Multivariable Exam 3 Review: Thomas 16.5

By Tyler Clarke in Calculus on 2025-4-10

Hello again, mes amis! We've only four sections left in the chapter, and this is one of them. In this section, we're taking a bit of a trip down memory lane, revisiting some old concepts with new insight: namely, parametric vector forms. If you don't already know how to handle parametric forms with one free variable (a line or path), you should probably learn about those externally; we're covering parametric forms with two free variables. Surfaces!

The general idea is that, just like you can represent any line or path with a function `r(u) = [x(u), y(u), z(u)]`, you can represent any surface with a function `r(u, v) = [x(u, v), y(u, v), z(u, v)]`. `u` and `v` might be bounded, or might not. That's all there is to it!

For example, say we're given the plane `x + y + z = 1`. How can this be vector parameterized? Obviously doing this in one free variable won't work; it's not possible to represent a plane with a line, no matter how bendy. In two variables, however, it's quite possible. There are many, many possible ways to find the paramterization, but I'm going to use the most obvious one: set `x = u`, `y = v`, and we have `u + v + z = 1`: algebra turns this into `z = 1 - u - v`. So now we have three equations, relating each dimension to two variables. It's a parameterization! Sub'ing in gives us `r(u, v) = [u, v, 1 - u - v]`.

What about a more complicated shape? It turns out that most shapes are pretty intuitive: for instance, a cylinder with radius `rho` could be parameterized as `r(u, v) = [rho cos(u), rho sin(u), v]` where `v` is bounded to the height of the cylinder and `0 <= u <= 2 pi`. It's critical to remember, when parameterizing, the bounds of `u` and `v` are your choice - unless you've actually been provided bounds, you can set them to whatever you want; this means cylindrical (or even spherical!) coordinates are an option.

It's often useful to find the tangent plane at a given point on a surface. This is much, much easier if you have a parametric vector function. If we have a function `r(u, v)` describing a curve, and a point `(u, v)` where we want to construct the tangent plane (note: if you're given a point in `(x, y, z)`, you'll have to do some algebra to find `u` and `v`), we immediately know that the initial point `p` is `r(u, v)`. The normal vector is a bit harder to find: the function for it is `n = frac {dr} {du} times frac {dr} {dv}`. To understand why this works, you'll need to visualize `frac {dr} {du}` and `frac {dr} {dv}` as being directional vectors along the surface: assuming they both exist and are nonequal, both are aligned with the surface, and so the cross product between them must point directly out of or into the surface. Once we know `n` and `p`, of course, the plane is simply `n cdot [x, y, z] = n cdot p`.

Let's do an example. We've the sphere `r(u, v) = [4cos(u)sin(v), 4sin(u)sin(v), 4cos(v)]`, `0 <= u <= 2pi`, `0 <= v <= pi`. What's the tangent plane at the point `(2, 2, 2sqrt(2))`? (I might be a bit too lazy with these examples). We know that `4cos(u)sin(v) = 2`, `4sin(u)sin(v) = 2`, and `4cos(v) = 2sqrt(2)`. The first two are hairy, but that last one works out nicely to `v = frac {pi} 4`. Because we know `sin(frac {pi} 4) = frac {sqrt(2)} 2`, the equation for y becomes `4sin(u) = frac {4} {sqrt(2)}` - `sin(u) = frac {sqrt(2)} 2`. Obviously, then, `u = frac {pi} 4`.

Our initial point `p` was helpfully provided, so we don't have to do any work to find that. Now onto the normal vector! `frac {dr} {du} = [-4sin(u)sin(v), 4cos(u)sin(v), 0]`, and `frac {dr} {dv} = [4cos(u)cos(v), 4sin(u)cos(v), -4sin(v)]`. Calculating the cross product of those is a good exercise, but ultimately fairly simple; I recommend you do it, but if you don't want to, the answer is `[-16 cos(u) sin^2 (v), -16 sin(u) sin^2 (v), -16 sin(v) cos(v)]`. This is... pretty awful. Fortunately, we're almost done: all we have to do is plug in those values for `u` and `v`. Final result: `n = [-4sqrt(2), -4sqrt(2), -8]`. And we're done! Constructing the plane equation is really very easy; we end up with `-4sqrt(2)x - 4sqrt(2)y - 8z = -32sqrt(2)`, or, simplified, `sqrt(2)x + sqrt(2)y + 2z = 8sqrt(2)`.

Finally: after parameterizing a surface, it becomes quite simple to find the area. Plug in your bounds for `u` and `v` into a double integral, and simply apply the area element! For a general parameterized function, the area element is actually just `abs {frac {dr} {du} times frac {dr} {dv}}` - the magnitude of the normal vector to the tangent plane. In our sphere example above, that would be `sqrt((-16 cos(u) sin^2 (v))^2 + (-16 sin(u) sin^2 (v))^2 + (-16 sin(v) cos(v))^2)` - which simplifies to `16sin(v)`, magically enough. The integral is thus `int_0^{2pi} int_0^{pi} 16sin(v) dv du = 64pi`. This matches up perfectly with the formula for the surface area of a sphere, `4pi r^2`. Great! Let's add that to the element of integration table:

An Increasingly Exhaustive Table of Area Elements

• `dV = dx dy dz` (or other orderings thereof)
• `dV = r dz dr d theta` (or other orderings thereof)
• `dV = rho^2 sin(phi) d rho d phi d theta` (or other orderings thereof)
• `dV = Jac(r(u, v)) dx dy` (or other orderings thereof; same pattern in 3d)
• `dV = |frac {dr}{dt}| dt`
• `dA = abs {frac {dr} {du} times frac {dr} {dv}} du dv` for a surface parameterization `r(u, v)`.

I've intentionally ignored implicit surfaces because they don't seem to be showing up in this course. If you actually have the Thomas textbook, I'd recommend reading through that part, for interestingness value if nothing else. Until 16.6, arrivederci and good luck!