Question

Find the arc length of the polar curve r=1-cos(theta) on the interval 0 13 years ago

Answer 1

i get down to \[\int\limits_{0}^{\pi} \sqrt{2-2\cos(\theta)} d \theta \]

Answer 2

but don't know what to do from there

Answer 3

Arc length = \(\displaystyle\int_0^\pi\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\;d\theta\).

\(r=1-\cos\theta\\
\dfrac{dr}{d\theta}=\sin\theta\)

\[\begin{align*}L&=\int_0^\pi\sqrt{(1-\cos\theta)^2+\sin^2\theta}\;d\theta\\
&=\int_0^\pi\sqrt{2-2\cos\theta}\;d\theta\\
&=\sqrt2\int_0^\pi\sqrt{1-\cos\theta}\;d\theta\\
&=\sqrt2\int_0^\pi\sqrt{1-\cos\theta}\cdot\frac{\sqrt{1+\cos\theta}}{\sqrt{1+\cos\theta}}\;d\theta\\
&=\sqrt2\int_0^\pi\frac{\sqrt{1-\cos^2\theta}}{\sqrt{1+\cos\theta}}\;d\theta\\
&=\sqrt2\int_0^\pi\frac{\sqrt{\sin^2\theta}}{\sqrt{1+\cos\theta}}\;d\theta&&\text{note }\sqrt{\sin^2\theta}=\sin\theta \text{ for }\theta\in[0,\pi]\\
&=\sqrt2\int_0^\pi\frac{\sin\theta}{\sqrt{1+\cos\theta}}\;d\theta\end{align*}\]
Also, note that the denominator is 0 when \(\theta=\pi\), so you'll have to treat this as an improper integral.

Answer 4

You'll rewrite the integral as
\[\lim_{b\to\pi}\int_0^b\frac{\sin\theta}{\sqrt{1+\cos\theta}}\;d\theta\]
A u-sub will help from here on out.