Line integrals

Recall

A vector field is a function that takes in a vector and spits out a vector, and a scalar field is a function that takes in a vector and spits out a scalar.

Over a scalar field

Say we have some scalar field $f : \mathbb R^3 \to \mathbb R$ and some curve $C$ parametrized by $\mathbf r(t) : [a,b] \to \mathbb R^3$ . We want to integrate $f$ along $C$ :

\int_C f d \mathbf s

Since the curve is parametrized by $\mathbf r$ , we can write an equivalent integral:

\int_a^b f(\mathbf r(t)) d\mathbf r

Where $d\mathbf r$ is the length of a differential slice of the curve. We can calculate this length as $d\mathbf r = \lVert \mathbf r'(t) \rVert dt$ . This gives us the general equation for a line integral over a scalar field:

\int_C f ds = \int_a^b f(\mathbf r(t)) \lVert \mathbf r'(t) \rVert dt

Recall

If $\mathbf r(t) = \begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}$ , $\lVert \mathbf r'(t) \rVert = \sqrt {(x')^2 + (y')^2 + (z')^2}$

Over a vector field

Say we have some vector field $\mathbf F = (P, Q, R)$ and a curve $C$ parametrized by $\mathbf r : [a,b] \to \mathbb R^3$ . We want to integrate the dot product of $\mathbf F$ along $C$ (both of these forms mean the same thing):

\int_C \mathbf F \cdot d\mathbf r = \int_C Pdx + Qdy + Rdz

$d\mathbf r$ is the differential change in position along the curve. Note that in this case it is a vector, simply $\mathbf r'(t) dt$ . This gives us the final equation:

\int_C \mathbf F \cdot d\mathbf r = \int_a^b \mathbf F(\mathbf r(t)) \cdot \begin{pmatrix} x'(t)\\y'(t)\\z'(t) \end{pmatrix} dt

In general, you can follow these steps to compute a line integral over a vector field:

Choose a parametrization $\mathbf r(t) = (x,y,z), t \in [a,b]$ .
Substitute $x, y, z$ in the integrand $(\mathbf F)$ .
Also substitute $dx = x' dt$ , $dy = y' dt$ , $dz = z' dt$ and $ds = \lVert r' \rVert dt$ , $d\mathbf r = \mathbf r' dt$ .
The result will be a single variable integral that you can compute directly.

Fundamental theorem of calculus

Let’s say we have a curve $C$ that goes from points $\mathbf A$ to $\mathbf B$ , and a scalar function $f : \mathbb R^3 \to \mathbb R$ . Then the fundamental theorem of calculus says:

\int_C \nabla f \cdot d\mathbf r = f(\mathbf B) - f(\mathbf A)

Proof

Let $\mathbf r : [a,b] \to \mathbb R^3$ parametrize $C$ .

\int_C \nabla f \cdot d\mathbf r = \int_a^b \nabla f(\mathbf r(t)) \cdot \mathbf r'(t) dt

Let $G(t) = f(\mathbf r(t))$ . By the chain rule, $G'(t) = Df_{\mathbf r(t)} \cdot D\mathbf r(t)$ . (where $D\mathbf g$ is the total derivative or Jacobian matrix of $g$ )