Ways to prove the area of an ellipse formula One can prove the ellipse area formula $A=\pi a b$ ($a$, $b$ the major and minor semi-axis) either by integration or by the stretched-circle argument. See for instance here: https://proofwiki.org/wiki/Area_of_Ellipse
Are there any other proofs of this formula?
 A: Another way could be via Cavalieri's principle. Suppose we have an ellipse with semi-axes $a$ and $b$ and construct a circle with radius $\sqrt{ab}$. Draw a pair of lines $FF'$, $EE'$ tangent to the circle at the endpoints of a diameter (see figure below) and rotate the ellipse so that it has those lines as tangents. If we show that any chord $CD$ of the circle, parallel to $FF'$, has the same length as the corresponding chord $C'D'$ of the ellipse, then circle and ellipse have the same area $\pi ab$.
Let's consider first of all the diameter $AB$ of the circle parallel to $FF'$: the corresponding diameter $A'B'$ of the ellipse forms with $E'F'$ a pair of conjugate diameters. It is well known that the parallelogram formed by semi-diameters $O'B'$ and $O'F'$ has area $ab$: as the height of such parallelogram with respect to base $O'B'$ is $\sqrt{ab}$ we deduce then $O'B'=\sqrt{ab}$. It follows that $AB=A'B'$.
Let's now prove the analogous equality for any other couple of corresponding chords $CD$ and $C'D'$. The equation of conjugate diameters states that
$$
{O'M'^2\over O'F'^2}+{C'M'^2\over O'A'^2}=1,
$$
but on the other hand we also have:
$$
{O'M'\over O'F'}={OM\over OF}={OM\over \sqrt{ab}}.
$$
Inserting this into the previous equality, and taking into account that $O'A'=\sqrt{ab}$ we get:
$$
C'M'=\sqrt{ab-OM^2}=CM,
$$
where we also used Pythagoras' theorem applied to triangle $OMC$. This completes the proof, because $M'$ is the midpoint of $C'D'$.

A: Let's get creative. For any $R>1$ we have
$$ f_R(\theta)=\frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}(-1)^n \frac{e^{ni\theta}}{R^{n+1}} $$
hence by Parseval's identity
$$\int_{0}^{2\pi}\frac{d\theta}{(R^2+1)+2R\cos\theta}= \int_{0}^{2\pi}f_R(\theta)\overline{f_R}(\theta)\,d\theta = 2\pi\sum_{n\geq 0}\frac{1}{R^{2n+2}}=\frac{2\pi}{R^2-1}$$
and by multiplying both sides by $(R^2+1)$ we get
$$ \int_{0}^{2\pi}\frac{d\theta}{1+\frac{2R}{R^2+1}\cos\theta}=2\pi\frac{R^2+1}{R^2-1}. $$
Through a substitution we get that for any $B>1$ the identity
$$ \int_{0}^{2\pi}\frac{d\theta}{B+\cos\theta} = \frac{2\pi}{\sqrt{B^2-1}} $$
holds. By differentiating both sides with respect to $B$ we get
$$\forall B>1,\quad \int_{0}^{2\pi}\frac{d\theta}{(B+\cos\theta)^2}=\frac{2\pi B}{(B^2-1)^{3/2}}.\tag{1} $$
Let us assume that an ellipse has semiaxis $a>b$. How can we write its polar equation with respect to a focus?
By imposing that the property $PF_1+PF_2=2a$ holds, i.e. by imposing that
$$ \rho(\theta) + \sqrt{\rho^2(\theta)\sin^2(\theta)+(\rho(\theta)\cos(\theta)-2c)^2} = 2a, $$
$$ \rho^2(\theta)-4c\rho(\theta)\cos(\theta)+4c^2 = 4a^2-4a\rho(\theta)+\rho^2(\theta), $$
$$ -c\rho(\theta)\cos(\theta) = b^2-a\rho(\theta), $$
$$ \rho(\theta) = \frac{b^2/a}{1-e\cos\theta} \tag{2}$$
where $e=\frac{c}{a}=\sqrt{1-\frac{b^2}{a^2}}$. The area enclosed by the ellipse is 
$$ A(a,b)=\frac{1}{2}\int_{0}^{2\pi}\rho^2(\theta)\,d\theta = \frac{b^4}{2a^2}\int_{0}^{2\pi}\frac{d\theta}{(1-e\cos\theta)^2}\stackrel{\theta\mapsto\pi+\theta}{=}\frac{b^4}{2a^2 e^2}\int_{0}^{2\pi}\frac{d\theta}{\left(\frac{1}{e}+\cos\theta\right)^2}\tag{3}$$
and by invoking $(1)$ we have
$$ A(a,b) = \frac{2\pi b^4}{2a^2(1-e^2)^{3/2}}=\large{\color{red}{\pi a b}}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\mathcal{A} & =
\iint_{x^{2}/a^{2}\ +\ y^{2}/b^{2}\ =\ 1}\dd x\,\dd y =
ab\iint_{x^{2}\ +\ y^{2}\ =\ 1}\dd x\,\dd y =
ab\int_{0}^{2\pi}\int_{0}^{1}r\,\dd r\,\dd\phi
\\[5mm] & =
\bbox[15px,#ffc,border:1px solid navy]{\pi ab}
\end{align}
