Integration: area enclosed by graph of $x^4 + y^4 = 1$ I'm attempting to find the area enclosed by the graph $x^4 + y^4 = 1$ as shown below.
My approach was to rearrange the equation so it is in terms of $y = f(x)$ and integrate one of the top two quadrants with respect to $x$ and then multiply by $4$ to get the area for the whole shape.  I've never tried to integrate this kind of graph before and I'm not sure If I've done it correctly.  Any input or assistance would be much appreciated.  Thanks.

 A: In general, the curve
$$y^{1/a}+x^{1/a}=1$$
is called a superellipse, and it's area is given by
$${4\cdot \Gamma(\alpha+1)^2 \over \Gamma(2\alpha +1)}$$
Here $\Gamma$ is Euler's Gamma function, defined for $\Re(\alpha)>0$ as 
$$\Gamma(\alpha)= \int_0^{\infty}e^{-\mu}\mu^{\alpha}\frac{d \mu}{\mu}$$
A closely related function is Euler's Beta function, given by
$$B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ 
for $\Re(a),\Re(b) > 0$.
For the details, you can see this question.
A: Essentially the solution you have is good, but I would like to have this one anyways.
The required area is by symmetry
$$ A = 4 \int_0^1 (1-x^4)^{\frac{1}{4}} \hspace{4pt} \mathrm{d}x$$
Substitute $u = x^4 \hspace{4pt} \Rightarrow  x^3 = u^{\frac{3}{4}} \hspace{4pt}, \mathrm{d}u = 4 x^3 \mathrm{d}x $
$$ 
\begin{align*}
A &= 4 \int_0^1 \frac{(1-u)^{\frac{1}{4}} \hspace{4pt}\mathrm{d}u}{4 u^{\frac{3}{4}}}\\
 &= \int_0^1 u^{\frac{-3}{4}} (1-u)^{\frac{1}{4}}  \hspace{4pt}\mathrm{d}u\\
 &= \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)}\\
&=  \frac{2 \times 3.287}{\sqrt{\pi}} \\
&\approx 3.708
\end{align*}
$$ 
