3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.