I want to write the following double integral

$$\int_0^2 \int_{0}^{\sqrt{1-(x-1)^2}} \frac{x+y}{x^2+y^2} \, \mathrm d x \mathrm d y$$

in polar coordinates. The region is a circle centered at $(1,0)$ and with radius $1$.

I am having problems with finding the bounds for $r$. When we are on the circle,

$$(r \cos\theta -1 )^2 + r^2 \sin^2 \theta = 1$$

which implies that $r^2 -2 r \cos \theta +1 =0$. At his point I am stuck. Can you please explain how to find the polar equivalent of this integral? Thank you.

  • $\begingroup$ Your bounds for $r$ are allowed to be a function of $\theta$, so can you just use the quadratic formula to solve for $r$? What does the give you? $\endgroup$ Commented Apr 7, 2019 at 17:25
  • $\begingroup$ What about centering the circle? $x'=x-1$ $\endgroup$
    – Wilem2
    Commented Apr 7, 2019 at 17:50

1 Answer 1



$\theta$ goes from $-\frac{\pi}2$ to $\frac{\pi}2$.

We have $$r^2-2r\cos \theta = 0$$

(You forgot to cancel out the $1$ on both sides).

That is the upper limit of $r$ is $r= 2\cos \theta$.

That is

$$\int_{0}^\frac{\pi}{2}\int_0^{2\cos \theta} \frac{r\cos \theta + r\sin \theta}{r^2}\cdot r\, \, drd\theta$$

  • $\begingroup$ Thank you. By the way, the lower limit of $\theta$ must be $0$, right? $\endgroup$
    – Ninja
    Commented Apr 7, 2019 at 18:42
  • 1
    $\begingroup$ ah, yes, you are right, we are integrating only the upper disc. $\endgroup$ Commented Apr 7, 2019 at 18:47

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