# Polar to cartesian limit change.

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 $$y$$.

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

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$$\int_{1/2}^1\int_{\sqrt{1-y^2}}^{\sqrt3y}x\,\mathrm dx\,\mathrm dy.$$