I was fooling around with a graphing calculator, and I noticed a pattern in the functions of $$x^a+y^a=1$$ where $a$ is an even number. As $a$ increases, the graph begins to look like a square (if you split the function into $f_1(x)$ and $f_2(x)$, where, in $f_1(x)$, $y \geq 0$, and where, in $f_2(x),y \leq 0$). This means that, if $x \to \infty$, the graph of $f(x)$ will be that of a square, and thus, the area between the two functions will be 4. How can I begin to prove $$\lim_{a \to \infty}\int\limits_{-1}^1\pm\sqrt[a]{1-x^a}dx=4$$ for $a$ is even, or the general solution for $r$ - that is, $$\lim_{a \to \infty}\int\limits_{-r}^r\pm\sqrt[a]{r^2-x^a}dx=4r^2$$ for $a$ is even? I'm more interested in hints than in a complete proof. I'm not sure there is one, because of the requirement that $a$ be even.
Note: I'm not positive this is correct, because all I have to go on is a pictorial pattern. If I am in fact wrong, I'd be glad if someone could correct me.
Notice that for $a=2n$ $$\int^{1}_{-1}\sqrt[a]{1-x^a}\,dx=\int^{1}_{-1}\sqrt[2n]{1-x^{2n}}\,dx=2\int^{1}_{0}\sqrt[2n]{1-x^{2n}}\,dx$$ Substitute $y=x^{2n}$ then $\,dy=2nx^{2n-1}\,dx$ and one can rewrite the above integral as $$2\int^{1}_{0}\sqrt[2n]{1-x^{2n}}\,dx=\frac{1}{n}\int^{1}_{0}\sqrt[2n]{1-y}\cdot y^{\frac{1-2n}{2n}}\,dy$$ Now using the definition of Beta function i.e. $$B(w,z)=\int^{1}_{0}u^{w-1}(1-u)^{z-1}\,du$$ for $\Re{(w)}>0$ and $\Re{(z)}>0$ we can rewrite the last integral expression as follows $$\frac{1}{n}\int^{1}_{0}\sqrt[2n]{1-y}\cdot y^{\frac{1-2n}{2n}}\,dy=\frac{1}{n}\int^{1}_{0}(1-y)^{1/2n}\cdot y^{\frac{1-2n}{2n}}\,dy=\frac{1}{n}\int^{1}_{0}(1-y)^{\frac{1}{2n}+1-1}\cdot y^{\frac{1}{2n}-1}\,dy=\frac{1}{n}B(\frac{1}{2n}+1,\frac{1}{2n})$$ Using $$B(w,z)=\frac{\Gamma{(w)}\Gamma{(z)}}{\Gamma{(w+z)}}$$ we have $$\frac{1}{n}B(\frac{1}{2n}+1,\frac{1}{2n})=\frac{1}{n}\cdot \frac{\Gamma{(\frac{1}{2n}+1)}\Gamma{(\frac{1}{2n})}}{\Gamma{(1+\frac{1}{n})}}$$ In the limit we have $$\lim_{n\to\infty}\frac{1}{n}\cdot \frac{\Gamma{(\frac{1}{2n}+1)}\Gamma{(\frac{1}{2n})}}{\Gamma{(1+\frac{1}{n})}}=\lim_{n\to\infty}\frac{2}{2n}\cdot \frac{\Gamma{(\frac{1}{2n}+1)}\Gamma{(\frac{1}{2n})}}{\Gamma{(1+\frac{1}{n})}}=\frac{\Gamma{(0+1)}}{\Gamma{(1+0)}}\cdot \lim_{n\to\infty}\frac{2}{2n}\Gamma{(\frac{1}{2n})}=\frac{\Gamma{(1)}}{\Gamma{(1)}}\cdot 2=2$$ This is the value of the area of the upper half of the "square" so the total area would be $4$. The same technique can be used for any $r$ other than $1$.