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Find the length of $r=1+\sin{\theta}$.

I got to $\sqrt{2} \int\limits_{0}^{2\pi}\sqrt{1+\sin{\theta}} \,\mathrm{d}\theta$.

And the first way I used to solve the integral was substitution of $1+\sin{\theta}=u$.
Thus $ \cos\theta \,\mathrm{d}\theta = \mathrm{d}u$. and I used $\cos\theta=+\sqrt{1-\sin^2\theta}$ to find the interval which is $[-\pi/2,\pi/2]$. Considering the original interval of $[0,2\pi]$, I multiplied the result by $2$ for the symmetry to make the length of the interval $2\pi$, without knowing if it's correct.

The second way is $\frac{1+\cos{\phi}}{2}=\cos^2{(\phi/2)}$, substituting $\phi=\theta-\pi/2$,
thus $\sqrt{1+\sin{\theta}}=\sqrt{2\cos^2{(\theta/2-\pi/4)}}=+\sqrt{2}\cos{(\theta/2-\pi/4)}$.
From $-\pi/2 <\theta/2-\pi/4<\pi/2$, I got the interval $[-\pi/2,3\pi/2]$.
I end up with an interval length of $2\pi$, so I just found the final result from this interval.

So, I used two method to solve the integral and needed to reconsider the interval for integral.
Though I got a result, I'm not fully understanding the way to find the interval.

Could you tell why these methods are correct or incorrect?

Edit: I should have mentioned I found the answer of $8$. It's not the way to get the answer that I'm asking. I'm confused about interval.

Björn Friedrich
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nik
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  • How did you get $\sqrt{1+\sin\theta}$ under the integral sign? You should have $$ \sqrt{1+\left(\frac{dr}{d\theta}\right)^2} = \sqrt{1+\cos^2\theta} $$ if I am not mistaken. And why do you have $\sqrt 2$ in front of the integral? – MSDG Sep 08 '18 at 11:52
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    Shouldn't it be $\sqrt{r^2+(\frac{dr}{d\theta})^2}$? – nik Sep 08 '18 at 12:10
  • Ah, yes of course. Sorry! – MSDG Sep 08 '18 at 12:26

1 Answers1

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Length of the curve is

$$I=\int_{0}^{2 \pi}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta = \int_{0}^{2 \pi}\sqrt{r^2+\cos^2\theta} \hspace{3pt}d \theta$$ $$I=\int_{0}^{2 \pi}\sqrt{1+2\sin \theta +sin^2\theta+\cos^2\theta} \hspace{3pt}d \theta$$ $$I=\sqrt2\int_{0}^{2\pi}{\sqrt{1+\sin{\theta}}}\;d\theta$$ $$I=\sqrt{2}{\displaystyle\int_{0}^{2\pi}}\sqrt{2}\cos\left(\dfrac{2\theta-{\pi}}{4}\right)\,\mathrm{d}\theta$$

$$I=2{\displaystyle\int_{0}^{2\pi}}\cos\left(\dfrac{2\theta-{\pi}}{4}\right)\,\mathrm{d}\theta$$

$u=\dfrac{2\theta-{\pi}}{4}$ thus $\mathrm{d}\theta=2\,\mathrm{d}u$

$$I=4\class{steps-node}{\displaystyle\int}\cos\left(u\right)\,\mathrm{d}u=-4\sin u+c=-4\sin(\dfrac{2\theta-{\pi}}{4})+c$$

As $\theta$ goes from 0 to $2 \pi$

$$I=-4\sin(\dfrac{3\pi}{4})+4\sin(\dfrac{-\pi}{4})=-4\sqrt2$$

Deepesh Meena
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