# Integral conversion to polar coordinates - bounds $\int\limits_0^1 \int\limits_0^1\sqrt{x^2+y^2}\ dxdy$ [duplicate]

I have an integral

$$\int_0^1 \int_0^1\sqrt{x^2+y^2}\ dxdy$$

and its result is $$\approx0.765...$$

I convert it to polar coordinates and get

$$\int_a^b \int_c^dr\ drd\phi$$

But how can I compute $$a,b,c,d$$?

• There is no trivial way to represent a square in polar coordinates. May 21, 2016 at 10:35
• $dx\ dy\ne dr\ d\phi$. May 21, 2016 at 10:43
• A square in polar coordinates should be divided into two parts; the first part (for example) $a=0, b_1=\frac \pi 4, c=0, d= \frac 1{\cos \theta}$, the second can be left you as an exsice.
– Paul
May 21, 2016 at 10:49
• @Paul Thank you, I understand, where you got these bounds... but if I use $$b_2 = 5\pi/4$$ I got wrong final result May 21, 2016 at 11:21
• Note that $a_2=\frac\pi 4$ and $b_2$ should be $\frac \pi 2$ @MartinPerry
– Paul
May 21, 2016 at 11:41

Note $dxdy=rdrd\theta$ here. so we have, $$\int_0^1 \int_0^1\sqrt{x^2+y^2}\ dxdy = \int_0^\frac\pi4\int_0^\frac1{\cos \theta}r^2drd\theta+ \int_\frac\pi4^\frac\pi2\int_0^\frac1{\sin\theta}r^2drd\theta$$
• Thank you.. I have missed power at $r$ May 21, 2016 at 12:19
• That leads to $$\frac{1}{6}\left(\sqrt{2}+\log\cot\frac{\pi}{8}\right).$$ May 21, 2016 at 12:32
I do not believe that polar coordinates are the best approach, since they are not suitable for describing a square. Anyway, if you fix an angle $0 \leq \phi < 2\pi$, you must compute the length of the segment that joins the origin with the boundary of the square and with angular coefficient $\tan \phi$. This length depends on $\phi$, but I am unsure if the iterated integral becomes easy.
Your region in $x$-$y$ plane is a square in 1st quadrant. So, your $\theta$ limits (i.e your c and d) are $(0,\pi/2)$. However, you need to split up the integral at $\frac{\pi}{4}$. You can notice this by drawing arbitrary ray from starting from origin and see that it leaves out at $x=1$ (in region $0$ to $\frac{\pi}{4})$ and $y=1$ (in region $\frac{\pi}{4}$ to $\frac{\pi}{2})$). So, your limits for $r$ will be from $0$ to $\sec\theta$ and $0$ to $\csc\theta$ in first and second integrals respectively which can be found out by putting $x=r\cos\theta$ in $x=1$