\mathbf{S} = \hat{\mathbf{n}}\,dS$:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S (\mathbf{F} \cdot \hat{\mathbf{n}})\,dS.$

The flux measures the net flow of $\mathbf{F}$ through $S$ in the direction of $\hat{\mathbf{n}}$. Positive flux means net flow in the $\hat{\mathbf{n}}$ direction; negative flux means net flow opposite to $\hat{\mathbf{n}}$.

Let $\mathbf{F} = (0, 0, z)$ (a vertical field, stronger at greater heights) and let $S$ be the upper hemisphere of the unit sphere ($R = 1$) with the outward normal. From Example 1, $\mathbf{r}_\theta \times \mathbf{r}_\phi = -\sin\phi\,(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi)$.

Since we want the outward normal, we use $\mathbf{r}_\phi \times \mathbf{r}_\theta = \sin\phi\,(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi)$. On the sphere, $z = \cos\phi$, so $\mathbf{F}(\mathbf{r}(\theta,\phi)) = (0, 0, \cos\phi)$.

$\mathbf{F} \cdot (\mathbf{r}_\phi \times \mathbf{r}_\theta) = (0, 0, \cos\phi) \cdot \sin\phi\,(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi) = \sin\phi\cos^2\phi.$

The flux is:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^{2\pi}\int_0^{\pi/2} \sin\phi\cos^2\phi\,d\phi\,d\theta = 2\pi \int_0^{\pi/2} \sin\phi\cos^2\phi\,d\phi.$

With the substitution $u = \cos\phi$, $du = -\sin\phi\,d\phi$:

$2\pi \int_1^0 u^2\,(-du) = 2\pi \int_0^1 u^2\,du = 2\pi \cdot \frac{1}{3} = \frac{2\pi}{3}.$

Let $\mathbf{F}(x,y,z) = (x, y, z)$ — the position (or radial) field — and let $S$ be the sphere of radius $R$ with the outward normal. On the sphere, $\hat{\mathbf{n}} = \frac{1}{R}(x, y, z)$, so:

$\mathbf{F} \cdot \hat{\mathbf{n}} = \frac{1}{R}(x^2 + y^2 + z^2) = \frac{R^2}{R} = R.$

The normal component of $\mathbf{F}$ is the constant $R$ on the entire sphere. The flux is:

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S R\,dS = R \cdot \text{Area}(S) = R \cdot 4\pi R^2 = 4\pi R^3.$

We can verify this using the divergence theorem (Theorem 3, Section 7): $\nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$, so $\iiint_E 3\,dV = 3 \cdot \frac{4}{3}\pi R^3 = 4\pi R^3$. The two sides match — this is the divergence theorem at work.

Reversing the orientation of $S$ — replacing $\hat{\mathbf{n}}$ by $-\hat{\mathbf{n}}$ — negates the flux integral:

$\iint_{-S} \mathbf{F} \cdot d\mathbf{S} = -\iint_S \mathbf{F} \cdot d\mathbf{S}.$

This is the surface analog of the line integral identity $\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}$ from Topic 15, Remark 2. The scalar surface integral $\iint_S f\,dS$ is orientation-independent (it uses $\|\mathbf{r}_u \times \mathbf{r}_v\|$, which is always positive), but the flux integral is orientation-sensitive (it uses $\mathbf{r}_u \times \mathbf{r}_v$ directly, including its sign).

For a $C^1$ vector field $\mathbf{F} = (P, Q, R): D \to \mathbb{R}^3$, the curl of $\mathbf{F}$ is:

$\nabla \times \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\hat{\mathbf{i}} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)\hat{\mathbf{j}} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\hat{\mathbf{k}}.$

Using the determinant mnemonic:

$\nabla \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}.$

The $\hat{\mathbf{k}}$-component of $\nabla \times \mathbf{F}$ is $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ — exactly the 2D curl from Topic 15, Definition 9. The 3D curl extends the 2D curl by adding components for rotation about the $x$- and $y$-axes.

For a $C^1$ vector field $\mathbf{F} = (P, Q, R): D \to \mathbb{R}^3$, the divergence of $\mathbf{F}$ is the scalar field:

$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}.$

The divergence is the “scalar product” of the formal vector $\nabla = (\partial_x, \partial_y, \partial_z)$ with $\mathbf{F}$. It measures the net outflow per unit volume at each point.

Let $\mathbf{F}$ be $C^1$ at a point $\mathbf{p}$. Let $B_r(\mathbf{p})$ be the ball of radius $r$ centered at $\mathbf{p}$ and $S_r(\mathbf{p})$ its boundary sphere (with outward normal). Then:

$\nabla \cdot \mathbf{F}(\mathbf{p}) = \lim_{r \to 0} \frac{1}{\frac{4}{3}\pi r^3} \oiint_{S_r(\mathbf{p})} \mathbf{F} \cdot d\mathbf{S}.$

The divergence is the flux per unit volume in the limit of infinitesimally small enclosing surfaces.

By the divergence theorem (Theorem 3, which we will prove in Section 7), $\oiint_{S_r} \mathbf{F} \cdot d\mathbf{S} = \iiint_{B_r} \nabla \cdot \mathbf{F}\,dV$. By the Mean Value Theorem for triple integrals (Topic 13), there exists $\mathbf{p}_r \in B_r(\mathbf{p})$ such that:

$\iiint_{B_r} \nabla \cdot \mathbf{F}\,dV = (\nabla \cdot \mathbf{F})(\mathbf{p}_r) \cdot \frac{4}{3}\pi r^3.$

Dividing by the volume and taking $r \to 0$: as $r \to 0$, $\mathbf{p}_r \to \mathbf{p}$, and continuity of $\nabla \cdot \mathbf{F}$ gives $(\nabla \cdot \mathbf{F})(\mathbf{p}_r) \to (\nabla \cdot \mathbf{F})(\mathbf{p})$. $\square$

Let $\mathbf{F}$ be $C^1$ at a point $\mathbf{p}$, and let $\hat{\mathbf{n}}$ be a unit vector. Let $C_r$ be the circle of radius $r$ centered at $\mathbf{p}$ in the plane perpendicular to $\hat{\mathbf{n}}$, oriented by the right-hand rule. Then:

$(\nabla \times \mathbf{F})(\mathbf{p}) \cdot \hat{\mathbf{n}} = \lim_{r \to 0} \frac{1}{\pi r^2} \oint_{C_r} \mathbf{F} \cdot d\mathbf{r}.$

The component of the curl along $\hat{\mathbf{n}}$ is the circulation per unit area in the plane perpendicular to $\hat{\mathbf{n}}$, in the limit of infinitesimally small loops. This generalizes Topic 15, Proposition 2 from 2D (where $\hat{\mathbf{n}} = \hat{\mathbf{k}}$ always) to arbitrary directions in 3D.

Let $f$ be a $C^2$ scalar field and $\mathbf{F}$ a $C^2$ vector field on an open domain in $\mathbb{R}^3$. Then:

1. $\nabla \times (\nabla f) = \mathbf{0}$ — the curl of a gradient is always zero.

2. $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ — the divergence of a curl is always zero.

3. $\nabla \times (\nabla \times \mathbf{F}) = \nabla(\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}$ — the curl-curl identity, where $\nabla^2 \mathbf{F}$ is the vector Laplacian (Laplacian applied component-wise).

Identity (1) says gradient fields are curl-free — this is the 3D version of the exactness condition $\partial P/\partial y = \partial Q/\partial x$ from Topic 15. Identity (2) says curl fields are divergence-free. Together, they form an exact sequence:

$C^\infty(D) \xrightarrow{\nabla} \text{Vec}(D) \xrightarrow{\nabla \times} \text{Vec}(D) \xrightarrow{\nabla \cdot} C^\infty(D),$

where the composition of any two consecutive arrows is zero. In the language of differential forms, this is the de Rham complex $\Omega^0 \xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \Omega^3$ with $d^2 = 0$ (→ Smooth Manifolds on formalML).

Let $\mathbf{F}(x, y, z) = (yz,\; xz,\; xy)$. We compute:

$\nabla \times \mathbf{F} = \left(\frac{\partial(xy)}{\partial y} - \frac{\partial(xz)}{\partial z}\right)\hat{\mathbf{i}} + \left(\frac{\partial(yz)}{\partial z} - \frac{\partial(xy)}{\partial x}\right)\hat{\mathbf{j}} + \left(\frac{\partial(xz)}{\partial x} - \frac{\partial(yz)}{\partial y}\right)\hat{\mathbf{k}}$

$= (x - x)\hat{\mathbf{i}} + (y - y)\hat{\mathbf{j}} + (z - z)\hat{\mathbf{k}} = \mathbf{0}.$

The curl vanishes because $\mathbf{F} = \nabla(xyz)$ — it is a gradient field, and Identity (1) from Theorem 1 guarantees $\nabla \times (\nabla f) = \mathbf{0}$.

The divergence is $\nabla \cdot \mathbf{F} = \frac{\partial(yz)}{\partial x} + \frac{\partial(xz)}{\partial y} + \frac{\partial(xy)}{\partial z} = 0 + 0 + 0 = 0$. This field is both curl-free and divergence-free — it has no rotation and no sources or sinks.

Let $\mathbf{F}(x, y, z) = (-y, x, 0)$ — the 3D extension of the rotation field from Topic 15.

$\nabla \times \mathbf{F} = \left(\frac{\partial 0}{\partial y} - \frac{\partial x}{\partial z}\right)\hat{\mathbf{i}} + \left(\frac{\partial(-y)}{\partial z} - \frac{\partial 0}{\partial x}\right)\hat{\mathbf{j}} + \left(\frac{\partial x}{\partial x} - \frac{\partial(-y)}{\partial y}\right)\hat{\mathbf{k}} = (0, 0, 2).$

The curl is $(0, 0, 2)$, pointing in the $\hat{\mathbf{k}}$ direction with magnitude 2. The $\hat{\mathbf{k}}$-component is $\frac{\partial x}{\partial x} - \frac{\partial(-y)}{\partial y} = 1 + 1 = 2$, exactly the 2D curl from Topic 15, Example 15. The 3D curl encodes the same rotation information, plus the fact that the rotation axis is vertical.

The divergence is $\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0$ — rotation without expansion.

Let $S$ be an oriented, piecewise-smooth surface in $\mathbb{R}^3$ bounded by a simple, closed, piecewise-smooth curve $C = \partial S$, with orientation induced by the right-hand rule. Let $\mathbf{F} = (P, Q, R)$ be a $C^1$ vector field on an open region containing $S$. Then:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.$

The circulation of $\mathbf{F}$ around the boundary $C$ equals the flux of the curl $\nabla \times \mathbf{F}$ through the surface $S$.

We prove Stokes’ theorem for the case where $S$ is the graph of a $C^2$ function $z = g(x, y)$ over a region $D \subseteq \mathbb{R}^2$. The general case follows by decomposing an arbitrary surface into graph patches using a partition of unity.

Setup. Parameterize $S$ as $\mathbf{r}(x, y) = (x, y, g(x,y))$ for $(x, y) \in D$. The boundary curve $C = \partial S$ lies above the boundary $\partial D$. If $\partial D$ is parameterized by $(x(t), y(t))$ for $t \in [a, b]$, then $C$ is parameterized by $(x(t), y(t), g(x(t), y(t)))$.

Left side (line integral). We compute $\oint_C P\,dx + Q\,dy + R\,dz$. On $C$, $dz = g_x\,dx + g_y\,dy$ (by the chain rule), so:

$\oint_C P\,dx + Q\,dy + R\,dz = \oint_C (P + Rg_x)\,dx + (Q + Rg_y)\,dy,$

where all functions are evaluated at $(x, y, g(x,y))$. This is now a 2D line integral around $\partial D$.

Right side (surface integral). We compute $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$. From Example 3, $\mathbf{r}_x \times \mathbf{r}_y = (-g_x, -g_y, 1)$. Writing $\nabla \times \mathbf{F} = (R_y - Q_z,\; P_z - R_x,\; Q_x - P_y)$, the flux of the curl is:

$\iint_D [(R_y - Q_z)(-g_x) + (P_z - R_x)(-g_y) + (Q_x - P_y)]\,dA,$

where all partial derivatives of $P$, $Q$, $R$ are evaluated at $(x, y, g(x,y))$.

Showing equality via Green’s theorem. We apply Green’s theorem (Topic 15, Theorem 5) to the 2D line integral:

$\oint_{\partial D} (P + Rg_x)\,dx + (Q + Rg_y)\,dy = \iint_D \left[\frac{\partial(Q + Rg_y)}{\partial x} - \frac{\partial(P + Rg_x)}{\partial y}\right]\,dA.$

We now expand the integrand on the right. We must use the chain rule carefully, because $P$, $Q$, $R$ depend on $z = g(x,y)$.

