Calculate the double integral $\iint_D {(1+x^2 + y^2)ln(1+x^2+y^2)dxdy} $ where $D = \{(x,y) \in \mathbb R^2 | \frac{x}{\sqrt3} \leq y \leq x , x^2 + y^2 \leq 4\}$.

I heard there is a way called Polar Coordinates but the more I looked and read about it the more I did not understand.

But I started drawing $D$ and wolfram gave this:

enter image description here

But doesn't $D$ also include the opposite direciton of this? And if so and if not, how would I calculate it with "Polar Coordinates?" I know Polar Coordinates is a wide subject and I am sorry for asking it this way, but I did not understand scholar papers.

  • 2
    $\begingroup$ The inequality $x/\sqrt{3} \leqslant x$ implies $x \geqslant 0$. Note that $D$ is a circular sector, that means it's easy to parameterise in polar coordinates. The integrand is rotationally symmetric, so that is also easily expressible in polar coordinates. $\endgroup$ Jul 4, 2013 at 13:35
  • $\begingroup$ Just put $x^2+y^2=r^2$ and $dx dy=rd\theta d\phi$ $\endgroup$
    – user5402
    Jul 4, 2013 at 13:37

2 Answers 2


Taking @Daniel Fisher's comment to its logical conclusion, the lines imply a circular sector in which $\theta \in [\pi/6,\pi/4]$. Thus the area integral is

$$\int_{\pi/6}^{\pi/4} d\theta \, \int_0^2 dr \, r \, (1+r^2) \, \ln{(1+r^2)}$$

  • $\begingroup$ are you sure $\theta \in [\pi / 4, \pi / 3]$? I mean, $arctan(1 / \sqrt3) = \pi / 6$ $\endgroup$
    – TheNotMe
    Jul 4, 2013 at 13:43
  • $\begingroup$ @TheNotMe: thanks for catching that; got her backwards. $\endgroup$
    – Ron Gordon
    Jul 4, 2013 at 13:44

Looking at your sketch and remembering that $x^2 + y^2 = r^2$, it can be seen, in polar coordinates that your angle starts at $\frac{\pi}{4}$ and moves up to $\frac{\pi}{6}$, whereas your radius, goes from 0 to 2, thus gving the following limits:

$\frac{\pi}{6}\leq\theta\leq\frac{\pi}{4}$, $0\leq r\leq 2$

Your integral can now be rewritten in polar coordinates as \begin{align}I &=\int\limits_{\theta_0}^{\theta_1} \int \limits_{r_0}^{r_2}f(x).r drd \theta \\ &=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{4}} \int \limits_{0}^{2}(1+r^2)ln(1+r^2)r drd \theta\end{align}which can be solved more easily than Cartesian coordinates.


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