I graphed it and also found that it was a cardioid (or a sideways heart). Ns am gaining stuck on the arc length.

this is what ns have:

$$r=(\frac 12(1+\cos\theta) $$

$$\frac drd\theta= -\frac 12\sin\theta $$

what I get for under the square source is :

$$\frac 14+\frac 12\cos\theta+\frac 14\cos^2\theta+ \frac 14\sin^2\theta $$

I ended up gaining stuck with $\frac12+\frac12\cos\theta $

But ns don"t think this is right, where did i go wrong?




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Things look at fine. By the double-angle identity you have already used, the square root is $|\cos(\theta/2)|$. Integrate. Note the absolute value sign.




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\beginalign\color#66f\Large L&=2\int_0^\pi\root\pars\dd r^2 + r^2\pars\dd\theta^2=2\int_0^\pi\root\bracks\totald\rm r\pars\theta\theta^2 + \rm r^2\pars\theta\,\dd\theta\\<3mm>&=2\int_0^\pi\root1 \over 4\,\sin^2\pars\theta + \cos^4\pars\theta \over 2\,\dd\theta\\<3mm>&=2\int_0^\pi\root1 \over 4\,\sin^2\pars\theta+ \bracks1 + \cos\pars\theta \over 2^2\,\dd\theta=\int_0^\pi\root2\bracks1 + \cos\pars\theta\,\dd\theta\\<3mm>&=\int_0^\pi\root4\cos^2\pars\theta \over 2\,\dd\theta=2\int_0^\pi\verts\cos\pars\theta \over 2\,\dd\theta=4\int_0^\pi/2\cos\pars\theta\,\dd\theta = \color#66f\Large 4\endalign