Area enclosed by x^4 + y^4 = 1 - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-19T16:54:06Z http://mathoverflow.net/feeds/question/26823 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26823/area-enclosed-by-x4-y4-1 Area enclosed by x^4 + y^4 = 1 fuzzylintman 2010-06-02T14:14:04Z 2010-06-02T16:25:59Z <p>Trying to solve for the area enclosed by <code>$x^4+y^4=1$</code>. 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. </p> <p>Anyway, the question becomes more general, since we <em>think</em> that </p> <p><code>$lim_{n\to\infty}\int_0^1{(1-x^n)^{1/n}} = {1\over4}$</code> &nbsp;&nbsp;&nbsp;(it approaches a square / becomes linear)</p> <p>can anyone confirm that this is true or not?</p> http://mathoverflow.net/questions/26823/area-enclosed-by-x4-y4-1/26825#26825 Answer by Wadim Zudilin for Area enclosed by x^4 + y^4 = 1 Wadim Zudilin 2010-06-02T14:21:22Z 2010-06-02T14:30:03Z <p>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 $z\to 0$.</p> http://mathoverflow.net/questions/26823/area-enclosed-by-x4-y4-1/26828#26828 Answer by Thomas Kragh for Area enclosed by x^4 + y^4 = 1 Thomas Kragh 2010-06-02T14:54:18Z 2010-06-02T16:25:59Z <p>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):</p> <p>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.</p> <p>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.</p>