according to wolframalpha $\ell(\gamma) = 6$
but I found $0$ and I still haven't figured out where's my mistake.
my attempt :
$$\gamma(t) = (\cos^3t,\sin^3t) \implies \gamma'(t) = 3(-\sin t\cos^2 t,\cos t\sin^2 t) $$ so $$||\gamma'(t)||^2 =9(\sin^2 t\cos^4 t+\cos^2 t\sin^4 t) = 9(\sin^2 t(\cos^2 t(\sin^2 t+\cos^2 t))) = 9\sin^2 t \cos^2 t$$ $$||\gamma'(t)|| = 3\sin t \cos t=\frac32\sin(2t)$$ $$\ell(\gamma)=\int_0^{2\pi}||\gamma'(t)||dt =\frac32\int_0^{2\pi}\sin(2t)dt = \frac34[-\cos(2t)]_0^{2\pi} =0$$
can someone please point out where I did something wrong ?
Note that this curve will lie in all the four quadrants. To calculate the length of this curve it is enough to calculate the length in the first quadrant. Now use the formula for length, $l= \int_{t=0}^{t=2\pi}\sqrt{1+ (\frac{dy}{dx})^2}dx$
So by symmetry we have, \begin{align*} l &= 4\int_0^{\pi/2}\sqrt{\bigg(\frac{d}{dt}(\cos^3t)\bigg)^2+\bigg(\frac{d}{dt}(\sin^3 t)\bigg)^2}dt \\ &=4\int_0^{\pi/2}3\sqrt{\cos^4t\sin^2t+ \sin^4t\cos^2t} \ dt \\ &= 12\int_0^{\pi/2}\cos t\sin t \ dt = 6 \end{align*}