# Area of $x^{10}+y^{10}\leq 1$

Whilst looking at someone's vector calculus problem, they mentioned that, making use of Green's Theorem, they had to express the line integral of the boundary of $$x^{10}+y^{10}\leq 1$$ in terms of its area. The thing is they gave you the area as computed with Mathematica to be $$4\Gamma^2(11/10)/\Gamma(6/5)\approx 3.943.$$

And I was wondering how you'd prove this: $$\mathcal{A}=\iint_{x^{10}+y^{10}\leq1}dydx=2\int_{-1}^1\sqrt[10]{1-x^{10}}dx=\dfrac{4\Gamma^2(11/10)}{\Gamma(6/5)}.$$

• It is a method due to Dirichlet; it is in Whittaker and Watson, A Course of Modern Analysis. First edition was 1902, on pages 191-193. I found an online scan ... let me see if I can find it again. Well, this is the third edition, ia801601.us.archive.org/6/items/courseofmodernan00whit/… Jul 27, 2023 at 21:41
• page n8mbered 258, pdf page called 272 Jul 27, 2023 at 21:50
• Are you trying to do this with Green’s Theorem? Do the line integral $$\frac12\int_C -y\,dx+x\,dy,$$ parametrizing $C$ by $x=(\cos t)^{1/5}$, $y=(\sin t)^{1/5}$. Jul 27, 2023 at 23:50
• We could also apply what you said: Choosing $\mathbf{F}=x\boldsymbol{j}$, we get that $$\iint_A (\partial_x F_y-\partial_y F_x)dydx=\iint_A dydx,$$ and by Green's theorem we know that is also equal to $$\oint_{\partial A}\mathbf{F\bullet dr}=\oint_{\partial A}xdy$$ and by parametrizing $x$ and $y$ like you said, we get $$\frac{1}{5}\int^{2\pi}_0 \cos^{\frac{6}{5}}t\sin^{-\frac{4}{5}}tdt,$$ which has somewhat the form of the beta function: $$\mathfrak{B}(x,y)=2\int_0^{\frac{\pi}{2}}\cos^{2x-1}\phi\sin^{2y-1}\phi d\phi.$$ And from here IDK how to proceed. Jul 28, 2023 at 0:39
• I am not sure why but when computing the integral numerically, I don't get what I'm supposed to... Jul 28, 2023 at 1:11

We can tackle a more general question: that of the unit $$\ell^p$$ ball in $$\mathbb{R}^n$$, i.e. the volume of the body given by $$\{ (x_1,\cdots,x_n) \in \mathbb{R}^n \mid |x_1|^p + |x_2|^p + \cdots + |x_n|^p \le 1 \}$$ Note that your case is $$n=2$$ and $$p=10$$.
Let $$V_{n,p}$$ denote the volume of this ball. Then, analogous to your transformation from a multi-integral to a single integral, we see $$V_{n,p} = 2 V_{n-1,p} \int_{0}^1 (1-x^p)^{(n-1)/p} \, \mathrm{d}x$$ To see this, note that $$|x_1|^p + |x_2|^p + \cdots + |x_n|^p \le 1 \iff |x_2|^p + \cdots + |x_n|^p \le 1 - |x_1|^p$$ and rewrite as $$V_{n,p} = \int_{-1}^1 \mathrm{d}x_1 \int_{ \sum_2^n |x_i|^p \le 1 - |x_1|^p} \mathrm{d} x_2 \cdots \mathrm{d} x_n$$ with the right integral as the volume of an $$(n-1)$$-dimensional ball.
Now, with the substitution of $$u=x^p$$, we see that $$x=u^{1/p} \qquad \mathrm{d}u = px^{p-1} \, \mathrm{d} x \qquad \mathrm{d} x = \frac 1 p u^{\frac 1 p - 1}$$ and hence $$\int_{0}^1 (1-x^p)^{(n-1)/p} \, \mathrm{d}x = \frac 1 p \int_{0}^1 u^{\frac 1 p - 1} (1-u)^{(n-1)/p} \, \mathrm{d}u$$ The beta function is defined by $$B(x,y) := \int_0^1 t^{x-1} (1-t)^{y-1} \, \mathrm{d} t \equiv \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ so we hence see $$V_{n,p} = \frac 2 p V_{n-1,p} \cdot \frac{ \Gamma(\frac 1 p ) \Gamma( \frac{n-1}{p} + 1 )}{\Gamma( \frac n p + 1 )}$$ Apply the recursion property of the gamma function, $$\Gamma(z+1) = z \Gamma(z)$$, to then get $$V_{n,p} = 2 V_{n-1,p} \cdot \frac{ \Gamma(\frac 1 p + 1 ) \Gamma( \frac{n-1}{p} + 1 )}{\Gamma( \frac n p + 1 )}$$ This gives us a recursive relation; iterating backwards to $$V_{1,p} = 2$$, we then get the desired result of $$V_{n,p} = 2^n \frac{\Gamma^n(\frac 1 p + 1)}{\Gamma(\frac n p + 1)}$$ With $$n=2$$ and $$p=10$$, we see your desired result: $$V_{2,10} = 4 \frac{\Gamma^2(11/10)}{\Gamma(6/5)}$$
This is a corollary to the relationship between the gamma and beta functions:$$\mathrm{B}(x,y) \doteq \int_0^1 t^{x-1}(1-t)^{y-1} dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$ To see this we use the substitution $$u = t^n$$ to find that $$\int_0^1 (1-t^n)^{1/n}\,dt = \int_0^1 (1-u)^{1/n} u^{-1+1/n}\,du = \frac{1}{n}\mathrm{B}(1+1/n,1/n) = \frac{\Gamma(1+1/n)\Gamma(1/n)}{n \Gamma(1+2/n)} = \frac{\Gamma(1+1/n)^2}{\Gamma(1+2/n)}.$$ The final step uses the identity $$\Gamma(x+1) = x \Gamma(x)$$. Interestingly, if we expand this formula in powers of $$1/n$$ we find out how fast the area converges to that of the $$2 \times 2$$ square: $$4 \frac{\Gamma(1+1/n)^2}{\Gamma(1+2/n)} = 4 - \frac{2\pi^2}{3n^2} + O(n^{-3}).$$