Integrate the Circumference of an Ellipse to Find the Area

Question

For a circle of radius R, one can find the area by integrating the circumference equation in the interval $(0, R)$,

$$\text{Area} = \int^R_0 2\pi r\ dr = \pi R^2$$

My intuition for this is that we're doing a continuous sum over all circles with radius in the range $(0, R)$, this family of circles fills up the whole space and gives us the area.

Is there a way to do this for an ellipse?

The circumference of an ellipse with semi-major $a$ and semi-minor $b$ is:

$$\text{Circumference} = 4\int^{\pi/2}_0 \sqrt{a^2 \cos^2(\theta) + b^2\sin^2(\theta)}\ d\theta$$

I tried to consider an ellipse with semi-major axis $A$ and semi-minor $B$ and a family of ellipses with semi-major $At$ and semi-minor $Bt$ such that we can scale the ellipse by a factor $t$.

I then considered that the collection of ellipses we need to "fill" our area are those where $t$ is in the interval $(0,1)$. Considering this, I tried integrating over this interval:

$$\text{Area} \stackrel{?}{=} 4\int^1_0\int^{\pi/2}_0 \sqrt{A^2t^2 \cos^2(\theta) + B^2t^2\sin^2(\theta)}\ d\theta\ dt$$

I'm pretty sure this isn't correct though (the area of the ellipse should be $\pi AB$).

I think I sort-of understand why it doesn't work. When you scale a circle, the space between any point before and after the scale is the same for all points. Ellipses don't do that, which I think is why my "filling" intuition here needs something extra.

My differential geometry is a bit rusty but I feel like there should be a way to make this work by using the correct element for integration. I'm just not sure how to get there.

Answer 1

Observe that an ellipse, with minor axis $a$ and major axis $b$, can be view as the part of the plane $z=y\tan\beta $ inside the cylinder $x^2+y^2=a^2$, where $\beta$ is the angle it forms with the $xy$-plane, satisfying $\cos\beta = a/b$.

Now, we could consider that the ellipse is made up from lots of ellipse rings that, when projected onto the $xy$-plane, become corresponding circles. Those circles can then be integrated easily.

The surface integral for the ellipse is then given by

$$ S=\int_0^a \int_0^{2\pi} f(r,\theta) rdr d\theta$$

where the projection, or the scaling, factor is actually fairly simple,

$$ f(r,\theta)=\sqrt{1+\left(z_y^{’}\right)^2}=\sec\beta= \frac{b}{a}$$

The area of the ellipse becomes

$$ S=\int_0^a \frac{2\pi b}{a}rdr \tag{1}$$

As seen, the area of the ellipse rings that fill up the whole ellipse is scaled as $2\pi r(b/a)$. This may also be viewed equivalently as the ‘circumference’ of each ellipse ring.

As expected, the surface integral (1) yields

$$S=\pi ab$$

Answer 2

You can use the map $$g:\quad(t,\theta)\mapsto\left\{\eqalign{x&=a\, t\cos\theta \cr y&= b\, t\sin\theta\cr}\right.\qquad(0\leq t\leq 1, \ 0\leq\theta\leq 2\pi)$$ as parametrization of the elliptical disc $E$. For constant $t$ you obtain smaller ellipses embedded in $E$, and for constant $\theta$ you obtain rays from $(0,0)$ to the given boundary ellipse. In order to compute the area of $E$ you need the Jacobian $$J_g(t,\theta)=\det\left[\matrix{x_t&x_\theta\cr y_t&y_\theta\cr}\right]= ab\, t$$ and then obtain $${\rm area}(E)=\int_E 1\>{\rm d}(x,y)=\int_{\hat E}1\>\bigl|J_g(t,\theta)\bigr|\>{\rm d}(t,\theta)=\int_0^{2\pi}\int_0^1 ab\>t\>dt\>d\theta=\pi a b\ .$$