Multivariable Exam 3 Review: Thomas 16.2

By Tyler Clarke in Calculus on 2025-4-3

Welcome back! We continue chapter 16 with a discussion of vector fields. Essentially, a vector field is a function with the same input dimensions as output dimensions. In `R^2` and `R^3`, we visualize this as a bunch of vectors permeating all of space. If you've taken physics, you've probably encountered vector fields - the electric and magnetic fields are in fact vector fields! A way to think about vector fields is as the velocity of a given infinitesimal region of a fluid: it's at a position, and the velocity vector is pointing in some direction with some magnitude. This is actually a very good way to represent a flowing fluid, for our purposes.

An example of a vector field we've used extensively thus far is the gradient function of a surface! At every point along the surface, it outputs a vector representing the slope at that point.

One of the most useful operations we can perform on a vector field is a line integral (remember 16.1?). The integral of a vector field over a line is a fairly abstract concept, but it has real-world uses: imagine we have a function for the force at any given point, and we want to know the total force over a line. Vector fields! The form for a vector field integral is simply the dot product of the vector field and the tangent vector to the point on our line: given a line `r(t)` in a vector field `f(x, y) = [u, v]`, we have `int f(x, y) * dr`. This is one of those very elegant things that has interesting consequences.

Let's do an example. Straight from the textbook, we have `f(x, y, z) = [z, xy, -y^2]` and `r(t) = [t^2, t, sqrt(t)]`, where `a <= t <= b`. Getting `f(x, y, z)` in terms of `t` is simple enough: just substitute the components of `r`: `f(t) = [sqrt(t), t^3, -t^2]`. `dr` is going to be `frac {dr} {dt} * dt`, so let's find `frac {dr} {dt}`: `frac d {dt} [t^2, t, sqrt(t)] = [2t, 1, frac 1 {2sqrt(t)}]`. So our integral is `int_a^b [sqrt(t), t^3, -t^2] * [2t, 1, frac 1 {2sqrt(t)}] dt`. Evaluating the dot product gives us `int_a^b 2t^{3/2} + t^3 - frac {t^{3/2}} {2} dt`. I'll leave further simplification as an exercise to the reader. In the case where `a = b`, this is also known as the circulation.

In some cases, we may want to know only the line integral in a single direction: in our wire-under-force example, this would mean finding only the x-component or y-component or z-component of the force. In these cases, it's as simple as zeroing the unused components: if your vector field is `F(x, y, z) = [m(x, y, z), n(x, y, z), o(x, y, z)]`, and we want to integrate over the y-axis, our vector field becomes `F(x, y, z) = [0, n(x, y, z), 0]`. Simple enough! If you prefer a vector solution, then taking the unit vector `vec u` along the axis we care about, we have `F_{vec u}(x, y, z) = vec u * (F(x, y, z) * vec u)`. Let's do an example. What is specifically the y-component of the line integral `f(x, y, z) = [z, xy, -y^2]`? Given that `vec y = [0, 1, 0]`, we have `[0, 1, 0] * ([x, xy, -y^2] * [0, 1, 0]) = [0, xy, 0]`. We already know that `dr = [2t, 1, frac 1 {2sqrt(t)}] dt`. `int_a^b [0, xy, 0] * [2t, 1, frac 1 {2sqrt(t)}]` simplifies to `int_a^b xy dt`. We've already determined `xy` = `t^3`, so this becomes simply `int_a^b t^3 dt`.

Now onto flux. The flux across a simple closed curve (that is, a closed curve that does not cross over itself) in the xy plane is similar to the line integral, except it finds the value that crosses the curve, rather than the value along the curve: if the line integral of a vector field is total force along the curve, then the flux over a vector field is the total force perpendicular to the curve. Essentially, `int_a^b f(x, y, z) * dn` rather than `int_a^b f(x, y, z) * dr`, where `n` is the outward normal vector of `r`. What is an outward normal vector? For a closed curve, it's a vector pointing perpendicular to the curve and facing away from the curve (that is, traveling in the direction of the normal vector will lead you out of the boundary of the curve). For a 2d curve, you can find this normal vector with the cross product of tangent vector and the perpendicular axis - for instance, for a curve parallel to the xy plane, `T xx k`. Note that this may sometimes point inwards, and sometimes point outwards! It's up to you to flip the direction, if necessary. Let's do an example: let's find the flux of the line `r(t) = [t, 1 - t]` in the circulating vector field `f(x, y) = [-y, x]`. We can easily determine that `dr = [1, -1] dt`, and that cross `k` is simply `[1, -1]` (are the dimensions here a problem? No. To find the cross product, I briefly extended `dr` into three dimensions with a `0` in the z-component, and then cut off the z-dimension at the end. This is a perfectly valid operation in this specific situation.)

Hence, `dn = [1, -1] dt`, meaning our integral is `int [-y, x] * [1, -1] dt` (no bounds necessary - we aren't going to evalute a final result). A quick substitution and a dot product gives us `int -1 dt`, the indefinite integral of which is `-t + C`. Simple, and geometrically accurate.

Section 16.3 will go into quite a bit more detail, as we'll be dealing with the problem of conservative vs nonconservative paths. For now, da svidanya and bonne nuit!