Line and Surface Integrals – Definitions, Theorems, and Applications

Line Integrals

A line integral of a scalar field $f(x,y,z)$ over a curve $C$ is
$${\int }_{C}fds={\int }_a^{b}f(\mathbf{r}(t))\Vert {\mathbf{r}}^{\prime }(t)\Vert dt,$$ where $\mathbf{r}(t)$ parametrizes $C$.
For a vector field $\mathbf{F}(x,y,z)=⟨P,Q,R⟩$, the work integral is
$${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_a^b\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)dt.$$

Parameterization and Orientation

1. Choose a parametrization $\mathbf{r}(t)$ of $C$.
2. Compute ${\mathbf{r}}^{\prime }(t)$.
3. Evaluate $\mathbf{F}(\mathbf{r}(t))$ or $f(\mathbf{r}(t))$.
4. Integrate over the parameter interval.

The direction of traversal (orientation) matters for vector line integrals; reversing the curve changes the sign.

flowchart TD
    A["Choose curve C"] --> B["Parametrize: r(t)"]
    B --> C["Compute r'(t)"]
    C --> D["Evaluate F(r(t)) · r'(t)"]
    D --> E["Integrate over t-interval"]

Example 1 – Work along a semicircle

Compute
$$W={\int }_C(ydx+xdy)$$ where $C$ is the upper half of the circle $x^2+y^2=1$ from $(1,0)$ to $(-1,0)$.

Parametrization:
$$\mathbf{r}(t)=⟨\cos t,\sin t⟩,t\in [0,\pi ].$$ Then
$$dx=-\sin tdt,dy=\cos tdt.$$

Integral:
$$W={\int }_0^{\pi }(\sin t(-\sin t)+\cos t(\cos t))dt={\int }_0^{\pi }(-{\sin }^{2}t+{\cos }^{2}t)dt={\int }_0^{\pi }\cos 2tdt.$$ $$W={\frac{\sin 2t}{2}|}_0^{\pi }=0.$$

Interpretation: The net work done by the vector field $⟨y,x⟩$ along the semicircle is zero because the field is conservative (see below).

Conservative Fields and Potential Functions

A vector field $\mathbf{F}$ is conservative if there exists a scalar potential $\varphi$ such that $\mathbf{F}=\nabla \varphi$.
For a conservative field, the line integral depends only on endpoints: $${\int }_C\mathbf{F}\cdot d\mathbf{r}=\varphi (\mathbf{r}(b))-\varphi (\mathbf{r}(a)).$$

Checking Conservativeness

In ${\mathbb{R}}^3$, $\mathbf{F}=⟨P,Q,R⟩$ is conservative iff
$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x},\frac{\partial P}{\partial z}=\frac{\partial R}{\partial x},\frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}.$$

Surface Integrals

A surface integral of a scalar field $f$ over a parametric surface $\mathbf{r}(u,v)$ is
$${\iint }_{S}fdS={\iint }_{D}f(\mathbf{r}(u,v))\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv,$$ where $D$ is the parameter domain.

For a vector field $\mathbf{F}$, the flux across $S$ is
$${\iint }_S\mathbf{F}\cdot \mathbf{n}dS={\iint }_D\mathbf{F}(\mathbf{r}(u,v))\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)dudv,$$ with $\mathbf{n}$ the unit normal.

Example 2 – Flux through a spherical patch

Find the flux of $\mathbf{F}=⟨x,y,z⟩$ across the part of the sphere $x^2+y^2+z^2=1$ lying in the first octant ($x,y,z\ge 0$).

Parameterization: Use spherical coordinates
$$\mathbf{r}(\theta ,\varphi )=⟨\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi ⟩,$$ with $\theta \in [0,\frac{\pi }{2}]$, $\varphi \in [0,\frac{\pi }{2}]$.
Then
$${\mathbf{r}}_{\theta }=⟨-\sin \varphi \sin \theta ,\sin \varphi \cos \theta ,0⟩,{\mathbf{r}}_{\varphi }=⟨\cos \varphi \cos \theta ,\cos \varphi \sin \theta ,-\sin \varphi ⟩.$$ $${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }=⟨{\sin }^2\varphi \cos \theta ,{\sin }^2\varphi \sin \theta ,\sin \varphi \cos \varphi ⟩.$$ Its magnitude is $\Vert {\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }\Vert =\sin \varphi$.

