Steinmetz solid#Tricylinder

File:Steinmetz-cc.svg

File:Steinmetz-cc2-ag.svg

A bicylinder generated by two cylinders with radius {{mvar|r}} has the volume

$$V\; =\; \backslash frac\{16\}\{3\}\; r^3,$$

and the surface area{{mathworld | urlname = SteinmetzSolid | title = Steinmetz Solid}}{{cite journal

| author = Moore, M.

| title = Symmetrical intersections of right circular cylinders

| jstor = 3615957

| journal = The Mathematical Gazette

| volume = 58

| issue = 405

| pages = 181–185

| year = 1974

| doi = 10.2307/3615957}}

$$A\; =\; 16\; r^2.$$

The upper half of a bicylinder is the square case of a domical vault, a dome-shaped solid based on any convex polygon whose cross-sections are similar copies of the polygon, and analogous formulas calculating the volume and surface area of a domical vault as a rational multiple of the volume and surface area of its enclosing prism hold more generally.{{cite journal

|doi = 10.2307/27641977

|author1 = Apostol, Tom M.

|author2 = Mnatsakanian, Mamikon A.

|title = Solids circumscribing spheres

|journal = American Mathematical Monthly

|volume = 113

|year = 2006

|issue = 6

|pages = 521–540

|url = http://www.mamikon.com/USArticles/CircumSolids.pdf

|mr = 2231137

|jstor = 27641977

|access-date = 2007-03-25

|archive-url = https://web.archive.org/web/20120207004921/http://www.mamikon.com/USArticles/CircumSolids.pdf

|archive-date = 2012-02-07

|url-status = dead

}} In China, the bicylinder is known as móu hé fāng gài (牟合方蓋), literally "two square umbrella"; it was described by the third-century mathematician Liu Hui.{{cite book |last1=Wang |first1=Jianpang |last2=Fan |first2=Lianghuo |last3=Xu |first3=Binyan |title=School Mathematics Textbooks In China: Comparative Studies And Beyond |date=2021 |publisher=World Scientific |page=476}}

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

$$\backslash begin\{align\}$$

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{align}

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 {{math|1 : 2 : 3}}. For one half of a bicylinder a similar statement is true:

• The relations of the volumes of the inscribed square pyramid $(a=2r,\backslash \; h=r,\backslash \; V=\backslash tfrac\{4\}\{3\}r^3),$ the half bicylinder $(V=\backslash tfrac\{8\}\{3\}\; r^3)$ and the surrounding squared cuboid $(a=\; 2r,\backslash \; h=r,\backslash \; V=4r^3)$ are {{math|1 : 2 : 3}}:$$\backslash begin\{array\}\{ccccc\}$$

\frac{4}{3}r^3 &:& \frac{8}{3}r^3 &:& 4r^3 \\[2pt]

1 &:& 2 &:& 3

\end{array}

Consider the equations of the cylinders:

$$\backslash begin\{align\}$$

x^2+z^2 &= r^2 \\

x^2+y^2 &= r^2

\end{align}

The volume will be given by:

$$V\; =\; \backslash iiint\_V\; \backslash mathrm\{d\}z\backslash ,\backslash mathrm\{d\}y\backslash ,\backslash mathrm\{d\}x$$

With the limits of integration:

$$\backslash begin\{array\}\{rcccl\}$$

-\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:

$$\backslash begin\{align\}$$

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{align}

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

File:Klostergew.svg

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

and the total surface area is:

$$A\; =\; 8B=16r^2.$$

To derive the volume of a bicylinder (white), one can enclose it within a cube (red). When a plane, parallel to the axes of the cylinders, intersects the bicylinder, it forms a square. This plane's intersection with the cube results in a larger square. The area difference between these two squares corresponds to four smaller squares (blue). As the plane traverses through the solids, these blue squares form square pyramids with isosceles faces at the cube's corners. The apexes of these pyramids are located at the midpoints of the cube's four edges. Moving the plane through the entire bicylinder results in a total of eight pyramids.

File:Sphere volume derivation using bicylinder.jpg|Zu Chongzhi's method (similar to Cavalieri's principle) for calculating a sphere's volume includes calculating the volume of a bicylinder.

File:Bicylinder and cube sections related by pyramids.png|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:

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

and then we can calculate that the bicylinder volume is

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

[[File:Steinmetz-ccc.svg|450px|thumb|Generating the surface of a tricylinder:

At first two cylinders (red, blue) are cut. The so generated bicylinder is cut by the third (green) cylinder.]]

The intersection of three cylinders with perpendicularly intersecting axes generates a surface of a solid with vertices where 3 edges meet and vertices where 4 edges meet. The set of vertices can be considered as the edges of a rhombic dodecahedron. The key for the determination of volume and surface area is the observation that the tricylinder can be resampled by the cube with the vertices where 3 edges meet (s. diagram) and 6 curved pyramids (the triangles are parts of cylinder surfaces). Similar considerations can determine the volume and the surface area of the curved triangles as it is done for the bicylinder above.

The volume of a tricylinder is

$$V\; =\; 8(2\; -\; \backslash sqrt\{2\})\; r^3$$

and the surface area is

$$A\; =\; 24(2\; -\; \backslash sqrt\{2\})\; r^2.$$

With four cylinders, with axes connecting the vertices of a tetrahedron to the corresponding points on the other side of the solid, the volume is

$$V\_4\; =\; 12\; \backslash left(\; 2\backslash sqrt\{2\}\; -\; \backslash sqrt\{6\}\; \backslash right)\; r^3\; \backslash ,$$

With six cylinders, with axes parallel to the diagonals of the faces of a cube, the volume is:

$$V\_6\; =\; \backslash frac\{16\}\{3\}\; \backslash left(\; 3\; +\; 2\backslash sqrt\{3\}\; -\; 4\backslash sqrt\{2\}\; \backslash right)\; r^3\; \backslash ,$$

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{{Reflist}}

• [https://archive.today/20130124181306/http://sketchup.google.com/3dwarehouse/details?mid=b170ed71b798d500a6b083aba0e2034a A 3D model of Steinmetz solid in Google 3D Warehouse]

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Category:Euclidean solid geometry