I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating

$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$

So we should be able to find the area of the ellipse by

$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$

$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$

$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$

$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$


First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.

What am I doing wrong?

  • $\begingroup$ One can obtain the answer without doing much thinking: to obtain the ellipse, we stretch the circle $a$ units through one axis and $y$ units through the other, which amounts to $ab$ times the area of the unit circle, which is $\pi$, so $ab\pi$. Alternatively, the transformation $(x,y)\mapsto (ax,by)$ has Jacobian $ab$. $\endgroup$ – Pedro Tamaroff Sep 14 '13 at 3:20
  • $\begingroup$ @Eric Auld: see math.stackexchange.com/questions/1933765/… $\endgroup$ – 926reals Nov 9 '16 at 3:02

There are already a lot of good answers here, so I'm adding this one primarily to dazzle people w/ my Mathematica diagram-creating skills.

As noted previously,

$x(t)=a \cos (t)$

$y(t)=b \sin (t)$

does parametrize an ellipse, but t is not the central angle. What is the relation between t and the central angle?:

cool, huh?

Since y is bSin[t] and x is aCos[t], we have:

$\tan (\theta )=\frac{b \sin (t)}{a \cos (t)}$


$\tan (\theta )=\frac{b \tan (t)}{a}$

Solving for t, we have:

$t(\theta )=\tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

We now reparametrize using theta:

$x(\theta )=a \cos (t(\theta ))$

$y(\theta )=b \sin (t(\theta ))$

which ultimately simplifies to:

$x(\theta)=\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$

$y(\theta)=\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$

Note that, under the new parametrization, $y(\theta)/x(\theta) = tan(\theta)$ as desired.

To compute area, we need $r^2$ which is $x^2+y^2$, or:

$r(\theta )^2 = (\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2+ (\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2$

(note that we could take the square root to get r, but we don't really need it)

The above ultimately simplifies to:

$r(\theta)^2 = \frac{1}{\frac{\cos ^2(\theta )}{a^2}+\frac{\sin ^2(\theta )}{b^2}}$

Now, we can integrate $r^2/2$ to find the area:

$A(\theta) = (\int_0^\theta \frac{1}{\frac{\cos ^2(x )}{a^2}+\frac{\sin ^2(x )}{b^2}} \, dx)/2$

which yields:

$A(\theta) = \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

good for $0\leq \theta <\frac{\pi }{2}$

Interestingly, it doesn't work for $\theta =\frac{\pi }{2}$ so we can't test the obvious case without using a limit:

$\lim_{\theta \to \frac{\pi }{2}} \, \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

which gives us $a*b*Pi/4$ as expected.



Putting $x=r\cos\theta,y=r\sin\theta$

$$\frac {x^2}{a^2}+\frac{y^2}{b^2}=1,$$


  • $\begingroup$ Great. So I see that $\vec{r}(t) = (a\cos{t}, b\sin{t})$ does not parameterize the ellipse by the angle. Can you spot my error in evaluating the integral as I set it up? $\endgroup$ – Eric Auld Sep 14 '13 at 3:18
  • 1
    $\begingroup$ @EricAuld, clearly you can see the value of $r$ here in terms of $\theta$. We can't set polar equation in term of $\theta$ in that way. You can see some other methods here : en.wikipedia.org/wiki/Ellipse#Area $\endgroup$ – lab bhattacharjee Sep 14 '13 at 3:23

Your question has been answered, so now we look at how to find the area, using your parametrization, which is a perfectly good one.

The area is the integral of $|y\,dx|$ (or alternately of $|x\,dy|$. over the appropriate interval.

We have $y=b\sin\theta$ and $dx=-a\sin\theta\,d\theta$. So the area is $$\int_0^{2\pi} |-ab\sin^2\theta|\,d\theta.$$ Using $\sin^2\theta=\frac{1-\cos 2\theta}{2}$, we find that the area is $$\int_0^{2\pi} ab\frac{1-\cos 2\theta}{2}\,d\theta.$$ This is indeed $\pi ab$.

  • $\begingroup$ Thanks, Andre. So the integral I set up was the wrong one. But can you see where is the error in my evaluation of the integral? Wolfram Alpha yields a different answer... $\endgroup$ – Eric Auld Sep 14 '13 at 12:43
  • 1
    $\begingroup$ Your integral is right. But we can't really blame poor Alpha for disagreeing, implicitly you told it $b=\sqrt{2}$. $\endgroup$ – André Nicolas Sep 14 '13 at 17:14
  • $\begingroup$ Thank you! That was really bothering me... $\endgroup$ – Eric Auld Sep 14 '13 at 18:10
  • $\begingroup$ You are welcome. These kinds of minor typos can be the hardest to find, I got lucky. $\endgroup$ – André Nicolas Sep 14 '13 at 18:16

Here you go - this person even made your mistake, then someone else corrected it.


  • $\begingroup$ Awesome! So I was wrong in assuming that that integral represented the area. Can you see how I have erred in evaluating the integral? $\endgroup$ – Eric Auld Sep 14 '13 at 3:14

This is another way to do it when one know the area of a circle: Consider the area of a circle with radius 1 in coordinates $(\xi, \eta)$ this is:

$$ \int d\xi d \eta = \pi $$

now if you define new coordinates in your ellipse equation $\xi = \frac{x}{a}, \quad \eta= \frac{y}{b}$ you obtain a circle of radius one: $\xi^2 + \eta^2 =1$

The area of the ellipse you want is $ \int dx dy = ab \int d\xi d\eta = \pi ab$.


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