# Converting the double integral in polar coordinates to Cartesian coordinates

I want to convert

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int_1^{\csc{\theta}}r^2\cos{\theta}drd\theta$$

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.

• $r=\csc{\theta}\Leftrightarrow y=1$, but this integral is much easier to compute in polar coordinates. Commented Oct 2, 2022 at 9:28
• 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 Commented Oct 2, 2022 at 9:34
• 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 Commented Oct 2, 2022 at 9:38
• 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 $x^2+y^2=1$
– user1055322
Commented Oct 2, 2022 at 9:41

$$r=1\implies x^2+y^2=1.$$ $$r=\csc\theta\implies r\sin\theta=1\implies y=1.$$ $$\theta=\frac{\pi}{6}\implies \tan\theta=\frac{1}{\sqrt{3}}\implies r\cos\theta=\sqrt{3}r\sin\theta\implies x=\sqrt{3}y.$$ $$\theta=\frac{\pi}{2}\implies \tan\theta=\infty\implies x=0.$$
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 $$dxdy$$.
• Got it! Thanks! $$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int_1^{\csc{\theta}}r^2\cos{\theta}drd\theta=\int_{\frac{1}{2}}^1\int_{\sqrt{1-y^2}}^{\sqrt{3}y}xdxdy$$