Surface Integrals and Stokes' Theorem

Why Surface Integrals Are Needed

A flat plate with variable density required a double integral. A surface in space—a three-dimensional curve—requires a surface integral. Physical problems leading to them: the mass of a surface with variable density, heat flow through an unevenly heated shell, fluid flow through a membrane, electric flux through a surface (Gauss's law). Surface integrals are the language of field physics.

Parametrically Defined Surfaces

A surface $S$ is defined by a mapping $r: D \subset \mathbb{R}^2 \to \mathbb{R}^3$:

$ r(u,v) = (x(u,v), y(u,v), z(u,v)), \quad (u,v) \in D. $

Tangent vectors: $r_u = \partial r / \partial u = (\partial x/\partial u, \partial y/\partial u, \partial z/\partial u)$ and $r_v = \partial r / \partial v$—lie in the tangent plane to the surface.

Normal vector: $N = r_u \times r_v$—is perpendicular to the tangent plane. Its length $|N| = |r_u \times r_v|$ is the "stretching factor" of parametrization. Element of area: $dS = |r_u \times r_v| du dv$.

Examples of parametrizations:

• Sphere of radius $R$: $r(\varphi, \theta) = (R \sin \varphi \cos \theta, R \sin \varphi \sin \theta, R \cos \varphi)$, $\varphi \in [0, \pi], \theta \in [0, 2\pi]$. $|N| = R^2 \sin \varphi$.
• Cylinder $r = R$: $r(\theta, z) = (R \cos \theta, R \sin \theta, z)$. $|N| = R$.
• Graph $z = f(x,y)$: $r(x,y) = (x, y, f(x,y))$. $|N| = \sqrt{f_x^2 + f_y^2 + 1}$.

Surface Integral of the First Kind

$ \iint_S f , dS = \iint_D f(r(u,v)) \cdot |r_u \times r_v| , du , dv $

Meaning: we integrate the scalar function $f$ over the area of the surface. It does not depend on orientation.

Example—mass of a hemisphere. Surface density $\rho(x,y,z) = z$, hemisphere $S$: $x^2 + y^2 + z^2 = R^2$, $z \geq 0$.

Parametrization: $\varphi \in [0, \pi/2], \theta \in [0, 2\pi], z = R \cos \varphi, dS = R^2 \sin \varphi , d\varphi , d\theta$.

$ M = \iint_S z , dS = \int_0^{2\pi} \int_0^{\pi/2} (R \cos \varphi) R^2 \sin \varphi , d\varphi , d\theta = 2\pi R^3 \int_0^{\pi/2} \cos \varphi \sin \varphi , d\varphi = 2\pi R^3 \cdot \left[\frac{\sin^2 \varphi}{2}\right]_0^{\pi/2} = \pi R^3. $

Surface Integral of the Second Kind (Flux)

An oriented surface: we pick a unit normal $n = N/|N|$ (direction of “outer” or “inner” traversal). Then:

$ \iint_S \mathbf{F} \cdot dS = \iint_S \mathbf{F} \cdot n , dS = \iint_D \mathbf{F}(r(u,v)) \cdot (r_u \times r_v) , du , dv $

Physical meaning—flux of field $\mathbf{F}$ through $S$: amount of substance/charge/energy passing through surface $S$ per unit time when the velocity/flux field is $\mathbf{F}$.

Example—flux of field $\mathbf{F} = (0, 0, z)$ through the upper hemisphere. The normal is directed "outwards" (upward from the center). For the sphere parametrization: $r_u \times r_v$ points outward for $\theta \in [0,2\pi], \varphi \in [0,\pi/2]$.

$\mathbf{F} \cdot (r_u \times r_v) = (0, 0, R \cos \varphi) \cdot (R^2 \sin^2 \varphi \cos \theta, R^2 \sin^2 \varphi \sin \theta, R^2 \sin \varphi \cos \varphi)\ldots$ (working calculations give) $= R^3 \sin \varphi \cos^2 \varphi$.

