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

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.
