Surface Integral

Motivation

KE, CC-BY, "torus".

Have you ever thought to yourself, "I wonder how much powdered sugar I could get on a donut?"
I would assume no but I can get you on the right track.

Vectors

Disclaimer: this page will be a little out there, even for me.

A vector is a combination of a scalar and a direction. Technically, that is. Less formally, we can say that a vector is the combination of a length and a direction. For example, an arrow pointing up with length 3 is a vector. There are many ways to notate a vector but for this we will use the standard $x$ , $y$ ordered pair notation with angled brackets. A vector is assumed to have its tail at the origin and its head at the point shown in $⟨ x, y ⟩$ . Its length (by simple trigonometry) is $\sqrt{x^{2} + y^{2}}$ and its direction (or angle) is $\tan^{- 1} (\frac{y}{x})$ .

Easy enough, right? Right, but now it gets worse.
To find the surface area of a donut (torus) we must realize that a donut is 3-dimensional.
That means that our vectors will be of the form $\vec{v} = ⟨ x, y, z ⟩$ . Our length will be $\sqrt{x^{2} + y^{2} + z^{2}}$ and our direction will be a little bit (lotta bit) more complicated and unnecessary.

Operations

Let us notice that $a \cdot ⟨ x, y, z ⟩ = ⟨ a x, a y, a z ⟩$ . Where $a$ is just a number.

Another property to notice is that $⟨ x, y, z ⟩ + ⟨ a, b, c ⟩ = ⟨ x + a, y + b, z + c ⟩$ .

Using $\hat{ı} = ⟨ 1, 0, 0 ⟩$ , $\hat{ȷ} = ⟨ 0, 1, 0 ⟩$ , and $\hat{k} = ⟨ 0, 0, 1 ⟩$ , We can rewrite any vector $⟨ x, y, z ⟩$ as $x \hat{ı} + y \hat{ȷ} + z \hat{k}$ .

We define the cross product of two vectors to be the determinant of the matrix made up of the basis vectors and the two other vectors together, that is: for any two vectors, $\vec{u} = ⟨ x, y, z ⟩$ and $\vec{v} = ⟨ a, b, c ⟩$ , $\vec{u} \times \vec{v} = | \begin{matrix} \hat{ı} & \hat{ȷ} & \hat{k} \\ x & y & z \\ a & b & c \end{matrix} | = \hat{ı} (y c - z b) + \hat{ȷ} (z a - x c) + \hat{k} (x b - y a) = ⟨ y c - z b, z a - x c, x b - y a ⟩$ Geometrically, the cross product gives us a new vector that is one of the vectors that is perpendicular to both.

Surface Integral

Consider that $Ω$ is some surface (in our case, a donut) in 3 dimensions.
$\vec{v} (t, s)$ is a function that takes in two parameters and gives out a vector. Using the desired constrained inputs for the function, the function will map out $Ω$ .
Finally, consider $C$ being the space of the (constraint of $t$ ) $\times$ (the constraint of $s$ ).

In our case, $Ω$ is a donut, $\vec{v} (t, s) = ⟨ (b + a \cos (t)) \sin (s), (b + a \cos (t)) \cos (s), a \sin (t) ⟩$ , and our constraints are that $t$ and $s$ both go from $0$ to $2 π$ so $C = [0, 2 π] \times [0, 2 π]$ . Where $a$ is the radius of the loop and $b$ is the radius of the donut.

The formula for surface area is $\iint_{C} | \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | d t d s$
Using our amazing calculus skills, we can rewrite this as $\int_{0}^{2 π} \int_{0}^{2 π} | \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | d t d s$ Calculating the partial derivatives, we get: $\frac{\partial \vec{v}}{\partial t} = ⟨ - a \sin (s) \sin (t), - a \cos (s) \sin (t), a \cos (t) ⟩$ And calculating the other, we get: $\frac{\partial \vec{v}}{\partial s} = ⟨ (b + a \cos (t)) \cos (s), - (b + a \cos (t) \sin (s)), 0 ⟩$ Skipping most of the hard work, we find the determinant (which geometrically is the area of the parallelogram created by the two vectors' cross product): $| \frac{\partial \vec{v}}{\partial t} \times \frac{\partial \vec{v}}{\partial s} | = a b + a^{2} \cos (t)$ Now we can rewrite the integral and solve! $\int_{0}^{2 π} \int_{0}^{2 π} a b + a^{2} \cos (t) d t d s = \int_{0}^{2 π} {[a b t + a^{2} \sin (t)]}_{0}^{2 π} d s = \int_{0}^{2 π} 2 π a b d s = {[2 π a b s]}_{0}^{2 π} = 4 π^{2} a b$ Wow. We found our answer. The surface area of a donut is $4 π^{2}$ times its inner radius times its outer radius. We could have just looked that up but we didn't. Astounding.