I need to evaluate an integral over the $D=\{x^2+y^2 >1; \frac{x^2}{a^2}+\frac{y^2}{b^2}<1\}$, but I can't find the limits of integration simply by changing to polar coordinates.


  • $\begingroup$ It seems almost obvious that either $\;a\ge1\;$ or $\;b\ge1\;$. What else are you given? $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 12:05

1 Answer 1


In polar coordinates, the region can be represented by a whole $2\pi$ turn of $\theta$.

For the radius, it is a function of instant $\theta$ values.

$\int_0^{2\pi} \int_1^{U(\theta)} F(r,\theta)*r*dr*d\theta$

where $r = U(\theta)$ can be found as follows:

Suppose that a point on the ellipse has angle $\theta$, then we have $\frac{r^2cos^2(\theta)}{a^2} + \frac{r^2sin^2(\theta)}{b^2} = 1$

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

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

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


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