deadlyboringmath.us/site/posts/multivariable-thomas-16.1.html

[!]
[=post-]
<p>
    We're finally into vector calculus! This starts out interesting, and gets even cooler. 16 has been my favorite chapter so far.
</p>
<p>
    We're diving into this with the line integrals of scalar functions. Essentially, these are integrals over a path of a function defined at every point
    along that path - in physical terms, the line integral of density is the mass of a path. This technique, however, is quite a bit more general than that:
    you can also find work done, arc length, etc, in various different configurations. Rather like with cylindrical and spherical areas, the interesting bit here
    is the new <i>area element</i>: it's the arc length, `dV = |r'| dt = sqrt((frac{dx}{dt})^2 + (frac{dy}{dt})^2 + (frac{dz}{dt})^2) dt` (the same is true in multiple dimensions;
    just add more terms to the pythagorean identity). Note that the final result is in `dt`: this is because `r(t)` is a <i>parameterization</i> of the curve,
    and so this works out to a single variable integral over the `t`-bounds. Our element of integration table now looks like this:
</p>
<h3>A Significantly Less Brief Table of Area Elements</h3>
<ul>
    <li>
        `dV = dx dy dz` (or other orderings thereof)
    </li>
    <li>
        `dV = r dz dr d theta` (or other orderings thereof)
    </li>
    <li>
        `dV = rho^2 sin(phi) d rho d phi d theta` (or other orderings thereof)
    </li>
    <li>
        `dV = Jac(r(u, v)) dx dy` (or other orderings thereof; same pattern in 3d)
    </li>
    <li>
        `dV = |frac {dr}{dt}| dt`
    </li>
</ul>
<p>
    Let's do a few quick examples. First up: given a density function `d(x, y) = xy` and a line from `(1, 0)` to `(2, 3)`, what is the mass along the line?
    The first step here is to parameterize. Because this is a line, our parameterization will look like `vec r(t) = vec m + vec bt`. Our starting point is
    `vec b = (1, 0)`, and the difference between the endpoint and the starting point is `(1, 3)`, so we have `vec r(t) = (1, 0) + (t, 3t)`. Next up are the bounds,
    but these are very easy: `vec r(t) = (1, 0)` when `t=0`, and `vec r(t) = (2, 3)` when `t=1`, so our bounds are `0 <= t <= 1`. Now we need the density function
    in terms of `t`. Because `r_x(t) = 1 + t`, and `r_y(t) = 3t`, we can substitute in quite easily: `d(t) = xy = 3t + 3t^2`. Finally, we need to get the area
    element. `vec r'(t) = (1, 3)`, so it's `sqrt(1^2 + 3^2)dt = sqrt(10)dt`. This gives us a final integral `3sqrt(10) int_0^1 t + t^2 dt`, which is quite easy to evaluate
    to `3sqrt(10) * frac 5 6`.
</p>
<p>
    Another example is to find the mass of a circular arc in the first quadrant, with radius 2 and density `x^2 + y^2`. Remember that the parameterization of
    a circle is `vec c(t) = \[rcos(t), rsin(t)]`, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(t) = \[2cos(t), 2sin(t)]` for `0 <= t <= frac pi 2`.
    The density function here is convenient because it's just `r^2`, based on the pythagorean identity. If you substitute in `2cos(t)` for `x` and `2sin(t)` for y, you will get
    `2^2 = 4`. Constant density is always nice! Finally, we need to find the element of integration. `vec r'(t) = \[-2sin(t), 2cos(t)]`, so `|r'|dt` = `sqrt(4sin^2(t) + 4cos^2(t))` -
    `2 dt`. Thus, we have the integral `int_0^{frac pi 2} 8 dt`; this obviously works out to `4pi`.
</p>
<p>
    A convenient property of line integrals is that they're additive: if you know the line integral over a bunch of connected segments of a path, the line integral of the
    whole path is just the sum of those segments. This is a fairly obvious property, and I won't go too far into detail: just remember that additivity is true even for
    fairly strange paths (as long as they're <i>conservative</i> - we'll get into what that means in a few sections), so if you know for instance `int_0^1 f(t) dt = 3`, `int_1^3 f(t) dt = 7`,
    and `int_2^3 f(t) dt = 4`, you know that `int_0^2 f(t) dt = int_0^1 f(t) dt + int_1^3 f(t) dt - int_2^3 f(t) dt = 3 + 7 - 4 = 6`.
</p>
<p>
    The first and second moments of a given line integral work the same way as in other integrals. For first moment about an axis, multiply the density function by that axis;
    for second moment about an axis, multiply by the squared distance to that axis. For example, if the mass of a path is `int_0^5 f(t) dt`, the first moment about the x-axis is
    `int_0^5 x f(t) dt` (substitution with the parameterization is necessary!), and the second moment (moment of inertia) is `int_0^5 (y^2 + z^2)f(t) dt` (with substitution, of course).
</p>
[/]
[=title "Multivariable Exam 3 Review: Thomas 16.1"]
[=author "Tyler Clarke"]
[=date "2025-4-3"]
[=subject "Calculus"]
[#post.html]