Area Enclosed by $x^{2n} +y^{2n}=2n$ Recently I was just playing with the idea of shapes with higher coefficients of x and y.
Using Desmos, I plotted graphs of the form $x^{2n}+ y^{2n} = 1$ All these of,course pass through the four points $\{(1,0),(0,1),(0,-1),(-1,0)\}$.
As one increases the value of $n$, The graph sort of pushes towards the lines  $x,y=1,-1$. This is reasonable as none of the variables can actually exceed unity.
But when instead of 1 we define the shape as $x^{2n}+ y^{2n} = 2n$, the graph bulges outwards, but with increasing values of $n$, shrinks back, visually and intuitively I think the area enclosed by the graph would become 4 as ${n\to \infty} $. But I could not mathematically prove it using integration or otherwise.
Also, is there any name for such shapes and do we use them anywhere for some practical benefits?
And can we manipulate the graphs to change the vertices of the limiting "square", if there is one?
PS: Just to clarify my knowledge base: I am a Class 12th student in India and preparing for JEE ADVANCED, so I am familiar with basic concepts of finding areas using calculus.
 A: I can assume that you want to prove that
$x=1,y=1,x=-1,y=-1$ are asymptotic boundaries of your curve (when $n \to \infty$) for $|x|,|y| \leq 1$
Take
$$ x^{2n}=2n-y^{2n} $$
$$ x=\pm \sqrt[2n]{2n-y^{2n}}$$
First we know that
$$\lim_{n \to \infty} n^{\frac1{n}}=1$$
Equally
$$\lim_{n \to \infty} (n \pm 1)^{\frac1{n}}=1$$
This is to say
$$ x=\pm\lim_{n \to \infty} \sqrt[2n]{2n-y^{2n}}=\pm 1$$
The same if you take
$$ y=\pm \sqrt[2n]{2n-x^{2n}}$$
$$ y=\pm\lim_{n \to \infty} \sqrt[2n]{2n-x^{2n}}=\pm 1$$
That is for $|x|,|y| \leq 1$. Now you need to prove that we cannot have $|x|,|y| > 1$ at infinity. But this is not possible for the same reasoning as above that claims that it is strictly $|x|,|y| = 1$ at infinity. It would have to exist one point where both $|x|>1$ and $|y|>1$. However, none of the approaching curves (when $2n < +\infty$) has this property. Therefore, its asymptote cannot have this property either.
The conclusion about the asymptotic area follows.
You could attack the area directly but the reasoning is more or less the same.
$$S_{n}=2\int_{-1-\delta}^{1+\delta}\sqrt[\frac{1}{2n}]{2n-x^{2n}}dx$$
However, we have just confirmed that $\delta \to 0$ and $\sqrt[\frac{1}{2n}]{2n-x^{2n}} \to 1$ as $n \to +\infty$. It is giving
$$S_{\infty}=2\int_{-1}^{1}dx=4$$
A: method of Dirichlet, in Whittaker and Watson, the area is
$$ \frac{ 4 \; (2n)^{\frac{1}{n}} \; \Gamma \left(1 + \frac{1}{2n}  \right)^2  }   { \Gamma \left(1 + \frac{1}{n}  \right) } $$
When $n=1$  we need to know $ \Gamma \left( \frac{3}{2}  \right) = \frac {\sqrt \pi }{2}. $ In this case we have a circle of radius $\sqrt 2,$  area is $2 \pi $
