reversing the order of this double integral in polar coordinates

Question

Lets say I want to find the area of this shape:

One way which works in giving the correct area is the following:

$dA=rd\theta dr$

$\int_{0}^{\pi}\int_{0}^{2Rsin\theta }rdrd\theta$

I can visualise this as first creating a line of length $2Rsin\theta $ and then sweeping it over $\pi $ rads to give the full area.

However if i were to first integrate over $\theta $(reverse the order) what would my new limits be? I tried $\int_{0}^{2R}\int_{0}^{arcsin(r/2R) }rd\theta dr$. But it does not give me the same answer.

What are the right limits if integrating over $d\theta$ first?

Answer 1

If you draw the integration region in the $\theta - r$ plane you will see that inverting the order of integration leads to $$ \int_0^{2R} \int_{\arcsin\left(\frac{r}{2R}\right)}^{\pi-\arcsin\left(\frac{r}{2R}\right)} r \, d\theta dr $$

You need to consider different branches of $\arcsin$ when you cross $\frac{\pi}{2}$.

Answer 2

I think that would be

$$\int _0^{2 R}\int _{\pi -\arcsin\left(\frac{r}{2 R}\right)}^{\arcsin\left(\frac{r}{2 R}\right)}r\,d\theta\,dr=R^2\,\pi$$