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Chapter 8 • Theory & Derivations

Unit 8: Surface Integrals, Stokes' Theorem & Gauss' Divergence Theorem

Parametric surfaces, area differentials, surface integrals of scalar and vector fields, Stokes' curl circulation theorem, Gauss' divergence theorem, physical applications, and the grand unification of differential forms.

§8.1 Parametric Surfaces, Tangent Planes and Surface Area

Just as a space curve is traced out by a vector function of a single parameter $t$, a two-dimensional surface in $\mathbb{R}^3$ is described by a vector function of two parameters $u$ and $v$:

$$\vec{r}(u, v) = x(u, v)\hat{i} + y(u, v)\hat{j} + z(u, v)\hat{k} = \langle x(u, v), y(u, v), z(u, v) \rangle$$

where $(u, v)$ varies over a planar parameter region $D \subset \mathbb{R}^2$.


1. Tangent Vectors and Normal Vectors

Holding $v = v_0$ constant, $\vec{r}(u, v_0)$ traces a grid curve on the surface whose tangent vector is:

$$\vec{r}_u = \frac{\partial\vec{r}}{\partial u} = \left\langle \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right\rangle$$

Similarly, holding $u = u_0$ constant gives the grid curve tangent vector:

$$\vec{r}_v = \frac{\partial\vec{r}}{\partial v} = \left\langle \frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} \right\rangle$$

The normal vector to the tangent plane at $(u_0, v_0)$ is the cross product:

$$\mathbf{\vec{n} = \vec{r}_u \times \vec{r}_v}$$

The surface is called smooth if $\vec{r}_u \times \vec{r}_v \neq \vec{0}$ everywhere.


2. The Surface Area Differential and Total Area

An infinitesimal parameter rectangle $\Delta u \times \Delta v$ maps to a curved patch on the surface approximated by the tangent parallelogram spanned by $\vec{r}_u \Delta u$ and $\vec{r}_v \Delta v$.

The area of this parallelogram is:

$$\Delta S \approx |(\vec{r}_u \Delta u) \times (\vec{r}_v \Delta v)| = |\vec{r}_u \times \vec{r}_v| \Delta u \Delta v$$

In differential form:

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

Total Surface Area:

$$\mathbf{A(S) = \iint_D |\vec{r}_u \times \vec{r}_v| \, dA}$$
Explicit Cartesian Surfaces $z = g(x, y)$:

Parameterize with $u = x, v = y$: $\vec{r}(x, y) = \langle x, y, g(x, y) \rangle$. $\vec{r}_x = \langle 1, 0, g_x \rangle$, $\vec{r}_y = \langle 0, 1, g_y \rangle$. $\vec{r}_x \times \vec{r}_y = \langle -g_x, -g_y, 1 \rangle$.

$$|\vec{r}_x \times \vec{r}_y| = \sqrt{1 + g_x^2 + g_y^2}$$
$$\mathbf{A(S) = \iint_D \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy}$$

§8.2 Surface Integrals of Scalar Fields and Vector Flux

1. Surface Integrals of Scalar Fields

If $f(x, y, z)$ is a continuous scalar function defined on a smooth surface $S$:

$$\mathbf{\iint_S f(x, y, z) \, dS = \iint_D f(\vec{r}(u, v)) |\vec{r}_u \times \vec{r}_v| \, du \, dv}$$

If $f(x, y, z) = \rho(x, y, z)$ is surface mass density, the integral computes the total mass of the curved shell.


2. Oriented Surfaces and Unit Normals

A surface $S$ is orientable (two-sided) if it is possible to define a continuous unit normal vector field $\hat{n}$ across the entire surface (excluding non-orientable manifolds like the Möbius strip).

For a parameterized smooth surface, the unit normal is:

$$\mathbf{\hat{n} = \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|}}$$

3. Surface Integrals of Vector Fields (Flux Integrals)

Let $\vec{F}$ be a continuous vector field on an oriented surface $S$ with unit normal $\hat{n}$.

