3d volume of the set of $(2\times2)$-matrices of rank $\leq1$ and norm $\leq1$ The problem:Let $M$ be the set of all $2\times2$ matrices $A$ with rank less than or equal to 1 and $\left|A\right|\le1$,where $|A|^2$ denotes the sum of squares of the entries of $A$.Find the three-dimensional volume of $M$.
My thought to this question is as follows:
The condition of $A$ with rank $\le1$ means that $\begin{pmatrix} a_{11} &a_{12} \\ a_{21} &a_{22} \end{pmatrix}$  can be represented as $\begin{pmatrix} t &pt \\ kt &pkt \end{pmatrix}$,$where \, t, p, k \in \mathbb{R}$.
Also $\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}\le1$ since $|A|\le1$.
Let $f\colon \mathbb{R^3} \to \mathbb{R}$ be defined by the equation $f(t,p,k)=\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}$,where $t=a_{11}, pt=a_{12},kt=a_{21},pkt=a_{22}$.
In the above, let's define $g\colon \mathbb{R^3} \to \mathbb{R^4}$ by $g(t,p,k)=(a_{11},a_{12},a_{21},a_{22})=(t,pt,kt,pkt)$.
Then find the Jacobian matrix $D$ of $g$.
The ineqaulity $f(t,p,k)=\sqrt{a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2}\le1$ define a region $S$ in $\mathbb{R^3}$,then maybe we can parametrize it .Then find $\int_{S}\sqrt{\det{D^TD}}$ to find the volume of $\mathbb{R^3}$ in           $\mathbb{R^4}$.
Am I correct?Is there a better approach?
 A: Your plan is correct, but there is great danger that you will run into computational difficulties. Apart from this we have to make sure that we have an essentially one-one parametrization of $M$, or else we might obtain the double of the desired volume.
I propose to use the following parametrization of $M$:
$${\bf f}:\quad (t,u,v)\mapsto\quad  t\  \left[\matrix{\cos u\cr\sin u}\right]\bigl[\matrix{\cos v&\sin v\cr}\bigr]\ ,$$
with parameter domain $[0,1]\times[0,2\pi]\times[0,2\pi]$.
This leads to the matrix elements
$$a_{11}=t\cos u\cos v,\quad a_{12}=t\cos u\sin v,\quad a_{21}=t\sin u\cos v,\quad a_{22}=t\sin u\sin v\ ,$$
hence $a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2=t^2$. It is easy to check that ${\bf f}(t,u,v)={\bf f}(t',u',v')\>$ iff $\>t=t'$ and $(u,v)=(u',v')$, or $t=t'$ and $(u,v)=(u'\pm\pi,v'\pm\pi)$. This means that ${\bf f}$ produces a double covering of $M$.
We need the matrix
$$D:=\bigl[d{\bf f}(t,u,v)\bigr]=\left[\matrix{
{\partial a_{11}\over\partial t}&
{\partial a_{11}\over\partial u}&
{\partial a_{11}\over\partial v}\cr
{\partial a_{12}\over\partial t}&
{\partial a_{12}\over\partial u}&
{\partial a_{12}\over\partial v}\cr
{\partial a_{21}\over\partial t}&
{\partial a_{21}\over\partial u}&
{\partial a_{21}\over\partial v}\cr
{\partial a_{22}\over\partial t}&
{\partial a_{22}\over\partial u}&
{\partial a_{22}\over\partial v}\cr}\right]=\ldots\quad.$$
The Gramian simplifies considerably: One computes
$$D^\top D=
\left[\matrix{1&0&0\cr 0&t^2&0\cr0&0&t^2\cr}\right]\ ,$$
so that
$\sqrt{{\rm det}\bigl(D^\top D\bigr)}=t^2$. In this way we obtain
$${\rm vol}(M)={1\over2}\int_{[0,1]\times[0,2\pi]\times[0,2\pi]}t^2\>{\rm d}(t,u,v)={2\pi^2\over3}\ .$$
