I want to find the arc length of the equation: $ x^{2/3} + y^{2/3} = 4 $
My steps follow as:
$ y= f(x) = (4-x^{2/3})^{3/2} $
$ f'(x)= -x^{-1/3}(4-x^{2/3})^{1/2} $
$ [f'(x)]^2 = 4x^{-2/3}-1 $
$ [f'(x)]^2 + 1 = 4x^{-2/3} $
$ L = \int_\sqrt8^4 \sqrt{f'(x)^2 + 1} dx = 0.0955 $
Bounds of the upper integral are found by:
$ y= (4-x^{2/3})^{3/2} = x $
$ x= 2 $ , $ x = \sqrt8 $
The integral gives us the asteroid's 1/8th of length. We multiply by 8 and finally get the result 0.7646.
I solved this problem by imitating the steps exactly of a similar problem with $ x^{2/3} + y^{2/3} = 1 $ instead of this problem's $ x^{2/3} + y^{2/3} = 4 $.
Change of the number from 1 to 4 should not change the method of solving this question right? According to my professor's answer, I should be getting 8 as a result. I am wondering if I did something wrong or I wrote my professor's answer as wrong.
PS: This is my first stackexchange post, feel free to point out if I could have done something better in my post.
An alternative method is to use the parametric form of the astroid, i.e.,
$$ x=8\cos^3t\\ y=8\sin^3t $$
Employing symmetry we can say that
$$ \begin{align} s &=4\int_0^{\pi/2}\sqrt{\dot x^2+\dot y^2}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\sqrt{\cos^4t\sin^2t+\sin^4t\cos^2t}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t\sqrt{\cos^2t+\sin^2t}~dt\\ &=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t~dt\\ &=4\cdot8\cdot3\cdot\frac{1}{2}\\ &=48 \end{align} $$