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.

  • $\begingroup$ One trick is to scale your distance function differently in the direction of the major axis than in the direction of the minor axis, making all points of the ellipse be at an equal "distance" from the center. This has the effect of making the distance between two concentric ellipses uniform. This also changes the formula for the length of the circumference. Or you could get the exact same effect by projecting the ellipse onto a circle as in Quanto's answer. $\endgroup$
    – David K
    Aug 26, 2019 at 2:25

2 Answers 2


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

$$S=\pi ab$$


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 $${\rm area}(E)=\int_E 1\>{\rm d}(x,y)=\int_{\hat E}1\>\bigl|J_g(t,\theta)\bigr|\>{\rm d}(t,\theta)=\int_0^{2\pi}\int_0^1 ab\>t\>dt\>d\theta=\pi a b\ .$$


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