I have the following double integral in polar coordinates:

$$\int\limits_{{\pi}/{6}}^{{\pi}/{2}} \int_\limits1^{\csc \theta} r^2\cos \theta\, \mathrm dr\, \mathrm d\theta.$$ The question is to find the integral converting it into catesian coodinates.

I have done the following:

$r^2\cos \theta \mathrm dr\mathrm d\theta = x\mathrm dx\mathrm dy$ but I can't set the limits for $x$ and $y$.

  • $\begingroup$ Did you try to draw a diagram using $r = \csc\theta$ and $r = 1$? That should tell you the region you are integrating over. $\endgroup$
    – Math Lover
    May 28, 2021 at 17:27
  • $\begingroup$ $r = \csc\theta \implies r\sin\theta = 1 \implies y = 1$. $\endgroup$
    – Math Lover
    May 28, 2021 at 17:28

1 Answer 1


Let $R$ be the region that you are interested in and, for each $\theta\in\left[\frac\pi6,\frac\pi2\right]$, let $r_\theta$ be the ray whose origin os $(0,0)$ and which passes through $\bigl(\cos(\theta),\sin(\theta)\bigr)$.

You have that, for each $\theta\in\left[\frac\pi6,\frac\pi2\right]$, $r\in\left[1,\csc(\theta)\right]$, and so, for each $\theta$,$$x=r\cos(\theta)\in\left[\cos(\theta),\cot(\theta)\right]\quad\text{and}\quad y=r\sin(\theta)\in[\sin(\theta),1].$$So, for each $\theta$, the point $R\cap r_\theta$ which is closest to the the origin is $\bigl(\cos(\theta),\sin(\theta)\bigr)$, and the one which furthest from the origin is $\bigl(\cot(\theta),1\bigr)$. So, the region that you are interested in is bounded above by the line $y=1$ and it is bounded below by the lines $x^2+y^2=1$ (that is, the unit circle) and $y=\frac x{\sqrt 3}$ (which is the ray $r_{\pi/6}$); see the picture below.

So, your integral is equal to$$\int_{1/2}^1\int_{\sqrt{1-y^2}}^{\sqrt3y}x\,\mathrm dx\,\mathrm dy.$$

enter image description here


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