Euclidean volume of the unit ball of matrices under the matrix norm The matrix norm for an $n$-by-$n$ matrix $A$ is defined as $$|A| := \max_{|x|=1} |Ax|$$ where the vector norm is the usual Euclidean one. This is also called the induced (matrix) norm, the operator norm, or the spectral norm. The unit ball of matrices under this norm can be considered as a subset of $\Bbb R^{n^2}$. What is the Euclidean volume of this set? 
I'd be interested in the answer even in just the $2$-by-$2$ case.

I worked out the answer for the 2 by 2 case as well.
First, when dealing with 2 by 2 matrices in general, a convenient variable change is:
$$a\rightarrow\frac{w+x}{\sqrt{2}},d\rightarrow\frac{w-x}{\sqrt{2}},c\rightarrow\frac{y-z}{\sqrt{2}},b\rightarrow\frac{y+z}{\sqrt{2}}.$$
Then $a^2+b^2+c^2+d^2 = w^2+x^2+y^2+z^2$.  And the determinant $(ad-bc) = \frac{1}{2}(x^2+y^2-w^2-z^2)$.
(Aside: this set of coordinates lets you see for instance that the set of rank 1 matrices in the space of 2D matrices realized as $\mathbb{R}^4$ is a cone over the Clifford torus, since $x^2+y^2 = w^2+z^2$ on a sphere $x^2+y^2+w^2+z^2=r^2$ implies $x^2+y^2 = r^2/2$ and $w^2+z^2 = r^2/2$, which are scaled equations for a flat torus)
Let $r_1^2 = x^2+y^2, r_2^2 = w^2+z^2$.  (These are radial coordinates of a coordinate system consisting of two orthogonal 2D cylindrical coordinate systems).  Then the norm squared is:
$$\frac{1}{2}\left(r_1^2+r_2^2 + \sqrt{ (r_1^2+r_2^2)^2 - (r_1^2-r_2^2)^2 }\right)$$
When this is less than one, this corresponds to the region plotted below:

Note that each point in the $r_1,r_2$ picture corresponds to a different "torus", $x^2+y^2=r_1^2, w^2+z^2=r_2^2$.
We can now integrate over the shaded in region, $\int_{region} dw dx dy dz$.
This 4-D integral can be reduced to 2D using $r_1$ and $r_2$, since $dx dy = 2\pi r_1 dr_1, dw dz = 2\pi r_2 dr_2$:
$$(4\pi^2) \int_{region} dr_1 dr_2 r_1 r_2. $$
Now, note that we can rewrite $r_2$ in terms of $r_1$.  In particular, after some manipulation of our norm, the shaded-in region is defined by $r_2^2 \leq 2-2\sqrt{2}r_1+r_1^2=(\sqrt{2}-r_1)^2$.  Hence $r_2\leq \sqrt{2}-r_1$, and we can evaluate the $r_2$ integral:
$$4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 \int_{r_2=0}^{\sqrt{2}-r_1} r_2 dr_2 \\
    = 4\pi^2 \int_{r_1=0}^\sqrt{2} dr_1 r_1 (\sqrt{2}-r_1)^2/2\\
    = (4\pi^2) (1/6).$$
This yields $2\pi^2/3$, as Armin found.