Definition: The flux of $\vec{F}$ across $S$ is the surface integral of the normal component of $\vec{F}$:

$$\mathbf{\Phi = \iint_S \vec{F} \cdot d\vec{S} = \iint_S \vec{F} \cdot \hat{n} \, dS = \iint_D \vec{F}(\vec{r}(u, v)) \cdot (\vec{r}_u \times \vec{r}_v) \, du \, dv}$$

Notice that the magnitude $|\vec{r}_u \times \vec{r}_v|$ in the unit normal cancels directly with the area element $dS = |\vec{r}_u \times \vec{r}_v| du dv$!

Physical Meaning:

If $\vec{F} = \rho \vec{v}$ is the mass flow rate per unit area of a moving fluid, $\iint_S \vec{F} \cdot d\vec{S}$ represents the total mass of fluid passing through surface $S$ per unit time.

§8.3 Stokes' Theorem, Gauss' Divergence Theorem & Grand Unification

1. Stokes' Theorem (Curl Circulation Theorem)

Sir George Gabriel Stokes (1854) established the higher-dimensional generalization of Green's theorem relating the circulation around a boundary curve in 3D space to the flux of curl through any spanning surface.

Theorem (Stokes' Theorem): Let $S$ be an oriented piecewise-smooth surface bounded by a simple, closed, piecewise-smooth space curve $C = \partial S$ whose orientation is positive relative to the normal vector $\hat{n}$ of $S$ (by the right-hand rule). If $\vec{F}$ is a vector field with continuous partial derivatives:

$$\mathbf{\oint_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}}$$
Fundamental Invariance (Surface Independence):

If $S_1$ and $S_2$ are any two distinct surfaces that share the exact same boundary curve $\partial S_1 = \partial S_2 = C$, then:

$$\iint_{S_1} (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_{S_2} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r}$$

The flux of curl through a surface depends solely on its boundary rim, not on its bulging shape!


2. Gauss' Divergence Theorem

Carl Friedrich Gauss (1813) proved the fundamental theorem connecting the flux of a vector field across a closed bounding surface to the volume integral of its divergence throughout the enclosed solid.

Theorem (The Divergence Theorem): Let $E$ be a simple solid region in $\mathbb{R}^3$, and let $S = \partial E$ be its boundary surface, oriented with outward-pointing unit normal $\hat{n}$. If $\vec{F}$ is a vector field with continuous partial derivatives on an open region containing $E$:

$$\mathbf{\iint_{\partial E} \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) \, dV}$$
Physical Interpretation:

The total net fluid volume exiting a closed container across its skin must equal the sum of all internal sources (divergence) within the container.

Major Applications in Physics:

1. Gauss' Law in Electromagnetism: $\iint_{\partial E} \vec{E} \cdot d\vec{S} = \frac{1}{\epsilon_0}\iiint_E \rho \, dV = \frac{Q_{enc}}{\epsilon_0} \implies \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$ (Maxwell's First Equation).

2. Fluid Continuity Equation: $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = 0$.


3. The Grand Unification: Generalized Stokes' Theorem

All the fundamental integral theorems of calculus across all dimensions are manifestations of one single, universal topological equation for differential forms on an oriented manifold $\Omega$ with boundary $\partial \Omega$:

$$\mathbf{\int_{\partial \Omega} \omega = \int_\Omega d\omega}$$

| Dimension | Manifold $\Omega$ | Boundary $\partial \Omega$ | Classical Theorem | | :---: | :---: | :---: | :---: | | $n = 1$ | Curve $[a, b]$ | Points $\{a, b\}$ | Fundamental Theorem of Calculus: $\int_a^b F'(x) dx = F(b) - F(a)$ | | $n = 2$ | Plane Region $D$ | Boundary Curve $C$ | Green's Theorem: $\oint_C P dx + Q dy = \iint_D (Q_x - P_y) dA$ | | $n = 2$ | Spatial Surface $S$ | Space Curve $C$ | Stokes' Theorem: $\oint_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$ | | $n = 3$ | Solid Volume $E$ | Closed Surface $S$ | Divergence Theorem: $\iint_{\partial E} \vec{F} \cdot d\vec{S} = \iiint_E (\nabla \cdot \vec{F}) dV$ |

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Solved Problem Example 8.1: Surface Area of a Paraboloid Cap Cut by a Plane

Find the exact surface area of the portion of the circular paraboloid:

$$z = 4 - x^2 - y^2$$

that lies above the $xy$-plane ($z \ge 0$).

Step 1: Express the surface and find the projection domain $D$ The surface is $z = g(x, y) = 4 - x^2 - y^2$. It intersects $z = 0$ when:

$$4 - x^2 - y^2 = 0 \implies x^2 + y^2 = 4$$

The projection domain $D$ in the $xy$-plane is a circular disk of radius $R = 2$:

$$D = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \le 4 \}$$

Step 2: Compute partial derivatives and surface area element $dS$

$$g_x = \frac{\partial z}{\partial x} = -2x, \qquad g_y = \frac{\partial z}{\partial y} = -2y$$
$$1 + g_x^2 + g_y^2 = 1 + (-2x)^2 + (-2y)^2 = 1 + 4(x^2 + y^2)$$

Thus:

$$dS = \sqrt{1 + 4(x^2 + y^2)} \, dA$$

Step 3: Convert to polar coordinates and integrate In polar coordinates, $x^2 + y^2 = r^2$ with $0 \le r \le 2$ and $0 \le \theta \le 2\pi$:

$$A = \iint_D \sqrt{1 + 4r^2} \, (r \, dr \, d\theta) = \left( \int_0^{2\pi} d\theta \right) \left( \int_0^2 r \sqrt{1 + 4r^2} \, dr \right)$$

Evaluate the $r$-integral with $u = 1 + 4r^2 \implies du = 8r \, dr \implies r \, dr = \frac{1}{8} du$: When $r = 0$, $u = 1$; when $r = 2$, $u = 1 + 4(4) = 17$:

$$\int_0^2 r \sqrt{1 + 4r^2} \, dr = \frac{1}{8} \int_1^{17} u^{1/2} \, du = \frac{1}{8} \left[ \frac{2}{3} u^{3/2} \right]_1^{17} = \frac{1}{12} (17^{3/2} - 1)$$

Multiply by $2\pi$:

$$A = 2\pi \cdot \frac{1}{12}(17\sqrt{17} - 1) = \mathbf{\frac{\pi}{6}(17\sqrt{17} - 1)} \approx \frac{\pi}{6}(70.093 - 1) \approx \mathbf{36.177}$$
Solved Problem Example 8.2: Verification of Stokes' Theorem over an Open Hemisphere

Verify Stokes' Theorem for the vector field:

$$\vec{F}(x, y, z) = -y\hat{i} + x\hat{j} - 2\hat{k}$$

where $S$ is the upper hemisphere $z = \sqrt{1 - x^2 - y^2}$ ($z \ge 0$), oriented with upward normal, and $C = \partial S$ is the unit boundary circle $x^2 + y^2 = 1, z = 0$ oriented counterclockwise.

Step 1: Evaluate the Line Integral $\oint_C \vec{F} \cdot d\vec{r}$ Parametrize the boundary circle $C$:

$$\vec{r}(t) = \langle \cos t, \; \sin t, \; 0 \rangle \quad (0 \le t \le 2\pi)$$
$$\vec{r}'(t) = \langle -\sin t, \; \cos t, \; 0 \rangle$$

Substitute into $\vec{F}$:

$$\vec{F}(\vec{r}(t)) = \langle -\sin t, \; \cos t, \; -2 \rangle$$

Compute the dot product:

$$\vec{F} \cdot \vec{r}'(t) = (-\sin t)(-\sin t) + (\cos t)(\cos t) + (-2)(0) = \sin^2 t + \cos^2 t = 1$$

Integrate:

$$\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} 1 \, dt = \mathbf{2\pi}$$

Step 2: Compute $\text{curl} \, \vec{F}$

$$\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & -2 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}\left(\frac{\partial}{\partial x}(x) - \frac{\partial}{\partial y}(-y)\right) = \hat{k}(1 - (-1)) = 2\hat{k}$$

Thus:

$$\nabla \times \vec{F} = \langle 0, \; 0, \; 2 \rangle$$

Step 3: Evaluate the Surface Integral $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$ Using surface independence (or projecting $S: z = \sqrt{1 - x^2 - y^2}$ onto $D: x^2 + y^2 \le 1$): For explicit surface $z = g(x, y)$:

$$d\vec{S} = \langle -g_x, \; -g_y, \; 1 \rangle \, dA$$

Now compute the dot product with $\nabla \times \vec{F} = \langle 0, 0, 2 \rangle$:

$$(\nabla \times \vec{F}) \cdot d\vec{S} = \langle 0, \; 0, \; 2 \rangle \cdot \langle -g_x, \; -g_y, \; 1 \rangle \, dA = 2 \, dA$$

Integrate over the unit disk $D$:

$$\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_D 2 \, dA = 2 \iint_D 1 \, dA = 2 \times (\pi \cdot 1^2) = \mathbf{2\pi}$$

Step 4: Conclusion

$$\oint_C \vec{F} \cdot d\vec{r} = 2\pi = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$$

Stokes' Theorem is verified with exact precision. $\blacksquare$

Solved Problem Example 8.3: Outward Flux Calculation via Gauss' Divergence Theorem

Use the Divergence Theorem to calculate the outward flux $\iint_S \vec{F} \cdot d\vec{S}$ of the vector field:

$$\vec{F}(x, y, z) = (x^3 + \tan(yz))\hat{i} + (y^3 + e^{xz})\hat{j} + (z^3 + \ln(1+x^2))\hat{k}$$

across the closed boundary surface $S$ of the solid cylinder $E$ defined by:

$$x^2 + y^2 \le 4, \qquad 0 \le z \le 3$$

Step 1: Compute $\text{div} \, \vec{F}$ Notice that evaluating the flux directly over the three distinct boundary patches of the cylinder (top disk, bottom disk, lateral cylinder) would be exceedingly difficult due to the non-elementary terms $\tan(yz), e^{xz}, \ln(1+x^2)$.

Compute the divergence:

$$\text{div} \, \vec{F} = \frac{\partial}{\partial x}(x^3 + \tan(yz)) + \frac{\partial}{\partial y}(y^3 + e^{xz}) + \frac{\partial}{\partial z}(z^3 + \ln(1+x^2))$$
$$= 3x^2 + 0 + 3y^2 + 0 + 3z^2 + 0 = 3(x^2 + y^2 + z^2)$$

Remarkably, all non-elementary terms vanish identically under differentiation!


Step 2: Apply the Divergence Theorem

$$\iint_S \vec{F} \cdot d\vec{S} = \iiint_E 3(x^2 + y^2 + z^2) \, dV$$

Step 3: Convert to Cylindrical Coordinates The solid cylinder $E$ in cylindrical coordinates is:

$$0 \le r \le 2, \qquad 0 \le \theta \le 2\pi, \qquad 0 \le z \le 3$$

Since $x^2 + y^2 = r^2$ and $dV = r \, dz \, dr \, d\theta$:

$$\iiint_E 3(r^2 + z^2) r \, dz \, dr \, d\theta = 3 \int_0^{2\pi} \int_0^2 \int_0^3 (r^3 + r z^2) \, dz \, dr \, d\theta$$

Step 4: Evaluate the Triple Integral The $\theta$-integral gives $2\pi$:

$$\Phi = 3(2\pi) \int_0^2 \left( \int_0^3 (r^3 + r z^2) \, dz \right) dr = 6\pi \int_0^2 \left[ r^3 z + r \frac{z^3}{3} \right]_{z=0}^{z=3} dr$$
$$= 6\pi \int_0^2 (3r^3 + 9r) \, dr$$

Evaluate the $r$-integral:

$$\int_0^2 (3r^3 + 9r) \, dr = \left[ \frac{3r^4}{4} + \frac{9r^2}{2} \right]_0^2 = \frac{3(16)}{4} + \frac{9(4)}{2} = 12 + 18 = 30$$

Multiply:

$$\Phi = 6\pi \times 30 = \mathbf{180\pi} \approx 565.487$$

The outward flux is exactly $180\pi$. $\blacksquare$