Multivariable Exam 3 Review: Thomas 16.1
By Tyler Clarke in Calculus on 2025-4-3
We're finally into vector calculus! This starts out interesting, and gets even cooler. 16 has been my favorite chapter so far.
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 area element: 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 parameterization 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:
A Significantly Less Brief Table of Area Elements
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`dV = dx dy dz` (or other orderings thereof)
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`dV = r dz dr d theta` (or other orderings thereof)
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`dV = rho^2 sin(phi) d rho d phi d theta` (or other orderings thereof)
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`dV = Jac(r(u, v)) dx dy` (or other orderings thereof; same pattern in 3d)
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`dV = |frac {dr}{dt}| dt`
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`.
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) = `, and the radial span of the first quadrant is from `0` to `frac pi 2`. Thus, `vec r(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`.
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 conservative - 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`.
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).
