The problem I want to solve is to calculate the filled area in the following diagram - so basically the area between the two circular arcs but with the red line cutting off one side. I think I have a solution but I'd like to be sure it's correct before going any further! enter image description here

Given this set up we should have the values of r1, r2, $\theta_1$, $\theta_2$ and should therefore be able to calculate $\theta_a$ quite easily. The area should then be given by: $$\int_{\theta = 0}^{\theta= \theta_2} d\theta \int_{r=r_0(\theta)}^{r=r_2} r \ dr$$ where $r_0(\theta) = r_1$ for $\theta \le \theta_1$ and to be determined for $\theta > \theta_1$. So this first part just gives the contribution $\frac{\theta_1}{2}(r_2^2-r_1^2)$. Now to get $r_0(\theta)$ for $\theta > \theta_1$ I refer to the following diagram: enter image description here

so $r_0(\theta_1+d\theta) \equiv r_1 + dr$ and $\theta_p = \pi - \theta_a$, $\theta_x = \pi - \theta_p-d\theta = \theta_a - d\theta$. Now applying the sine rule: $$ \frac{r_1}{\sin(\theta_a - d\theta)} = \frac{r_1+dr}{\sin(\pi-\theta_a)} = \frac{r_1+dr}{\sin(\theta_a)}$$ So writing $d\theta = \theta - \theta_1$ we have: $$r_0(\theta) = \frac{r_1 \sin(\theta_a)}{\sin(\theta_a-(\theta-\theta_1))}$$

So then the second contribution should be: $$\int_{\theta = \theta_1}^{\theta= \theta_2} d\theta \ \ _{r=r_0(\theta)}^{r=r_2}\left[ r^2/2 \right] = \frac{1}{2} \int_{\theta_1}^{\theta_2} (r_2^2 - \frac{r_1^2 \sin^2(\theta_a)}{\sin^2(\theta_a - \theta + \theta_1)}) d\theta$$

Which Mathematica tells me is: $$ \frac{1}{2} ( r_2^2 (\theta_2-\theta_1) - \frac{r_1^2\sin(\theta_2-\theta_1) \sin(\theta_a)}{\sin(\theta_a + \theta_1 - \theta_2)} ) $$

This seems reasonable enough, as $\theta_1 \to \theta_2$ this correction term goes to zero which it should, but still I don't trust myself enough to not have made a mistake so if anyone could check then that would be great.


I'm having a hard time following your post. Why don't you start with the sector of the annulus $\frac12(r_2^2-r_1^2)\theta_2$, then add the sector of the inner circle $\frac12r_1^2(\theta_2-\theta_1)$ and subtract the area of that skinny triangle $\frac12r_1r_2\sin(\theta_2-\theta_1)$ to get the blue area?

  • $\begingroup$ Yes I feel stupid now, this is a much easier way! They should in principle give the same results though! $\endgroup$ – Henry Jun 13 '16 at 15:44

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