Arc Length of an Astroid 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. 
 A: 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
$$
\begin{align}
s
&=4\int_0^{\pi/2}\sqrt{\dot x^2+\dot y^2}~dt\\
&=4\cdot8\cdot3\int_0^{\pi/2}\sqrt{\cos^4t\sin^2t+\sin^4t\cos^2t}~dt\\
&=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t\sqrt{\cos^2t+\sin^2t}~dt\\
&=4\cdot8\cdot3\int_0^{\pi/2}\cos t\sin t~dt\\
&=4\cdot8\cdot3\cdot\frac{1}{2}\\
&=48
\end{align}
$$
A: The function is indeed $f(x)=(4-x^{2/3})^{3/2}$; let's go slowly with the derivative:
$$
f'(x)=\frac{3}{2}(4-x^{2/3})^{1/2}\left(-\frac{2}{3}x^{-1/3}\right)=-\frac{(4-x^{2/3})^{1/2}}{x^{1/3}}
$$
Hence
$$
1+(f'(x))^2=1+\frac{4-x^{2/3}}{x^{2/3}}=4x^{-2/3}
$$
Now you want to compute the bounds. You get a quarter of the length with
$$
\int_0^8\sqrt{1+(f'(x))^2}\,dx=\int_0^8 2x^{-1/3}\,dx=2\Bigl[\frac{3}{2}x^{2/3}\Bigr]_0^8=12
$$
Your bounds are wrong. If you want an eighth, intersect the astroid with $y=x$, which yields $2x^{2/3}=4$, so $x=2^{3/2}=\sqrt{8}$, but the upper bound should be when $y=0$, that is, $x=8$.