Expanding $\frac{\partial}{\partial x}(Q + Rg_y)$:

$\frac{\partial Q}{\partial x} = Q_x + Q_z g_x, \quad \frac{\partial(Rg_y)}{\partial x} = R_x g_y + R_z g_x g_y + R g_{yx},$

so $\frac{\partial}{\partial x}(Q + Rg_y) = Q_x + Q_z g_x + R_x g_y + R_z g_x g_y + R g_{yx}$.

Expanding $\frac{\partial}{\partial y}(P + Rg_x)$:

$\frac{\partial P}{\partial y} = P_y + P_z g_y, \quad \frac{\partial(Rg_x)}{\partial y} = R_y g_x + R_z g_y g_x + R g_{xy},$

so $\frac{\partial}{\partial y}(P + Rg_x) = P_y + P_z g_y + R_y g_x + R_z g_y g_x + R g_{xy}$.

Subtracting. Since $g_{xy} = g_{yx}$ (by $C^2$ regularity), the $Rg_{xy}$ and $Rg_{yx}$ terms cancel. The $R_z g_x g_y$ terms also cancel. We are left with:

$\frac{\partial(Q + Rg_y)}{\partial x} - \frac{\partial(P + Rg_x)}{\partial y} = (Q_x - P_y) + (Q_z g_x - P_z g_y) + (R_x g_y - R_y g_x).$

Rearranging:

$= (Q_x - P_y) + (-g_x)(R_y - Q_z) + (-g_y)(P_z - R_x).$

Wait — let us verify. We have $(Q_z g_x - P_z g_y) + (R_x g_y - R_y g_x)$. Grouping by the $g$ factors:

$= -g_x(R_y - Q_z) - g_y(P_z - R_x) + (Q_x - P_y).$

This is exactly $(\nabla \times \mathbf{F}) \cdot (-g_x, -g_y, 1) = (\nabla \times \mathbf{F}) \cdot (\mathbf{r}_x \times \mathbf{r}_y)$.

So by Green’s theorem:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D (\nabla \times \mathbf{F}) \cdot (\mathbf{r}_x \times \mathbf{r}_y)\,dA = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.$

For a general oriented surface $S$ that is not a single graph, we decompose $S$ into graph patches $S_1, \ldots, S_k$ using a partition of unity. The interior boundary contributions from adjacent patches cancel (their normals point in opposite directions along the shared edge), leaving only the exterior boundary $C = \partial S$. $\square$

Let $\mathbf{F} = (-y, x, 0)$ and let $S$ be the upper hemisphere of the unit sphere, bounded by $C$: the unit circle in the $xy$-plane.

Line integral. The unit circle is $\mathbf{r}(t) = (\cos t, \sin t, 0)$ for $t \in [0, 2\pi]$, traversed counterclockwise. Then $\mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t, 0)$ and $\mathbf{r}'(t) = (-\sin t, \cos t, 0)$:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} [\sin^2 t + \cos^2 t]\,dt = 2\pi.$

Surface integral. From Example 9, $\nabla \times \mathbf{F} = (0, 0, 2)$. The flux of $(0, 0, 2)$ through the upper hemisphere with outward normal: using $d\mathbf{S} = \sin\phi\,(\sin\phi\cos\theta,\; \sin\phi\sin\theta,\; \cos\phi)\,d\phi\,d\theta$:

$\iint_S (0, 0, 2) \cdot d\mathbf{S} = \int_0^{2\pi}\int_0^{\pi/2} 2\sin\phi\cos\phi\,d\phi\,d\theta = 2 \cdot 2\pi \cdot \frac{1}{2} = 2\pi.$

Both sides equal $2\pi$. Stokes’ theorem confirmed.

Compute $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = (y^2, z^2, x^2)$ and $C$ is the triangle with vertices $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, oriented counterclockwise when viewed from the direction of the outward normal.

Strategy: computing the line integral directly would require parameterizing three edges and adding three integrals. Instead, we use Stokes’ theorem with the flat triangular surface $S$ spanning $C$.

The triangle lies in the plane $x + y + z = 1$, which we can write as $z = 1 - x - y$. So $g_x = -1$ and $g_y = -1$, and $\mathbf{r}_x \times \mathbf{r}_y = (1, 1, 1)$ (pointing outward, since the components are all positive — this is the outward normal to the plane $x + y + z = 1$).