Line Integral

Let $F$ be a vector field that is (piecewise) continuous on the path $c : [a, b] \to R^{3}$ where $c$ is $C^{1}$

Similar to the path integral a line integral of $F$ over $c$ is:

\int_{c} F \cdot d s = \int_{a}^{b} F (c (t)) \cdot c^{'} (t) d t

Unit Tangent

for paths $c$ where $c^{'} (t) \neq_{0}$ the tangent unit vector is

T (t) = \frac{c^{'} (t)}{| | c^{'} (t) | |}

Alternative Forms

Using the unit tangent
We can see that:

\begin{aligned} \int F \cdot d s & = \int_{a}^{b} [F (c (t)) \cdot T (t)] | | c^{'} (t) | | d t \\ = \int_{a}^{b} [F (c (t)) \cdot \frac{c^{'} (t)}{| | c^{'} (t) | |}] | | c^{'} (t) | | d t \\ = \int_{a}^{b} F (c (t)) \cdot c^{'} (t) d t \end{aligned}

Differential Form

\begin{matrix} \int_{c} F \cdot d s = \int_{c} F_{1} d x + F_{2} d y + F_{3} d z = \int_{a}^{b} (F_{1} \frac{d x}{d t} + F_{2} \frac{d y}{d t} + F_{3} \frac{d z}{d t}) d t \\ = \int_{a}^{b} (F_{1} (x (t)) \frac{d x}{d t} + F_{2} (y (t)) \frac{d y}{d t} + F_{2} (z (t)) \frac{d z}{d t}) d t \end{matrix}

where $F = [F_{1}, F_{2}, F_{3}]$ and $c (t) = [x (t), y (t), z (t)]$

$d s$ can be though of as $d x \hat{i} + d y \hat{j} + d z \hat{k}$

dr Notation

sometimes a line integral is written as

\int_{C} F \cdot d r

here $r$ is a position vector based at the origin and ending at $c (t)$ at time t
so we have $r (t) = x (t) \hat{i} + y (t) \hat{j} + z (t) \hat{k}$ in place of $c (t)$ where the line integral is

\int_{a}^{b} F (r (t)) \cdot \frac{d r}{d t} d t

Line Integrals of Gradient Fields

if $f$ is a gradient field then

$f = \nabla F ⟹ \frac{d}{d t} F (c (t)) = f (c (t)) \cdot c^{'} (t)$

by chain rule

Therefore

\int_{c} f \cdot d s = \int_{a}^{b} f (c (t)) \cdot c^{'} (t) d t = F (c (b)) - F (c (a)) by FTC

simplifying our computation

Parameterization of Curves

$c^{+}$ is the same curve as $c^{-}$ except in opposite directions. So say $c^{+} : [a, b] \to R^{2}$ and $c^{-} : [b, a] \to R^{2}$

\int_{c^{+}} F \cdot d s = - \int_{c^{-}} F \cdot d s

next Green's Theorem