What does the area of the figure formed by the implicit function $x^{n}+y^{n}=c$ approach as n approaches infinity?

Question

What is the area of the figure formed by the implicit function $x^{n} + y^{n} = c$ approach as $n$ approaches $\infty$? ($n$ is an even positive integer)

Of course, the area will likely be a function of $c$.

Was just wondering, no special reason.

Answer 1

The area $A$ can be written as,

$$A = \lim_{n \to \infty} 4\int_0^\sqrt[n]{c} \sqrt[n]{c-x^n}dx$$

Since $\sqrt[n]{c-x^n}$ achieves its maximum at $x=0$, we can bound it above,

$$A \le \lim_{n \to \infty} 4\int_0^\sqrt[n]{c} \sqrt[n]{c}dx = \lim_{n \to \infty} 4 c^{\frac 2 n} = 4$$

Maybe with some more work it can be shown that there's a simple lower bound that proves the limit is $4$.

Perhaps surprisingly this implies the area doesn't depend on $c$. If you wanted the area to be dependent on $c$, you could look at $x^n+y^n=c^n$ instead.

Answer 2

Playing with n=100,101 taking c=1 in an assumed equation of better physical dimension on Wolfram Alpha...

$$ x^n+y^n= c^n $$

So can you draw the such graphs for lower values of $n$ say $(7,8)$ and conclude the link between sharp vertex shape and larger $n$?

Ok, what is your experience with area integral for large $n$?