Find the volume enclosed by the surface $S := \{(x,y,z): x^2 + xy + y^2 + z^2 = 1\}$

Question

Find the volume enclosed by the surface $$S := \{(x,y,z): x^2 + xy + y^2 + z^2 = 1\}.$$

My attempt was this:

I moved the tricky $xy$-term over to the r.h.s. I now have $$x^2+y^2+z^2 = 1-xy,$$ which is a sphere of radius $\sqrt{1-xy}$.

The logical next steps would be to integrate the $1$ function, using a triple, iterated-integral, in spherical coordinates.

But the innermost integral will have an upper limit of $\sqrt{1-xy}$, which when I expand out the $x$ and $y$, will depend on all three spherical coordinates, $r$, $\theta$, and $\phi$.

What intermediate steps can I take first? Am I going in a completely wrong direction?

Thanks.

Answer 1

A surface doesn't have a volume. Presumably what is meant is the volume of the region enclosed by the surface, namely $R = \{(x,y,z): x^2 + x y + y^2 + z^2 \le 1\}$.

Hint: complete the square with respect to either $x$ or $y$. You'll find cross-sections that are disks...

Answer 2

The variables $x, y, z$ are coordinates, so it doesn't make sense to speak of a "sphere of radius $\sqrt{1 - x y}$".

One option is to eliminate the cross-term by making the linear change of variables, $$x = u + v, \qquad y = -u + v;$$ substituting gives that, in these coordinates, $S$ is given by $$u^2 + 3 v^2 + z^2 = 1.$$