What's the volume of $x^2+xy+y^2+u^2+uv+v^2=1$

Question

What's the volume of $x^2+xy+y^2+u^2+uv+v^2=1$?

I am fairly convinced that the answer should be $2\pi^2/3$

So hopefully this is a not-so-hard exercise in multi-variable calculus for someone who would enjoy the exercise. Its a little out of my depth. I can't quite see how the familiar techniques should generalize into into 4 dimensions. Also to give the answerer some clues about what I do know about calculus let me demonstrate that I CAN see how to do $x^2+xy+y^2=1$. I am really just showing that I remember some college calculus tools.

So for this simpler two dimensional case we should just look at this guy in polar coordinates.

$x^2+xy+y^2=r^2\big(1+2\sin(\theta)\cos(\theta) \big)$ after the standard change of coordinates.

$$r(\theta)=\frac{1}{\sqrt{1+\sin(\theta)\cos(\theta)}}$$

$$A=\frac12\int_0^{2\pi} r(\theta)^2\,d\theta=\frac{1}{2}\int_0^{2\pi} \frac{d\theta}{1+\sin(\theta)\cos(\theta)}=\frac{2\pi}{\sqrt{3}}$$ I am not quite sure how to generalize this approach in 4 dimension.

I can get the answer using a more convoluted elementary method.

I read here (Theorem 13) that $$R(k)=\operatorname{card} \{(x,y,u,v)\in \mathbb{Z^4}: x^2+xy+y^2+u^2+uv+v^2=k\}=12\sigma(n)-36\sigma(n/3)$$

Where $\sigma(x)$ refers to the sum of the divisors of $x$.

With this knowledge along with $$\sum_{n=1}^x \sigma(n) \approx \frac{\pi^2}{12}x^{2}$$

allows us to conclude that this volume should be $2\pi^2/3$. I talk about this Diophantine technique elsewhere so won't spend too long on this.

Can someone help me confirm this using calculus techniques?

I appreciate the help.

Answer 1

This sort of thing is easiest to do by using as little calculus as possible. First, we rewrite the left-hand side to be a sum of squares: $$ 1 = (x-y/2)^2 +\frac{3}{4} y^2 + (u-v/2)^2 + \frac{3}{4} v^2 $$

The volume is $$ \int_{ (x-y/2)^2 +\frac{3}{4} y^2 + (u-v/2)^2 + \frac{3}{4} v^2 \leq 1 } 1 \, dx \, dy \, du \, dv . $$ To change this into a volume we know, we let $X = x-y/2$, $Y = (\sqrt{3}/2) y$, $U = u-v/2$, $V = (\sqrt{3}/2) v$. Then the Jacobian is $$ \det{\frac{\partial(X,Y,U,V)}{\partial(x,y,u,v)}} \det{\begin{pmatrix} 1 & -1/2 & 0 & 0 \\ 0 & \sqrt{3}/2 & 0 & 0 \\ 0 & 0 & 1 & -1/2 \\ 0 & 0 & 0 & \sqrt{3}/2 \end{pmatrix}} = 3/4 , $$ and the integral becomes $$ \int_{X^2+Y^2 + U^2 + V^2 \leq 1 } \frac{4}{3} \, dX \, dY \, dU \, dV . $$ All we have to do now is find the volume of the unit sphere in $4$ dimensions. One can use a generalisation of polar coordinates, namely $$ \begin{align} X &= r\sin{\theta} \sin{\phi} \sin{\psi} \\ Y &= r\sin{\theta} \sin{\phi} \cos{\psi} \\ U &= r\sin{\theta} \cos{\phi} \\ V &= r\cos{\theta} , \end{align} $$ with $0<r<1$, $0<\theta<\pi$, $0<\phi<\pi$ and $0<\psi<2\pi$. A tedious computation reveals that the Jacobian is $r^3 \sin{\phi} \sin^3{\theta}$, and so the volume of the unit sphere is $$ \left( \int_0^1 r^3 \, dr \right) \left( \int_0^{\pi} \sin{\phi} d\phi \right) \left( \int_0^{\pi} \sin^2{\theta} d\theta \right) \left( \int_0^{2\pi} d\psi \right) . $$ All of these integrals are straightforward, and we find this volume is $\pi^2/2$, and hence the required volume is indeed $2\pi^2/3$.

There are easier ways to find the volume of a sphere, but they tend to use the Gamma-function.