I will be thankful if somebody solve this problem . I tried but i find it really complicated to integrate.

Find area of $ 3x^2 + 2y^2 \leqslant 9 $ , $ x\leqslant 1 $ using polar coordinates.

  • $\begingroup$ What did you try? Where did you get stuck? $\endgroup$
    – mickep
    Jun 25, 2017 at 17:17

2 Answers 2


The (single) integral solution to your problem is in the form


So we need to determine $r(\theta)$ as follows,

$$ \begin{align} r^2 &=x^2+y^2\\ &=x^2+\frac{9-3x^2}{2}\\ &=\frac{9-x^2}{2}\\ &=\frac{9-r^2\cos^2\theta}{2}\\ \end{align} $$

so that



$$A=\frac{1}{2}\int_0^{2\pi}\frac{9}{2+\cos^2\theta}~d\theta=\frac{3}{2}\sqrt{\frac{3}{2}}\tan^{-1}\left( \sqrt{\frac{2}{3}}\tan\theta\right)\biggr|_0^{2\pi}=3\sqrt{\frac{3}{2}}\pi\approx 11.5429$$


Let $x = \sqrt{3}r\cos \theta$ and $y = \frac{3}{\sqrt{2}}r \sin \theta$. Then we have that $dxdy = \frac{3\sqrt{3}}{\sqrt{2}}r drd\theta$.

The ellipse $3x^2 + 2y^2 = 9$ will give us that $r=1$ is the upper bound for $r$. For the lower bound of $r$ we have from the line that $r = \frac{1}{\sqrt{3}\cos \theta}$. Now we have to split the region in two parts. Obviously the intersections between the line and the ellipse are at $(1, \pm \sqrt3)$ and $0 \le \theta \le 2\pi$. Now on $[\frac{\pi}{3}, \frac{5\pi}{3}]$ $r$ varies between $0$ and $1$, while on $[0, \frac{\pi}{3}] \cup [\frac{5\pi}{3}, 2\pi]$ it varies between $0$ and $\frac{1}{\sqrt{3}\cos \theta}$. Hence:

$$A = \int_{- \frac{\pi}{3}}^{ \frac{\pi}{3}} \int^{\frac{1}{\sqrt{3}\cos \theta}}_{0} \frac{3\sqrt{3}}{\sqrt{2}}r drd\theta + \int_{ \frac{\pi}{3}}^{ \frac{5\pi}{3}} \int_{0}^{1} \frac{3\sqrt{3}}{\sqrt{2}}r drd\theta$$

  • $\begingroup$ Thank you , but I have to use single integrals . $\endgroup$
    – Lind
    Jun 25, 2017 at 17:21
  • $\begingroup$ @AnteMihaljević You can see that $x$ varies between $1$ and $\sqrt 3$, the upper bound is $\sqrt{9 - 3x^2}$ and the lower is $-\sqrt{9 - 3x^2}$. So we have that: $$A = \int_{1}^{\sqrt 3} 2\sqrt{9 - 3x^2} dx$$ $\endgroup$
    – Stefan4024
    Jun 25, 2017 at 17:25
  • $\begingroup$ @AnteMihaljević But I can't see how you can use polar coordinates. In fact polar coordinates is a change of variables in 2 dimensions. Maybe you were refering to trigonometric substitution, as here you can substitute $x = \sqrt{3}\cos \theta$ and simplify the integral $\endgroup$
    – Stefan4024
    Jun 25, 2017 at 17:26
  • $\begingroup$ $ x $ varies between $ -\sqrt 3 $ and $ 1 $ $\endgroup$
    – Lind
    Jun 25, 2017 at 17:32
  • $\begingroup$ @AnteMihaljević Oh, yeah, I got it wrong. I thought that you were referring to the right piece of the ellipse. $\endgroup$
    – Stefan4024
    Jun 25, 2017 at 17:42

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