How do I compute the area of Steinmetz solid? I want to compute the area of  the boundary of the solid $x^2+z^2\leq 1$ and $y^2+z^2\leq 1$ to compute his surface afterwards. But I somehow don't see how. First I thought about polar coordinates, then cylinder-coordinates and then splitting the solid up, but somehow nothing worked.
Could someone give me a hint? It would be really nice.
Thank you for your help
 A: I tried to parametrize one smooth surface out of a total of four pieces.
$$r(u,v) = (\cos u, v, \sin u)$$
where $ |u| \le \pi/2 $ and $ |v| \le \cos u $. The latter is derived from $ v^2 \le 1 - \sin^2 u$.
The derivatives of the surface are
$$r_u = (-\sin u, 0, \cos u)\\
r_v = (0, 1, 0)$$
Hence, the cross product of them is $ r_u \times r_v = -(\cos u, 0, \sin u) $ whose Euclidean norm is $ |r_u \times r_v| = 1 $.
By definition of surface area $S$, Wikipedia
\begin{array}{cl}
S &= \int_{-\pi/2}^{\pi/2}\int_{-\cos u}^{ \cos u} |r_u \times r_v| dv du\\
&=4\int_{0}^{\pi/2}\int_{0}^{ \cos u} 1 dv du\\
&=4\int_{0}^{\pi/2} \cos u du\\
&=4
\end{array}
Finally, the sum of all pieces of the same area is $4 \times 4 = 16$.
More comments: From the condition $x^2 + z^2 = 1$, we can easily get the main structure of $r(u,v)$. Now, what matters here is the domain of the function $r$, where $y^2 + z^2 \le 1$ comes into play. Substituting $y = v$ and $z = \sin u$ gives the valid domain of the function.
A: I would divide this surface into $4$ pieces, each parameterised by a disk:
\begin{eqnarray*}
&(1)&\qquad y^2+z^2\leq 1,\qquad x=\sqrt{1-z^2}\\
&(2)&\qquad y^2+z^2\leq 1,\qquad x=-\sqrt{1-z^2}\\
&(3)&\qquad x^2+z^2\leq 1,\qquad y=\sqrt{1-z^2}\\
&(4)&\qquad x^2+z^2\leq 1,\qquad y=-\sqrt{1-z^2}\\
\end{eqnarray*}
You can visualise this as $(1)$ being the right of the solid, $(2)$ the left, $(3)$ the back and $(4)$ the front.
Clearly the $4$ pieces cover the entire boundary, as for a point to be on the boundary of the solid, at least one of the equalities $y^2+z^2\leq 1$ or $x^2+z^2\leq 1$ must be attained.
Hermis14's parameterisation:

