Volume of $M:=\big\{(x,y,z)\in \mathbb{R}^3\,:\,ax^2+2bxy+cy^2\leq z \leq 3\big\}$

Question

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...

Answer 1

$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?

$\lambda_1\lambda_2 = ac-b^2$

Answer 2

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