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.