Proof of the volume formula

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

$${\displaystyle {\begin{aligned}V&=\int _{-r}^{r}(2x)^{2}\ \mathrm {d} z\\[2pt]&=4\cdot \int _{-r}^{r}x^{2}\ \mathrm {d} z\\[2pt]&=4\cdot \int _{-r}^{r}(r^{2}-z^{2})\ \mathrm {d} z\\[2pt]&={\frac {16}{3}}r^{3}.\end{aligned}}}$$

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 ${\displaystyle (a=2r,\ h=r,\ V={\tfrac {4}{3}}r^{3}),}$ the half bicylinder ${\displaystyle (V={\tfrac {8}{3}}r^{3})}$ and the surrounding squared cuboid ${\displaystyle (a=2r,\ h=r,\ V=4r^{3})}$ are 1 : 2 : 3:
  $${\displaystyle {\begin{array}{ccccc}{\frac {4}{3}}r^{3}&:&{\frac {8}{3}}r^{3}&:&4r^{3}\\[2pt]1&:&2&:&3\end{array}}}$$

  

Using Multivariable Calculus

Consider the equations of the cylinders:

$${\displaystyle {\begin{aligned}x^{2}+z^{2}&=r^{2}\\x^{2}+y^{2}&=r^{2}\end{aligned}}}$$

The volume will be given by:

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

With the limits of integration:

$${\displaystyle {\begin{array}{rcccl}-{\sqrt {r^{2}-x^{2}}}&\leqslant &z&\leqslant &{\sqrt {r^{2}-x^{2}}}\\[4pt]-{\sqrt {r^{2}-x^{2}}}&\leqslant &y&\leqslant &{\sqrt {r^{2}-x^{2}}}\\[4pt]-r&\leqslant &x&\leqslant &r\end{array}}}$$

Substituting, we have:

$${\displaystyle {\begin{aligned}V&=\int _{-r}^{r}\int _{-{\sqrt {r^{2}-x^{2}}}}^{\sqrt {r^{2}-x^{2}}}\int _{-{\sqrt {r^{2}-x^{2}}}}^{\sqrt {r^{2}-x^{2}}}\mathrm {d} z\,\mathrm {d} y\,\mathrm {d} x\\[2pt]&=8r^{3}-{\frac {8r^{3}}{3}}\\[2pt]&={\frac {16r^{3}}{3}}\end{aligned}}}$$

Proof of the area formula

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

cloister vault

$${\displaystyle B=\int _{0}^{\pi r}r\sin {\frac {\xi }{r}}\ \mathrm {d} \xi =r^{2}\cos {0}-r^{2}\cos {\pi }=2r^{2}}$$

and the total surface area is:

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

Alternate proof of the volume formula

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:

$${\displaystyle 8\times {\frac {1}{3}}r^{2}\times r={\frac {8}{3}}r^{3},}$$

and then we can calculate that the bicylinder volume is

$${\displaystyle (2r)^{3}-{\frac {8}{3}}r^{3}={\frac {16}{3}}r^{3}.}$$