Computing the volume inside a surface S, using a seemingly unrelated result, Consider the surface $$S = \{(x,y,z): x^2 + xy + y^2 + z^2 = 1\}$$.  What is the volume inside S?
This is actually part (b) of the question.  I'm not sure which approach to take.
But part (a) of the question asked to compute the length of C, where C is the graph of the function f(t) = $\frac {e^t + e^{-t}}{2}$ on the interval [0,$2\pi$].  I was able to solve part (a).
For these questions, almost always we need to use the result found in part (a) to answer part (b) ... but there doesn't seem to be a connection between the two parts of this question at all.
Thanks,
 A: Hint for the volume inside the surface $S$.
Find a linear transformation of the space such that the image of the surface $S$ is the sphere centered on the origin with radius equal to $1$. Then use change of variables theorem for integrals.
Giving some more details
$$(x,y,z) \to x^2 + xy +y^2 + z^2$$ is an inner product. Trying to find an orthogonal basis, you'll see that the linear application
$$A =
 \begin{pmatrix}
  \frac{1}{\sqrt{3}} & 1 & 0 \\
  -1 & \frac{1}{\sqrt{3}} & 0 \\
  0 & 0 & 1
 \end{pmatrix}$$ transforms the set $B= \{(x,y,z): x^2 + xy + y^2 + z^2 \le 1\}$ into the set $B^\prime= \{(u,v,w): u^2 + v^2 + w^2 \le 1\}$. $B\prime$ is the 3D unit ball. Then you can use the Substitution for multiple variables
$$\int_{\varphi(U)} f(R) dR=\int_Uf(\varphi(T))\vert \det (D \varphi)(T)\vert dT$$ with $U=B$, $\varphi=A$, $\varphi(U)=B^\prime$ and $f=1$ you get
$$\int_{B^\prime} 1dR=\frac{4}{3}\int_B 1 dT \text{ as } \det A =\frac{4}{3}$$
What you're looking for is $\displaystyle \int_B 1 dT$ which is equal to $\displaystyle \frac{3}{4}\int_B 1 dR = \frac{3}{4} \frac{4}{3} \pi^3 = \pi^3$ as the volume of a sphere of radius $r$ is equal to $\frac{4}{3} \pi r^3$
