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<title>Line Integrals SUMMARY</title>
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<h1>Line Integral summary</h1>
<p>By now, we've accumulated many strategies for calculating the line integral:
$$\int_C \myv F\cdot d\myv r$$
where
<ul><li>$\myv F$ is a vector function, with components $$\nonumber\myv F\equiv\myv
F(x,y,z)=P(x,y,z)\uv i+Q(x,y,z)\uv j+R(x,y,z)\uv k.$$
<li>$C$ is a label for a particular path / contour / curve in space.  Parameterized, the path is $\myv r(t)=\myc{ x(t), y(t), z(t)}$
<li>We can think of $d\myv r$ as a vector "step" along the path,  of length $|d\myv r|$ in a direction tangent to the path.
</ul>

<h2>Parameterize everything in sight</h2>
<p>Parameterize the path in terms of a variable $t$, that is:
$$C\equiv \myv r(t)=\myc{ x(t), y(t), z(t) }.\nonumber$$
<p>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)
$$d\myv r= \myv r'\,dt\nonumber.$$

<p>We can write our vector integral as:
$$ $$



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