TITLE: Find length of curve via integral
QUESTION [1 upvotes]: 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?

REPLY [2 votes]: 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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