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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?

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  • $\begingroup$ Two proofs isn't enough for you? Why? $\endgroup$ – TonyK Sep 11 '19 at 21:45
  • $\begingroup$ @TonyK Actually I wanted another proof that does not utilise any calculus concepts. $\endgroup$ – Dimitris Sep 12 '19 at 8:14
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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

$$ A(a,b) = \frac{2\pi b^4}{2a^2(1-e^2)^{3/2}}=\large{\color{red}{\pi a b}}. $$

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  • $\begingroup$ Thanks a lot for this proof. I also found this qc.edu.hk/math/Resource/AL/… $\endgroup$ – Dimitris Sep 11 '19 at 19:49
  • $\begingroup$ @Dimitris: glad you like it, this was more or less an exercise in style. It is more usual to exploit the area of the ellipse to compute integrals of the form $$\int_{0}^{2\pi}\frac{d\theta}{\Omega\pm\cos\theta}\quad\text{or}\quad \int_{0}^{2\pi}\frac{d\theta}{\left(\Omega\pm\cos\theta\right)^2}.$$ $\endgroup$ – Jack D'Aurizio Sep 11 '19 at 19:52
  • $\begingroup$ When using the the stretch-circle argument for the area of ellipse formula (that if someone wants to avoid calculus based concepts) how can one explain why this argument fails to provide the perimeter of the ellipse? Thanks again! $\endgroup$ – Dimitris Sep 11 '19 at 20:05
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    $\begingroup$ @Dimitris: affine maps preserve the ratio of areas, not the ratio of lengths! $\endgroup$ – Jack D'Aurizio Sep 11 '19 at 20:14
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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 $C'D'$.

enter image description here

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  • $\begingroup$ (+1) Neat! $\phantom{}$ $\endgroup$ – Jack D'Aurizio Sep 12 '19 at 1:27
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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