Letting $r = 8\sin^2 (\theta/2)$ for $0 \leq\theta\leq\pi$. How do you find the length $L$ of the graph of the polar equation?

Question

Hello am having difficulty solving can someone please explain how this would be solved. Thank you

Answer 1

$r = 8\sin^2\frac {\theta}{2} = 4 - 4\cos\theta$

And that is a cartioid.

$\int_0^{2\pi} (r^2 + (\frac {dr}{d\theta})^2)^\frac 12 d\theta$

$\int_0^{2\pi} (16 - 32 \cos\theta + 16 \cos^2\theta +16\sin^2\theta)^\frac 12 d\theta\\ \int_0^{2\pi} (32 - 32 \cos\theta)^\frac 12 d\theta\\ \int_0^{2\pi} 8(\frac{1 - \cos\theta}{2})^\frac 12 d\theta\\ \int_0^{2\pi} 8\sin\frac\theta2 d\theta\\ -16\cos\frac\theta2 |_0^{2\pi}\\ 32$

Answer 2

Hint. In general in polar coordinates you have $$ L=\int_{\large \theta_1}^{\large\theta_2}\sqrt{r^2+\left(\frac{dr}{d\theta} \right)^2}\:d\theta, $$ here $$ r=8\sin^2(\theta/2),\quad \frac{dr}{d\theta}=8\cos(\theta/2)\sin(\theta/2), \quad \theta_1=0,\quad \theta_2=\pi, $$ giving $$ r^2+\left(\frac{dr}{d\theta} \right)^2=64 \sin^2(\theta/2)\left(\cos^2(\theta/2)+\sin^2(\theta/2) \right)=64 \sin^2(\theta/2). $$ Can you take it from here?