How do I compute the area of Steinmetz solid?

Solution 1:

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.

Solution 2:

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: