Area Enclosed by $x^{2n} +y^{2n}=2n$

Question

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.

Answer 1

The Basic Problem

In the first quadrant, the curve $$\newcommand{\sgn}{\operatorname{sgn}} x^{2n}+y^{2n}=1\tag1 $$ can be parametrized by $$ (x,y)=\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)\tag2 $$

The area inside the curve is $$ \begin{align} &\frac12\int_0^{2\pi}(x,y)\times(x,y)'\,\mathrm{d}t\\ &=2\int_0^{\pi/2}(x,y)\times(x,y)'\,\mathrm{d}t\tag{3a}\\ &=2\int_0^{\pi/2}\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)\times\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)'\,\mathrm{d}t\tag{3b}\\ &=\frac2n\int_0^{\pi/2}\sin(t)^{1/n-1}\cos(t)^{1/n-1}\,\mathrm{d}t\tag{3c}\\ &=\frac2n\int_0^1t^{1/n-1}\left(1-t^2\right)^{\frac{1-n}{2n}}\frac{\mathrm{d}t}{\sqrt{1-t^2}}\tag{3d}\\ &=\frac1n\int_0^1t^{\frac{1-2n}{2n}}(1-t)^{\frac{1-2n}{2n}}\,\mathrm{d}t\tag{3e}\\ &=\frac{\Gamma\!\left(\frac1{2n}\right)\Gamma\!\left(\frac1{2n}\right)}{n\,\Gamma\!\left(\frac1n\right)}\tag{3f}\\ &=\frac{4\Gamma\!\left(1+\frac1{2n}\right)\Gamma\!\left(1+\frac1{2n}\right)}{\Gamma\!\left(1+\frac1n\right)}\tag{3g}\\ &=4-\frac{\pi^2}{6n^2}+O\!\left(\frac1{n^3}\right)\tag{3h} \end{align} $$ Explanation:
$\text{(3a)}$: the area is $4$ times the area in the first quadrant
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: compute the cross product
$\text{(3d)}$: substitute $t\mapsto\arcsin(t)$
$\text{(3e)}$: substitute $t\mapsto t^{1/2}$
$\text{(3f)}$: Beta integral
$\text{(3g)}$: $\Gamma\!\left(\frac1n\right)=n\,\Gamma\!\left(1+\frac1n\right)$
$\text{(3h)}$: $\Gamma'(1)=-\gamma$, $\Gamma''(1)=\frac{\pi^2}6+\gamma^2$

This converges pretty quickly to the square. The area outside the curve but inside the square is approximately $\frac{\pi^2}{6n^2}$.

---

Adjusting the Scales

In the first quadrant, the curve $$ x^{2n}+y^{2n}=2n\tag4 $$ can be parameterized by $$ (x,y)=(2n)^{\frac1{2n}}\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)\tag5 $$

That is, each coordinate has been scaled by $(2n)^{\frac1{2n}}$. This means the area has been scaled by $(2n)^{1/n}$. The area inside the curve is $$ \begin{align} (2n)^{1/n}\frac{\Gamma\!\left(\frac1{2n}\right)\Gamma\!\left(\frac1{2n}\right)}{n\,\Gamma\!\left(\frac1n\right)} &=(2n)^{1/n}\frac{4\Gamma\!\left(1+\frac1{2n}\right)\Gamma\!\left(1+\frac1{2n}\right)}{\Gamma\!\left(1+\frac1n\right)}\tag{6a}\\ &=4+\frac{4\log(2n)}{n}+\frac{2\log(2n)^2}{n^2}-\frac{\pi^2}{6n^2}+O\!\left(\frac{\log(n)^3}{n^3}\right)\tag{6b} \end{align} $$ This converges much more slowly to the square; the amount of area outside the square and inside the curve is approximately $\frac{4\log(2n)}{n}$.

---

Details on Computing the Area

The triangle with sides $(a,b)$ and $(c,d)$ is half of the parallelogram generated by $(a,b)$ and $(c,d)$, so we can compute its area as half of the determinant: $$ \frac12\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=\frac12(ad-bc)\tag7 $$ We can also compute the area of the triangle by setting it as the base for a prism with height $1$; generated by $(a,b,0)$, $(c,d,0)$, and $(0,0,1)$, and with volume computed using a triple product: $$ \frac12(a,b,0)\times(c,d,0)\cdot(0,0,1)=\frac12(ad-bc)\tag8 $$ Because of this and the fact that it is an antisymmetric bilinear form, the quantity $$ (a,b)\times(c,d)=ad-bc\tag9 $$ is sometimes called a cross-product, even though it is a scalar and not a vector.

By summing up triangular pieces of the area

we get the area of the closed curve traced out by $\vec{v}$ to be $$ \frac12\oint\vec{v}\times\mathrm{d}\vec{v}\tag{10} $$ In $(3)$, we have $\vec{v}=\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)$, so expanding $\text{(3c)}$ yields $$ \begin{align} &\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)\times\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)'\\ &=\left(\cos(t)^{1/n},\sin(t)^{1/n}\right)\times\frac1n\left(-\sin(t)\cos(t)^{1/n-1},\cos(t)\sin(t)^{1/n-1}\right)\tag{11a}\\ &=\frac1n\,\sin(t)^{1/n}\cos(t)^{1/n}\left(\frac{\cos(t)}{\sin(t)}+\frac{\sin(t)}{\cos(t)}\right)\tag{11b}\\ &=\frac1n\,\sin(t)^{1/n-1}\cos(t)^{1/n-1}\tag{11c} \end{align} $$ Explanation:
$\text{(11a)}$: take the derivative
$\text{(11b)}$: evaluate the cross product, pulling out a common factor
$\text{(11c)}$: $\frac{\cos(t)}{\sin(t)}+\frac{\sin(t)}{\cos(t)}=\frac1{\sin(t)\cos(t)}$

Answer 2

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

$$S_{\infty}=2\int_{-1}^{1}dx=4$$

Answer 3

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 $