Area enclosed by x^4 + y^4 = 1 - MathOverflow [closed]

Area enclosed by x^4 + y^4 = 1

2010-06-02T14:14:04Z

Trying to solve for the area enclosed by $x^4+y^4=1$ . A friend posed this question to me today, but I have no clue what to do to solve this. Keep in mind, we don't even know if there is a straightforward solution. I think he just likes thinking up problems out of thin air.

Anyway, the question becomes more general, since we think that

$lim_{n\to\infty}\int_0^1{(1-x^n)^{1/n}} = {1\over4}$ (it approaches a square / becomes linear)

can anyone confirm that this is true or not?

Answer by Wadim Zudilin for Area enclosed by x^4 + y^4 = 1

2010-06-02T14:21:22Z

I always prefer not to skip $dx$: $$ I_n=\int_0^1(1-x^n)^{1/n}dx. $$ After the change of variable $t=x^n$, the integral becomes the beta integral, $$ I_n=\frac1n\int_0^1(1-t)^{1/n}t^{1/n-1}dt =\frac1n\frac{\Gamma(1+1/n)\Gamma(1/n)}{\Gamma(1+2/n)} =\frac1n\frac{\Gamma(1/n)^2\cdot 1/n}{\Gamma(2/n)\cdot 2/n} \to1 \quad\text{as $n\to\infty$}, $$ as $1/\Gamma(z)\sim z$ as $z\to 0$.

Answer by Thomas Kragh for Area enclosed by x^4 + y^4 = 1

2010-06-02T14:54:18Z

I think this question smells of homework, but another answer, which to me totally obscures the geometric nature of the question has been posted, and I feel that this justifies the following answer (even if the question is closed):

The $l^p$ norms $\lvert(x,y)\rvert_p = (\lvert x\rvert^p+\lvert y \rvert^p)^{1/p}$ are norms and satisfies that if $\lvert(x,y)\rvert_p=1$ and $q>p$ then $\lvert(x,y)\rvert_q\leq 1$. So the unit "circles" of which you want to find the area grows.

It is also a fact that $\lvert (x,y) \rvert_p \to \max (\lvert x\rvert,\lvert y\rvert)$ as $p\to \infty$. So the unit circles converges to the square which is the boundary of $[-1,1]\times [-1,1]$. This implies by monotone convergence theorem that your integral converges to 1. Because the entire square has area 4.