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So, we have a function:

$$r=a\cos^3{\frac{\phi}{3}}$$

We need to get the arc's length on interval:

$$0 \leq \phi \leq \frac{\pi}{2}$$

So, using default formula: $$L = \int_{a}^{b}\sqrt{1+(f'(x))^2}dx$$

We got: $$L = \int_{0}^{\pi/2}\sqrt{1+(a\cos^2(\frac{\phi}{3})\sin(\frac{\phi}{3}))^2}d\phi $$

But how can we completely solve it?

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The arc-length in polar coordinates is $$ \mathrm{d}s^2=\mathrm{d}r^2+r^2\,\mathrm{d}\phi^2 $$ so it is \begin{align*} L&=\int_0^{\pi/2}\sqrt{\left(a\cos^3\frac\phi3\right)^2+\left(a\cos^2\frac\phi3\sin\frac\phi3\right)^2}\,\mathrm{d}\phi\\ &=\int_0^{\pi/2}a\cos^2\frac\phi3\,\mathrm{d}\phi \end{align*} which you can integrate easily.

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