I find myself frustrated with the solution of this problem since profit not find it, I'm stuck in the middle of the problem I can not solve the integral, I'm stuck in the solution of the integral is the same as follows:
$$r= \frac{6}{1+\cos\theta} \;\;\;\; 0 \leq \theta \leq \frac{\pi }{2} $$
P.d :
I would appreciate if anyone could recommend me some text or page in which you could learn to plot in polar and learn more about these problems.
Best Answer
Hint: The length of a polar curve $r(\theta)$ with starting angle $\alpha$ and ending angle $\beta$ is $$\int_\alpha^\beta \sqrt{r^2(\theta)+(r'(\theta))^2}$$
With your situation, it would be $$\int_0^{\pi/2} \sqrt{\left(\frac{6}{1+\cos\theta}\right)^2+\left(\left(\frac{6}{1+\cos\theta}\right)'\right)^2}\, d\theta$$ $$\frac{6}{1+\cos\theta}=3\sec^2(\theta/2)\text{ so}$$ $$\int_0^{\pi/2} \sqrt{9\sec^4(\theta/2)+\frac{576\sin^8\theta}{\sin^6(\theta/2)}}$$ Now try a weierstrass substitution ($t=\tan (\theta/2)$)