# Using polar coordinates to find the area of an ellipse

Considering an ellipse with the $$x$$ radius equal to $$a$$ and the $$y$$ radius equal to b$$:$$

I figured that some kind of parameterization might be:

$$x=a\cos\theta$$

$$y=b\sin\theta$$

and then polar $$r^2$$ is just $$x^2 + y^2$$

But then I tried to come up with some unit of infinitesimal area using triangles $$\left(\dfrac{d\theta r^2}{2}\right)$$ which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually

$$x=ar\cos\theta$$

$$y=br\sin\theta$$

But this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $$r$$ from $$0$$ to $$1$$ which I thought was eliminated because the equation for $$r$$ should be in terms of $$\theta$$ and the constants $$a$$ and $$b.$$

I would like some explanation of what I am doing wrong that would make some "physical" sense (or why physical intuition might fail for this problem)

## 8 Answers

Notice, since the ellipse: $x=a\cos\theta$ & $y=b\sin\theta$ is equally divided into four symmetrical regions hence, the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$ Now changing in Polar-coordinates by setting $y=b\sin\theta$ & $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$ $$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$ $$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$ $$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$

$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$

You were on the right track, but didn't pick the right infinitesimal triangle. It looks like you used a right triangle built on the inside of the ellipse, but as you discovered, that doesn't approximate the area swept out by the radius vector well enough. Take instead a point along the curve and another a small distance away along it. The area of the resulting triangle is, in Cartesian coordinates, $\frac12(x\,dy-y\,dx)$. Plugging in your parametrization of the ellipse gives $dA=\frac12ab\,d\theta$, which will clearly give the correct result when integrated.

More generally, when approximating integrals with Riemann sums, you need to take care that they converge to the right thing. This is the same type of error that occurs in the fallacious “stairstep” constructions which “prove” that $\pi=4$ or that $\sqrt2=2$. In this case, I expect that if you also computed the sum of the areas of the right triangles built on the outside of the ellipse that the difference between the inner and outer sums would not vanish in the limit.

• I would not say that the error is of the same type as the non-vanishing sums. The area estimate is off by a non unit factor because $d\theta$ is not the angle !
– user65203
Jun 14, 2021 at 12:43

Your parametrization only covers the edge of the ellipse. This is not enough if you try to find the area, which is the full interior. Hence the second form with $r$ (which is integrated from 0 to 1) is correct to find the area. Yours would be ok, if you try to only find the circumference.

• These explanations are plain wrong. Better delete.
– user65203
Jun 14, 2021 at 12:44
• ?? It is quite correct. $\text{Area} = \iint_{\text{Ellipse}} \mathrm d x \mathrm d y = \int_0^1 \int_0^{2 \pi} \mathrm d r \mathrm d \phi$ with the second choice of $x$ and $y$ in the original post (and only these), i.e. $x = a r \cos{\phi}, y = b r \sin{\phi}$. Point is that without $r$ (what the OP tried first) one does not get an area. This is qute correct and the systematic approach to such problems. Jun 14, 2021 at 14:52

Hint:

Use one of the polar equations for the ellipse: if the pole is at the centre of the ellipse, it is $$r^2=\frac{b^2}{1-e^2\cos^2\theta}, \quad\text{where }e =\sqrt{a^2-b^2}\enspace\text{is the eccentricity of the ellipse}$$ and integrate $$\int_0^\pi r^2\,\mathrm d\mkern1mu\theta.$$

• What you denote $e$ is not the eccentricity. And independently of that, your integral will not give the correct result.
– user65203
Jun 14, 2021 at 12:50
• @YvesDaoust: Really? As it's a bit old, I don't remember where I checked before posting. I'll have to verify again, but it's abit hard these days… Cheers! Jun 14, 2021 at 13:45

My answer is linked here.

My name is Marco [email protected].

• You don't need to provide your name here. Any comment under your answer will show up as an update. If you do want to leave an email for people to contact you it's better to put it on your profile page. Oct 7, 2017 at 19:01
• The OP is not working with the focal equation !
– user65203
Jun 14, 2021 at 12:07

I guess, the problem is in wrong approximation (look at light blue areas)

If use the formula for area of triangle $$\frac{1}{2}\left\| {{\bf{r}} \times {\bf{dr}}} \right\| = \frac{1}{2}rdr\sin \alpha % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaadaqbdaqaaiaahkhacqGHxdaTcaWHKbGaaCOC % aaGaayzcSlaawQa7aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaa % GaamOCaiaadsgacaWGYbGaci4CaiaacMgacaGGUbGaeqySdegaaa!4979!$$ you get the right result. Using of Green formula for area $$A = \frac{1}{2}\oint\limits_C {x \cdot dy - ydx} = \frac{1}{2}\oint\limits_C {\left\| {{\bf{r}} \times d{\bf{r}}} \right\|} = \frac{1}{2}\oint\limits_C {r\sin \phi dr} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % yqaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaa8qvaeaacaWG % 4bGaeyyXICTaamizaiaadMhacqGHsislcaWG5bGaamizaiaadIhaaS % qaaiaadoeaaeqaniablgH7rlabgUIiYdGccqGH9aqpdaWcaaqaaiaa % igdaaeaacaaIYaaaamaapufabaWaauWaaeaacaWHYbGaey41aqRaam % izaiaahkhaaiaawMa7caGLkWoaaSqaaiaadoeaaeqaniablgH7rlab % gUIiYdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaapufaba % GaamOCaiGacohacaGGPbGaaiOBaiabew9aMjaadsgacaWGYbaaleaa % caWGdbaabeqdcqWIr4E0cqGHRiI8aaaa!698A!$$ gives the same (right) result.

• The problem is not in the light blue areas. Also note that you notation that mixes $\alpha,\theta$ and $\phi$ is inconsistent.
– user65203
Jun 14, 2021 at 12:47

I am quite surprised that no one pointed out the following fact: the parametrisation in question does not contain the usual plane polar angle; what is present there is the eccentric anomaly. That is why you can't use the normal techniques for finding areas in plane polar coordinates, and you have to resort to (different forms of) Green's theorem in a plane.

• Your remark about the angle is correct. But there is no need for Green. (See my answer.)
– user65203
Jun 14, 2021 at 12:38

If we follow your reasoning, we get $$\frac12\int_0^{2\pi}(a^2\cos^2\theta+b^2\sin^2\theta)d\theta=\frac12\int_0^{2\pi}\frac{a^2(1+\cos2\theta)+b^2(1-\cos2\theta)}2d\theta \\=\frac\pi2(a^2+b^2),$$

which is wrong (except for a circle).

The mistake is to believe that $$d\theta$$ represents the angle of the infinitesimal triangles. This is wrong because $$\theta$$ is not the angle swept by a point on the ellipse. Instead we have

$$\tan\phi=\dfrac {b\sin\theta}{a\cos\theta}=\frac ba \tan\theta,$$ and

$$d\phi=\dfrac ba\sec^2\theta\,d\theta.$$

Now, $$\frac12\int_0^{2\pi}r^2\,d\phi=\frac b{2a}\int_0^{2\pi}(a^2\cos^2\theta+b^2\sin^2\theta)\sec^2\theta\,d\theta=\frac {ab}2\int_0^{2\pi}d\theta+\frac {b^3}{2a}\int_0^{2\pi}\tan^2\theta\,d\theta\\=\pi ab.$$

• This answer is excellent. Aug 3, 2023 at 3:16
• This answer is plainly wrong, despite a correct final result. Feb 20 at 7:19