Flux integrand:
$$\mathbf{F}(\mathbf{r})\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi })=⟨\sin \varphi \cos \theta ,\sin \varphi \sin \theta ,\cos \varphi ⟩\cdot ⟨{\sin }^2\varphi \cos \theta ,{\sin }^2\varphi \sin \theta ,\sin \varphi \cos \varphi ⟩={\sin }^3\varphi {\cos }^2\theta +{\sin }^3\varphi {\sin }^2\theta +\sin \varphi {\cos }^2\varphi ={\sin }^3\varphi +\sin \varphi {\cos }^2\varphi .$$ Thus the flux is
$$\Phi ={\int }_0^{\pi /2}{\int }_0^{\pi /2}({\sin }^3\varphi +\sin \varphi {\cos }^2\varphi )d\theta d\varphi .$$ Integrate over $\theta$: factor ${\int }_0^{\pi /2}d\theta =\frac{\pi }{2}$.
$$\Phi =\frac{\pi }{2}{\int }_0^{\pi /2}({\sin }^3\varphi +\sin \varphi {\cos }^2\varphi )d\varphi .$$ Compute the $\varphi$-integral: $${\int }_0^{\pi /2}{\sin }^3\varphi d\varphi =\frac{2}{3},{\int }_0^{\pi /2}\sin \varphi {\cos }^2\varphi d\varphi =\frac{1}{3}.$$ Hence
$$\Phi =\frac{\pi }{2}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{\pi }{2}\cdot 1=\frac{\pi }{2}.$$

Stokes’ Theorem

Stokes’ theorem relates a line integral around a closed curve $C$ to a surface integral over any surface $S$ bounded by $C$: $${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS.$$ Here $\mathbf{n}$ is the unit normal to $S$ consistent with the orientation of $C$.

Example 3 – Using Stokes’ Theorem

Let $\mathbf{F}=⟨z,x,y⟩$ and $C$ be the boundary of the part of the plane $z=1-x-y$ in the first octant.
Compute ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$.

Step 1 – Compute curl:
$$\nabla \times \mathbf{F}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\partial }_x& {\partial }_y& {\partial }_z\\ z& x& y\end{array}\right|=⟨1-0,1-0,1-1⟩=⟨1,1,0⟩.$$

Step 2 – Parameterize surface:
Use $x=u,y=v,z=1-u-v$ with $u,v\ge 0$ and $u+v\le 1$.
Then
$${\mathbf{r}}_u=⟨1,0,-1⟩,{\mathbf{r}}_v=⟨0,1,-1⟩,{\mathbf{r}}_u\times {\mathbf{r}}_v=⟨1,1,1⟩.$$ The unit normal $\mathbf{n}$ is $\frac{1}{\sqrt{3}}⟨1,1,1⟩$.

Step 3 – Surface integral:
$$(\nabla \times \mathbf{F})\cdot \mathbf{n}=⟨1,1,0⟩\cdot \frac{1}{\sqrt{3}}⟨1,1,1⟩=\frac{2}{\sqrt{3}}.$$ The area element $dS=\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv=\sqrt{3}dudv$.
Thus the integrand becomes $\frac{2}{\sqrt{3}}\cdot \sqrt{3}=2$.
Integrate over the triangular domain $u,v\ge 0,u+v\le 1$: $${\iint }_{S}2dudv=2\cdot \text{Area of triangle}=2\cdot \frac{1}{2}=1.$$ Hence
$${\oint }_C\mathbf{F}\cdot d\mathbf{r}=1.$$

Comparison of Line and Surface Integrals

Advantages & Disadvantages

In the real world

1. Google Maps Navigation – The shortest travel time between two points is found by minimizing a line integral of travel speed over a path.
   Product: Google Maps
   Idea used: Line integral of speed (inverse of velocity) along a route.

2. Ncell Signal Strength Mapping – The total received signal power over a cell tower’s coverage area is computed as a surface integral of the field strength over the ground plane.
   Product: Ncell base‑station coverage maps
   Idea used: Flux of electromagnetic field through a surface.

3. NEPSE Market Analysis – Risk assessment of a portfolio involves integrating a scalar risk density over a multi‑dimensional space of asset returns.
   Product: NEPSE risk dashboards
   Idea used: Surface integral of risk density over a defined region.

Exam tip

• Check conservativeness first: If $\nabla \times \mathbf{F}=0$ in a simply connected region, reduce the line integral to a potential difference.
• Use Stokes’ or Divergence theorem when the surface or volume is easier to describe than the boundary curve or surface.
• Pay attention to orientation: For closed curves, the positive orientation follows the right‑hand rule relative to the chosen normal.
• Parameterize correctly: For surfaces, ensure ${\mathbf{r}}_u\times {\mathbf{r}}_v$ points outward or in the required direction.
• Simplify integrands: Factor common terms, use symmetry, and reduce double integrals to single integrals when possible.

Summary

Unit 4 equips you with the tools to evaluate integrals over curves and surfaces, to recognize when a vector field is conservative, and to apply powerful theorems that convert difficult integrals into simpler ones. Mastery of these concepts is essential for solving problems in physics, engineering, and applied mathematics, and they form the mathematical backbone of many modern technologies.