I want to convert


into Cartesian coordinates.

I understand that $r\cos{\theta}=x$ and $rdrd\theta=dxdy$.

But how can I convert the upper limits and the lower limits?

I tried to use

$$\pi/6 \le f(r,\theta)$$

But it seems I can't find the upper limit since it is the function.

I know how to convert from Cartesian to polar.

But when it comes to polar to Cartesian, I am not sure how to draw that $r=\csc{\theta}$ in polar plane.

  • $\begingroup$ $r=\csc{\theta}\Leftrightarrow y=1$, but this integral is much easier to compute in polar coordinates. $\endgroup$ Oct 2, 2022 at 9:28
  • $\begingroup$ There is really no need to convert to Cartesian coordinates for this integral. If you want an answer on how to compute it in polar coordinates, just let me know $\endgroup$
    – Lorago
    Oct 2, 2022 at 9:34
  • $\begingroup$ It's probably for the exercise to convert. It is a useful skill to have and be able to do recognize an integral in both. Plus, I would consider the integral much easier in Cartesian if the $x$ direction were integrated first @AnneBauval $\endgroup$ Oct 2, 2022 at 9:38
  • $\begingroup$ Yes, it is an exercise in the Calculus book. So $r=1$ is a circle centered at $(0,0)$ with radius $1$? It seems that I can convert it into $x^2+y^2=1$ $\endgroup$ Oct 2, 2022 at 9:41

1 Answer 1


Lets do first the easy part first. Convert the bounds to cartesian coordinates:

$$r=1\implies x^2+y^2=1.$$ $$r=\csc\theta\implies r\sin\theta=1\implies y=1.$$ $$\theta=\frac{\pi}{6}\implies \tan\theta=\frac{1}{\sqrt{3}}\implies r\cos\theta=\sqrt{3}r\sin\theta\implies x=\sqrt{3}y.$$ $$\theta=\frac{\pi}{2}\implies \tan\theta=\infty\implies x=0.$$

Then we need to shade the region $D$ of the double integral on the cartesian plane, by drawing the circle $x^2+y^2=1$ and the lines $y=1$, $x=\sqrt{3}y$. Note that $D$ is $\frac{\pi}{6}\leq\theta\leq\frac{\pi}{2}$ and $1\leq r\leq \csc\theta.$ Unfortunately, this is the worst part of the question. But fortunately since circles and lines are the easiest graphs we can sketch them. We observe that $x^2+y^2=1$ cuts the line $x=\sqrt{3}y$ at a point $Q(\frac{\sqrt{3}}{2},\frac{1}{2})$.

Once you do that, we see that $D$ is $$\frac{1}{2}\leq y\leq 1$$ $$\sqrt{1-y^2}\leq x \leq \sqrt{3}y$$ And the order of integration is $dxdy$.

  • 1
    $\begingroup$ Got it! Thanks! $$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int_1^{\csc{\theta}}r^2\cos{\theta}drd\theta=\int_{\frac{1}{2}}^1\int_{\sqrt{1-y^2}}^{\sqrt{3}y}xdxdy$$ $\endgroup$ Oct 2, 2022 at 14:15

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