Targeted Audience: Year 1/2 Math/Physics majors. You may want to read a bit about wedge products and exterior derivatives

Multivariable Calculus is a prerequisite.

Statement

An Ellipse with semi-major axis a and semi-minor axis b has area \(\pi ab\)

Proof

There are many ways to approach this problem. Proofwiki provides 2 methods to do it, but they are considerably tedious and non-elegant.

Here we will use the powerful Green’s Theorem, which is closely related to Areas (Since it is the 2D version of Stoke’s theorem). This is rarely explained in lectures and is worth a discussion:

Consider Green’s Theorem (Stoke’s Theorem on \(xy\) plane):

\[\oint_C(P d x+Q d y)=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\]

We aim to find polynomials such that \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1\)

Say \(P(x,y)=-\frac{1}{2}y\) and \(Q(x,y)=\frac{1}{2}x\)

So we can estimate the area (denote by \(A\)) as:

\[A=\frac{1}{2} \oint_C(x dy-ydx)\]

Using the standard parameterization of an ellipse (see below), we can quickly calculate \(A\)

\[{\displaystyle (x,y)=(a\cos(t),b\sin(t))\quad {\text{for}}\quad 0\leq t\leq 2\pi }\] \[\begin{aligned} A & =\frac{1}{2} \int_C x d y-y d x=\frac{1}{2} \int_0^{2 \pi}[(a \cos t)(b \cos t) d t-(b \sin t)(-a \sin t) d t] \\ & =\frac{a b}{2} \int_0^{2 \pi} d t=\pi a b \end{aligned}\] \[\blacksquare\]

Collorary: A Circle with radius \(r\) has area \(\pi r^2\). The proof is trivial and is left as an exercise for the reader.

Generalized Stokes’ Theorem

For a sphere, consider: \(\alpha=x^1 d x^2 \wedge d x^3 .\)

Consider:

\[x^1=\sin\alpha\cos\phi\] \[x^2=\sin\alpha\sin\phi\] \[x^3=\cos\alpha\]

Thus: \(\alpha=\sin ^3 \alpha \cos ^2 \phi d \alpha \wedge d \phi\)

This is clearly a surface integral. Hence, we have:

\[\int_{\partial\Sigma} \alpha=\int_{\alpha=0}^\pi \int_{\phi=0}^{2 \pi} \sin ^3 \alpha \cos ^2 \phi d \alpha d \phi=\frac{4 \pi}{3} .\]

On the other hand, we have: \(d \alpha=d x^1 \wedge d x^2 \wedge d x^3\) and so

\[\int_{\Sigma} d \alpha=\int_{\Sigma} d x^1 \wedge d x^2 \wedge d x^3= \int_{0}^{2\pi} \int_{0}^{ \pi} \int_{0}^{1} (r^2 \sin \phi) dr d\alpha d\phi = \frac{4}{3} \pi\]

Hence we have showed that Generalized Stokes’ Theorem: \(\int_{\Sigma} d \alpha=\int_{\partial \Sigma} \alpha\)

works on a unit sphere.

Green’s Theorem (Special Case)

Considering Toby Lam’s Post. We can approach the same problem using wedge products. If you haven’t read his post, the problem we are considering is:

\[\text { Area of } D=\iint_D d x d y=-\int_C y d x\]

First, consider: \(d\alpha=dy \wedge dx\). We can easily see that \(\alpha=ydx\)

\[\text { Area of } D=\iint_\Sigma d x \wedge d y=-\iint_\Sigma d y \wedge d x=-\iint_\Sigma d\alpha=-\int_{\partial \Sigma} \alpha=-\int_{\partial \Sigma} ydx\]

For the general case, using Generalized Stokes’ Theorem , we can consider:

\[\int_{\partial D} P d x+Q d y=\iint_{D}dP \wedge dx+dQ \wedge dy\]

As:

\[d P=\sum_{i=1}^n \frac{\partial P}{\partial x^i} d x^i\]

We have:

\[\iint_{D}\frac{\partial P}{\partial y} d y \wedge dx+\frac{\partial Q}{\partial x} dx \wedge dy = \iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy\]

So:

\[\int_{\partial D} P d x+Q d y=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy\]

The astute reader can realized that this is Green’s Theorem - Stokes’ Theorem in 2D!