I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.

ellipse : $ x^2/a^2 + y^2/b^2 =1 $

Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $

At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem] $$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$

$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$

$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$ $$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$ $$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$ I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $

  • $\begingroup$ Could you include your steps showing how you got your answer? $\endgroup$
    – coffeemath
    Sep 13, 2018 at 12:52
  • 1
    $\begingroup$ Fix these things: The points have coordinates $x=ra\cos \theta, y=rb \sin \theta.$ The Jacobian is not $r.$ $\endgroup$
    – user376343
    Sep 13, 2018 at 13:06
  • $\begingroup$ Aren't polar coordinates in ellipse $ (acos(θ),bsin(θ)) $ ? $\endgroup$
    – Anuj
    Sep 13, 2018 at 13:11
  • 2
    $\begingroup$ Your $\theta$ is just some parameter; it is not the polar angle of the moving point. $\endgroup$ Sep 13, 2018 at 13:31
  • 1
    $\begingroup$ $(a\cos \theta,b\sin \theta)$ is a point on the boundary of the ellipse, why are you taking points on the ellipse? to find the area of the ellipse you have to take an elementary area inside the ellipse, which may be taken as $rdrd\theta$ $\endgroup$
    – Jimmy
    Sep 13, 2018 at 14:19

4 Answers 4


Set $x=ar\cos \theta, y=br\sin \theta.$ The Jacobian is $abr$ and we compute the area $$\mathcal {A}= \int_0^{\pi/2} \int_0^1 abr \;d r \;d\theta,$$ which is $\frac{ab\pi}{4}.$


The given ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.To transform it into polar coordinates,substitute $(x,y)=(r\cos\theta,r\sin\theta)$ to get $r=\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}$.

Take an elementary area $rdrd\theta$ inside the ellipse.

Then the area of the ellipse in the first quadrant is the sum of all such elementary areas

$\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{\dfrac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}} r \,dr \, d\theta$

$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2}{b^2\cos^2\theta + a^2\sin^2\theta} d\theta$

$=\dfrac12\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{a^2b^2\sec^2\theta}{b^2+ a^2\tan^2\theta} d\theta$

$=\dfrac{b^2}{2}\displaystyle\int_{\theta=0}^{\frac{\pi}{2}}\dfrac{\sec^2\theta}{\dfrac{b^2}{a^2}+ \tan^2\theta} d\theta$


$=\boxed{\boxed{\dfrac{\pi ab}{4}}}$

  • $\begingroup$ how did you come to that conclusion, as you substitute the limits into the tan(theta) part, but surely the answer would come out as $(\pi*a^2*b)/2$, or where am I going wrong with this? $\endgroup$ Apr 9, 2019 at 12:51
  • $\begingroup$ $\left.\dfrac{b^2}{2}\dfrac{a}{b}\tan^{-1}\left(\dfrac{a\tan\theta}{b}\right)\right|_{0}^{\frac{\pi}{2}}=\dfrac{ab}{2}\left[\tan^{-1}\left(\frac{a}{b}\cdot\infty\right)-\tan^{-1}\left(\frac{a}{b}\cdot 0\right)\right]=\dfrac{ab}{2}\left(\tan^{-1}\infty-\tan^{-1}0\right)=\dfrac{ab}{2}\left(\dfrac{\pi}{2}-0\right)=\dfrac{\pi ab}{4}$ $\endgroup$
    – Jimmy
    Apr 13, 2019 at 11:31

Your mistake is to believe that $\theta$ is the polar argument. It is not, because

$$\tan\phi=\frac yx=\frac ba\tan \theta\ne\tan\theta.$$

You can fix by taking the differential

$$(\tan^2\phi+1)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$

and substituting

$$\left(\frac{b^2}{a^2}\tan^2\theta+1\right)\,d\phi=\frac ba(\tan^2\theta+1)\,d\theta$$

so that the integral becomes

$$\frac12\int_0^{\pi/2}\frac{\dfrac ba(\tan^2\theta+1)}{\dfrac{b^2}{a^2}\tan^2\theta+1}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta \\=\frac{ab}2\int_0^{\pi/2}\frac{\sin^2\theta+\cos^2\theta}{b^2\sin^2\theta+a^2\cos^2\theta}(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta.$$

  • $\begingroup$ What does $\phi$ look like in relation to $\theta$ as seen in his diagram? $\endgroup$
    – DWade64
    Sep 13, 2018 at 14:40
  • $\begingroup$ @DWade64: roughly linear, plot it $\endgroup$
    – user65203
    Sep 13, 2018 at 14:47
  • $\begingroup$ Is the polar angle the angle counterclockwise from the $x$-axis? That's what $\theta$ is in his diagram. I'm not disputing your answer. I'm just trying to understand it. It probably comes down to understanding elliptical vs polar coordinates $\endgroup$
    – DWade64
    Sep 13, 2018 at 14:49
  • $\begingroup$ @DWade64: it's up to you, for this case it doesn't matter. $\endgroup$
    – user65203
    Sep 13, 2018 at 14:50

You can also use green's theorem. Set x=acos($\theta$), y=bsin($\theta$). Now we can set up the intergal. $A=\frac{1}{2}\int_{c} xdy-ydx. \\ dy=bcos(\theta), dx=-asin(\theta) \\ A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}abcos^2(\theta)+absin^2(\theta)\text{ }d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}ab \text{ }d\theta=\frac{\pi ab}{4}. \\ $


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