What is the volume inside $S$, which is the surface given by the level set $\{ (x,y,z): x^2 + xy + y^2 + z^2 =1 \}$? The solution given uses a linear algebraic argument that doesn't seem very instructive -- and may not even be correct, I think.  
We notice from the equation, that the surface is a quadratic form, level set = 1.  
$\bullet$ The solution rewrites it as $x^TAx$,
$\bullet$ Finds a symmetric matrix $A$ that gives the resulting quadratic form, 
$\bullet$ Computes the eigenvalues of $A$, 
And then (here's where the explanation doesn't really follow, I think) 
$\bullet$ A change of variables is made so that the surface becomes an ellipsoid,
$\bullet$ Finally using the volume of an ellipsoid formula, the final answer is given.
Is there another / better way of finding the volume of $S$?
Thanks,

First look at the elliptical cross-section in xy-plane:
$\{ (x,y): x^2 + xy + y^2 = 1 \}$
It is clear that this is symmetric in $x$ and $y$ so write
$x^2 + xy + y^2 = a(x+y)^2 + b(x-y)^2$ so by equating coefficients $a=\frac{3}{4}, b=\frac{1}{4}$
Therefore $\boxed{\dfrac{3}{4}(x+y)^2 + \dfrac{1}{4}(x-y)^2 = 1}$
Changing to new coordinate system: $X = \dfrac{1}{\sqrt{2}}\Big(x+y\Big), Y = \dfrac{1}{\sqrt{2}}\Big(x-y\Big)$ (where the $\frac{1}{\sqrt{2}}$ terms ensure that the new axes are of unit length), and writing in standard form:
$\dfrac{X^2}{({\sqrt{\frac{2}{3}}})^2} + \dfrac{Y^2}{({\sqrt{2}})^2} = 1$,  or for the full level set $\boxed{\dfrac{X^2}{({\sqrt{\frac{2}{3}}})^2} + \dfrac{Y^2}{({\sqrt{2}})^2} + \dfrac{z^2}{1^2} = 1}$
Hence, the level set is an ellipsoid with semi-axes of lengths $R_X = \sqrt{\frac{2}{3}}$ , $R_Y = \sqrt{2}$ and $R_Z = 1$.
So the volume is $\dfrac{4}{3}\pi \cdot R_X \cdot R_Y \cdot R_Z = \dfrac{8 \pi}{3 \sqrt{3}}$ 
