Line Integral summary

By now, we've accumulated many strategies for calculating the line integral: $$\int_C \myv F\cdot d\myv r$$ where

• $\myv F$ is a vector function, with components $$\nonumber\myv F\equiv\myv F(x,y,z)=P(x,y,z)\uv i+Q(x,y,z)\uv j+R(x,y,z)\uv k.$$
• $C$ is a label for a particular path / contour / curve in space. Parameterized, the path is $\myv r(t)=\myc{ x(t), y(t), z(t)}$
• We can think of $d\myv r\equiv \uv T d s$ as a vector "step" along the path, (along the arc) of length $|d\myv r|$ in a direction tangent to the path.

Parameterize everything in sight

Parameterize the path in terms of a variable $t$, that is: $$C\equiv \myv r(t)=\myc{ x(t), y(t), z(t) }.\nonumber$$

In this approach, think of the parameter $t$ as the time. Then the velocity vector for a particle moving along the path $\myv r(t)$ is: $$\nonumber \myv r'(t)=\myc{ x'(t), y'(t),z'(t)} =\myc{ \frac{dx}{dt}, \frac{dy}{dt},\frac{dz}{dt}}.$$ The velocity vector is tangent to the path. So, an increment of arclength $d\myv r$ along the path is (using displacement = velocity * time) $$d\myv r= \myv r'\,dt\nonumber.$$

We can write our vector integral as: $$ \int_{t_i}^{t_f}\myv F(x(t),y(t),z(t))\cdot\langle x'(t),y'(t),z'(t)\rangle\,dt$$