Steinmetz solid

For deriving the volume formula it is convenient to use the common idea for calculating the volume of a sphere: collecting thin cylindric slices. In this case the thin slices are square cuboids (see diagram). This leads to

${\displaystyle V=\int _{-r}^{r}(2x)^{2}\mathrm {d} z=4\cdot \int _{-r}^{r}x^{2}\mathrm {d} z=4\cdot \int _{-r}^{r}(r^{2}-z^{2})\mathrm {d} z={\frac {16}{3}}r^{3}}$.

It is well known that the relations of the volumes of a right circular cone, one half of a sphere and a right circular cylinder with same radii and heights are 1 : 2 : 3. For one half of a bicylinder a similar statement is true:

• The relations of the volumes of the inscribed square pyramid (${\displaystyle a=2r,h=r,V={\frac {4}{3}}r^{3}}$), the half bicylinder (${\displaystyle V={\frac {8}{3}}r^{3}}$) and the surrounding squared cuboid (${\displaystyle a=2r,h=r,V=4r^{3}}$) are 1 : 2 : 3.

Consider the equations of the cylinders:

${\displaystyle x^{2}+z^{2}=r^{2}}$

${\displaystyle x^{2}+y^{2}=r^{2}}$

The volume will be given by:

${\displaystyle V=\iiint _{V}\mathrm {d} z\mathrm {d} y\mathrm {d} x}$

With the limits of integration:

${\displaystyle -{\sqrt {r^{2}-x^{2}}}\leqslant z\leqslant {\sqrt {r^{2}-x^{2}}}}$

${\displaystyle -{\sqrt {r^{2}-x^{2}}}\leqslant y\leqslant {\sqrt {r^{2}-x^{2}}}}$

${\displaystyle -r\leqslant x\leqslant r}$

Substituting, we have:

${\displaystyle V=\int _{-r}^{r}\int _{-{\sqrt {r^{2}-x^{2}}}}^{\sqrt {r^{2}-x^{2}}}\int _{-{\sqrt {r^{2}-x^{2}}}}^{\sqrt {r^{2}-x^{2}}}\mathrm {d} z\mathrm {d} y\mathrm {d} x=8r^{3}-{\frac {8r^{3}}{3}}={\frac {16r^{3}}{3}}}$

The surface area consists of two red and two blue cylindrical biangles. One red biangle is cut into halves by the y-z-plane and developed into the plane such that half circle (intersection with the y-z-plane) is developed onto the positive ${\displaystyle \xi }$-axis and the development of the biangle is bounded upwards by the sine arc ${\displaystyle \eta =r\sin \left({\frac {\xi }{r}}\right),\ 0\leq \xi \leq \pi r}$. Hence the area of this development is

cloister vault

${\displaystyle B=\int _{0}^{\pi r}r\sin \left({\frac {\xi }{r}}\right)\mathrm {d} \xi =2r^{2}}$

and the total surface area is:

${\displaystyle A=8\cdot B=16r^{2}}$.

Deriving the volume of a bicylinder (white) can be done by packing it in a cube (red). A plane (parallel with the cylinders' axes) intersecting the bicylinder forms a square and its intersection with the cube is a larger square. The difference between the areas of the two squares is the same as 4 small squares (blue). As the plane moves through the solids, these blue squares describe square pyramids with isosceles faces in the corners of the cube; the pyramids have their apexes at the midpoints of the four cube edges. Moving the plane through the whole bicylinder describes a total of 8 pyramids.

• 
  Zu Chongzhi's method (similar to Cavalieri's principle) for calculating a sphere's volume includes calculating the volume of a bicylinder.
• 
  Relationship of the area of a bicylinder section with a cube section

The volume of the cube (red) minus the volume of the eight pyramids (blue) is the volume of the bicylinder (white). The volume of the 8 pyramids is: ${\displaystyle \textstyle 8\times {\frac {1}{3}}r^{2}\times r={\frac {8}{3}}r^{3}}$, and then we can calculate that the bicylinder volume is ${\displaystyle \textstyle (2r)^{3}-{\frac {8}{3}}r^{3}={\frac {16}{3}}r^{3}}$