# 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 ($d\theta r^2/2$) 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)

## 6 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 $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$ $$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$ $$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$ $$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$

$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$ My name is Marco mamello10@gmail.com.

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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.

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 $$\int_0^\pi r^2\,\mathrm d\mkern1mu\theta.$$

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 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.