$ \iint_S \mathbf{F} \cdot dS = \int_0^{2\pi} \int_0^{\pi/2} R^3 \sin \varphi \cos^2 \varphi , d\varphi , d\theta = 2\pi R^3 \left[-\frac{\cos^3 \varphi}{3}\right]_0^{\pi/2} = \frac{2\pi R^3}{3}. $

Stokes' Formula: the Connection Between Curl and Circulation

Stokes' theorem: Let $S$ be an oriented surface with a piecewise-smooth oriented boundary $\partial S$ (right-hand rule: if traversing $\partial S$, the surface is on the left). Then:

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

where curl $\nabla \times \mathbf{F} = \operatorname{rot} \mathbf{F} = (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})$.

Meaning: the circulation of the field along a closed contour equals the flux of the curl through the spanned surface. Curl is the "angular velocity of rotation" of the field: if a velocity field of a fluid has nonzero curl, the fluid "swirls" at that point.

Example. Compute $\oint_C \mathbf{F} \cdot dr$, where $\mathbf{F} = (-y^2, 2xz, 0)$ and $C$ is the unit circle in the plane $z = 0$.

Curl: $\nabla \times \mathbf{F} = (\partial(0)/\partial y - \partial(2xz)/\partial z, \partial(-y^2)/\partial z - \partial(0)/\partial x, \partial(2xz)/\partial x - \partial(-y^2)/\partial y) = (-2x, 0, 2z + 2y)$.

Span the disk $D$: $x^2 + y^2 \leq 1$, $z = 0$, normal $n = (0,0,1)$.

$(\nabla \times \mathbf{F}) \cdot n = 2z + 2y|_{z=0} = 2y$.

$ \iint_D 2y , dA = 2 \iint_D y , dA = 0 $ (by symmetry—odd function on a symmetric domain). ✓

Gauss–Ostrogradsky Formula: Divergence and Sources

Gauss–Ostrogradsky theorem: Let $V$ be a bounded region with a piecewise-smooth boundary $\partial V$, normal to $\partial V$ directed "outward". Then:

$ \iiint_V (\nabla \cdot \mathbf{F}) , dV = \iint_{\partial V} \mathbf{F} \cdot n , dS, $

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

Physical meaning: total flux of the field through a closed surface equals the volume integral of the divergence. $\operatorname{div} \mathbf{F}(x_0) > 0$—the point $x_0$ is a "source" of the field; $\operatorname{div} \mathbf{F} < 0$—"sink".

Example—verification. Field $\mathbf{F} = (x, y, z)$, $V$ is a sphere of radius $R$. $\operatorname{div} \mathbf{F} = 3$. Volume of the sphere: $4\pi R^3 / 3$. Flux: $3 \cdot 4\pi R^3 / 3 = 4\pi R^3$. Check by direct calculation: $\iint_{\partial V} \mathbf{F} \cdot n , dS = \iint_{\text{sphere}} R , dS = R \cdot 4\pi R^2 = 4\pi R^3$. ✓

Maxwell's Equations: A Unified Language

The four Maxwell equations in differential form use $\operatorname{div}$ and $\operatorname{rot}$:

$ \begin{aligned} \nabla \cdot E &= \rho / \varepsilon_0 \quad (\text{electric charges are sources of } E) \ \nabla \cdot B &= 0 \quad (\text{there are no magnetic monopoles}) \ \nabla \times E &= -\partial B / \partial t \quad (\text{Faraday's law}) \ \nabla \times B &= \mu_0 j + \mu_0 \varepsilon_0 \partial E / \partial t \quad (\text{Ampère–Maxwell law}) \end{aligned} $

Gauss's formula translates the first equation into Gauss's law: the total charge determines the flux of $E$ through any surface. Stokes' formula translates the third into Faraday's law in integral form: EMF in the contour $= -d\Phi/dt$.

Question for thought: the field of a point charge $\mathbf{F} = \mathbf{r} / r^3$ ($r$ = distance from the charge) has $\operatorname{div} \mathbf{F} = 0$ everywhere except at the origin. How can one rigorously compute the flux of this field through any surface enclosing the origin, using the Gauss–Ostrogradsky theorem?