Volume of $M:=\big\{(x,y,z)\in \mathbb{R}^3\,:\,ax^2+2bxy+cy^2\leq z \leq 3\big\}$ Let \begin{pmatrix} a & b \\ b & c \end{pmatrix}  be a positive definite matrix. How can I calculate the volume of $$M:=\big\{(x,y,z)\in \mathbb R^3\,:\,ax^2+2bxy+cy^2\leq z \leq 3\big\}\,?$$
I am not sure whe the property of the matrix comes into play...
 A: $ax^2+2bxy + cy^2 = \begin{bmatrix} x&y\end{bmatrix}\begin{bmatrix} a&b\\b&c\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix} \le z \le 3 $
Compare this matrix with what you show above.
M is an elliptical paraboloid.
Since this matrix is symmetric, it can be daigonalized with ortho-normal basis.
$\mathbf x^T P^T \Lambda P \mathbf x \le z \le 3$
and since P is ortho-normal it doesn't distort distances (or volumes), and we can say
$\mathbf x^T \Lambda \mathbf x \le z \le 3$
or 
$\lambda_1 x^2 + \lambda_2 y^2 \le  z$
Integrating in polar coordinates:
$\int_0^{2\pi}\int_0^3\int_0^{\sqrt z} \frac {r}{\sqrt{\lambda_1\lambda_2}} \ dr\ dz\ d\theta\\
\int_0^{2\pi}\int_0^3\frac {r^2}{2\sqrt{\lambda_1\lambda_2}} |_0^{\sqrt z}\ dz\ d\theta\\
\int_0^{2\pi}\int_0^3\frac {z}{2\sqrt{\lambda_1\lambda_2}} \ dz\ d\theta\\
\int_0^{2\pi}\frac {z^2}{4\sqrt{\lambda_1\lambda_2}} |_0^{3}\ d\theta\\
\frac {9\pi}{2\sqrt{\lambda_1\lambda_2}}
$
The product of eigenvalues?
$\lambda_1\lambda_2 = ac-b^2$
A: You can write $ax^2+2bxy+cy^2=\lambda\,u^2+\mu\,v^2$ for some $\lambda,\mu>0$, where $u$ and $v$ are linear functions in terms of $x$ and $y$ (depending on, of course, the parameters $a,b,c$).  We may assume that $\lambda\geq\mu$.  For a fixed $h\in\mathbb{R}_{\geq 0}$ (in this problem, $h=3$), I shall write
$$M:=\big\{(x,y,z)\in\mathbb{R}^3\,\big|\,ax^2+2bxy+cy^2\leq z\leq h\big\}$$ instead. 
Now, in the $(u,v,z)$-coordinates, the slice $(u,v,z)$ of $M$ at a fixed $z=\zeta\geq 0$ satisfies $$\lambda\,u^2+\mu\,v^2\leq \zeta\,.$$  This is an ellipse $E_\zeta$ with semiminor axis $\sqrt{\dfrac{\zeta}{\lambda}}$ and semimajor axis $\sqrt{\dfrac{\zeta}{\mu}}$.  The area of $E_\zeta$ equals $$\pi\left(\sqrt{\dfrac{\zeta}{\lambda}}\right)\left(\sqrt{\dfrac{\zeta}{\mu}}\right)=\dfrac{\pi\,\zeta}{\sqrt{\lambda\mu}}\,.$$
Hence, the volume of $M$ in the $(u,v,z)$-coordinate is thus
$$\int_0^h\,\dfrac{\pi\,z}{\sqrt{\lambda\mu}}\,\text{d}z=\frac{\pi\,h^2}{2\,\sqrt{\lambda\mu}}\,.$$
Now, assume that $T$ is the linear transformation sending $(x,y)$ to $(u,v)$; in other words,
$$\begin{bmatrix}u\\v\end{bmatrix}=T\,\begin{bmatrix}x\\y\end{bmatrix}\,.$$
By the Change-of-Variables Formula, we get that
$$\text{d}u\,\text{d}v=\big|\det(T)\big|\,\text{d}x\,\text{d}y\,.$$
That is, the volume of $M$ in the $(x,y,z)$-coordinates is equal to
$$\frac{1}{\big|\det(T)\big|}\,\left(\frac{\pi\,h^2}{2\,\sqrt{\lambda\mu}}\right)\,.$$
Here is the kick.  Show that
$$\big(\det(T)\big)^2\,\lambda\mu=\det\left(\begin{bmatrix}a&b\\b&c\end{bmatrix}\right)\,.$$
That is, the required volume is simply
$$\frac{\pi\,h^2}{2\,\sqrt{ac-b^2}}\,.$$

 Hint: We have $T^\top\,\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix}\,T=\begin{bmatrix}a&b\\b&c\end{bmatrix}$.

