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I am very confused about the area of an ellipse. I tried to derive it using two different methods. The cartesian coordinate method yielded $\pi ab $, as expected. Then using polar coordinates I got $\pi (a^2+b^2)/2$. I had never seen this form for the area before so at first I thought I did something wrong. But in one place here, I found that someone also had this expression. The person who have this result left it to the OP to show that these were equivalent. I have been trying to do so, but can't seem to get this result. In particular, I'd the above two were equivalent then

$a^2+b^2=2ab$

$(a-b)^2=0$

$a=b $

Which cannot be right! Help!

bzc
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Meep
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  • What you claim cannot be right is right. Yes, they are equivalent precisely when $a=b$, which means that otherwise they are not equivalent, which means that one of them is right and the other wrong. – Ivan Neretin May 06 '17 at 20:46
  • @IvanNeretin but the derivation of this second result was absolutely general for an ellipse. How can we get the area of a circle as the outcome when calculating the area of an ellipse? – Meep May 06 '17 at 21:06
  • Have you seen the answer from Derek (https://math.stackexchange.com/q/1891110) in the Math Stack Exchange you give ? One thing is sure : $\pi (a^2+b^2)/2$ is a wrong formula ! – Jean Marie May 06 '17 at 21:11
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    If you are referring to method 2 of Daryl's answer, the expression $A=\int_0^{2\pi}\int_0^{R(t)}r,dr,dt$ there is wrong. This expression only works when $t$ is an angle, not some general parametrization of the boundary. – achille hui May 06 '17 at 21:22
  • @achillehui OK. comment deleted – tomi May 06 '17 at 21:33
  • @achillehui why is t not the angle? – Meep May 06 '17 at 22:07

2 Answers2

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The difficulty here is expressing the ellipse in polar coordinates. Given that

$$y=b\sqrt{1-\frac{x^2}{a^2}}\\ r^2=x^2+y^2=x^2+b^2\left(1-\frac{x^2}{a^2} \right)=r^2\cos^2\theta\left(1-\frac{b^2}{a^2} \right)+b^2$$ and finally $$ r^2(\theta)=\frac{b^2}{1-\left(1-\frac{b^2}{a^2} \right)\cos^2\theta} $$

You can readily validate this in a polar plot. The area is given by

$$A=\frac{1}{2}\int_0^{2\pi}r^2d\theta$$

I cannot see my way to clear to evaluating this integral and Wolfram Alpha gives the indefinite integral as

$$\frac{1}{2}\int r^2d\theta=\frac{ab}{2}\tan^{-1}\left(\frac{a\tan\theta}{b} \right)$$

which appears to give an area of zero, except when $a=b$ and $A=\pi a^2$. Wolfram Alpha was unable to give the definite integral. However, I can attest to the fact that a numerical integration does indeed give $A=\pi ab$.

Cye Waldman
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  • (+1) As an antiderivative, $\frac{ab}{2}\tan^{-1}\left(\frac{a\tan\theta}{b}\right);$ has two jumps at $\theta = \frac{\pi}{2}$ and $\frac{3\pi}{2}$. The integral equals to $\left[\int_0^{\frac{\pi}{2}} + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} + \int_{\frac{3\pi}{2}}^{2\pi}\right]\frac{r^2}{2}d\theta = \frac{ab}{2}\left[ \left(\frac{\pi}{2} - 0\right) + \left(\frac{\pi}{2} - -!\frac{\pi}{2}\right)
    • \left(0 - -!\frac{\pi}{2}\right)

    \right] = \frac{ab}{2}(2\pi) = \pi ab$

     – achille hui May 07 '17 at 04:47
  • @achillehui Very nice job. Thanks! – Cye Waldman May 07 '17 at 13:44
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$a>0<b.$

Method 1. (Cartesian). Consider $f:\mathbb R^2\to \mathbb R^2$ where $f(x,y)=(x, yb/a).$ The function $f$ is continuous and 1-to-1 and maps any rectangle with sides parallel to the co-ordinate axes and area $A$ onto a rectangle with sides parallel to the co-ordinate axes and area $Ab/a.$ So $f$ maps any region with area $B$ onto a region with area $Bb/a.$ And $f$ maps the circular disc $\{(x,y): x^2+y^2\leq a^2\}$ onto $\{(x,y): b^2x^2+a^2y^2\leq a^2b^2\}.$ So the area of the ellipse is $(\pi a^2)b/a=\pi a b.$

Method 2. (Quasi-polar). The elliptical region is $\{ar\cos t, br\sin t): r\in [0,1]\land t\in [0,2\pi)\}.$ Note: If $a\ne b$ and $(x,y)=(ar\cos t, br\sin t)$ with $x\ne 0\ne y,$ then $t$ is NOT the polar angle of $(x,y).$ That is, $\tan t\ne y/x.$

The area of the ellipse $E$ is $$\int\int_Edxdy=\int_{t=0}^{2\pi}\int_{r=0}^1 J(r,t)dr dt$$ where $J(r,t)$ is the Jacobian of the change of variables $(x,y)=(ar\cos t,br\sin t).$ We have $J(r,t)=abr.$ So the area is $\int_0^{2\pi}(\int_0^1 abrdr)dt=\pi ab.$

DanielWainfleet
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