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For $r = 1 + \cos \theta, 0 \le \theta < 2\pi$ in 2D polar space calculate the length of $P$, the perimeter by: $$ \int_P \sqrt{(dx)^2 + (dy)^2} \tag{1} $$ by showing: $$ (dx)^2 + (dy)^2 = (dr)^2 + (rd\theta)^2.\tag{2} $$

I am unsure where to go with this. How do I show the 2nd statement and then use it to solve the integral?

Qmechanic
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H SJ
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1 Answers1

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Hint: $$r(\theta)~=~1+\cos(\theta)~=~2\cos^2\frac{\theta}{2},$$

so the perimeter is given by

$$ L~=~\int_0^{2\pi} \!{\rm d}\theta~\sqrt{x^{\prime}(\theta)^2+y^{\prime}(\theta)^2}~=~\int_0^{2\pi} \!{\rm d}\theta~\sqrt{r^{\prime}(\theta)^2+r(\theta)^2}~=~\int_0^{2\pi} \!{\rm d}\theta~2\cos\frac{\theta}{2} ~=~8.$$

Qmechanic
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