Converting the double integral in polar coordinates to Cartesian coordinates

Question

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.

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 $dxdy$.