development of a regular polygonal-based surface using

Development of a Regular Polygonal-basedSurface using GeoGebra

Shahid Saeed SiddiqiDepartment of Mathematics, University of Central Punjab, Lahore-Pakistan

Email: [email protected]

Abdul Rauf NizamiDepartment of Mathematics, University of Central Punjab, Lahore-Pakistan

Email: [email protected]

Abstract. If a solid cylinder with height equal to its radius, containinga hollow cone upside down, is sliced horizontally and the slices are s-tacked after converting them into discs with volumes equal to volumesof the slices, then the limiting surface becomes the hemisphere. In thepresent paper, the above idea has been extended by considering a reg-ular polygonal-based solid prism containing a corresponding pyramidupside down. The parametrization and the volume of the surface havebeen determined. The algorithms developed in the paper have been im-plemented using GeoGebra and the corresponding codes are also givenat the end.

Subject Classification (2010): Primary 00A05Keywords: Regular polygon, Elliptic arcs, Parametric representation

1. Introduction

To find the volume of a hemisphere, two different approaches are available,integral and geometric. In integral approach, the volume of hemisphere iscalculated using single and double integrals [1, 2]. Geometrically, the volumeof hemisphere is determined by showing the volume of hemisphere equal tothe volume of the cylinder outside the cone (as shown in Figure 1) [3]. Theidea was actually the cross-sections have the same area.

In this paper, the hemisphere has been developed using a different approach.The cylinder without the cone inside it has been sliced into m washers. Thesewashers, after converting to discs of volumes of corresponding washers, havebeen stacked as shown in Figure 2. The hemisphere is thus obtained as alimiting surface as m→∞.

2 Development of a Regular Polygonal-based Surface using GeoGebra

Figure 1

Figure 2

Initially, this idea has been extended by considering a square-based solidprism containing a corresponding pyramid upside down, as shown in Figure 3.Then the idea has been further extended to any regular polygonal-basedprism.

Figure 3. Main Solid

Development of a Regular Polygonal-based Surface using GeoGebra 3

2. Geometric Development of the Surface

To develop the structure shown in Figure 3, consider a pyramid with squarebase lying in the cuboid with the same square base and height equal to halfof the side of the square, as shown in Figure 4.

Figure 4. Pyramid in Cuboid

The cuboid with inserted pyramid has been sliced into n equal parts parallelto the xy-plane, as shown in Figure 5.

Figure 5. Sliced Cuboid

Moreover, each slice has been transformed to slab having the volume equal tothe volume of the original slice. One of the slices has been shown in Figure 6.

Figure 6. Slice with corresponding slab

Also, all the new slabs have been concentrically stacked with the same order,as shown in Figure 7.It may be noted that increasing the number of slices will smooth the surfacegenerated by these slabs. In the limiting case, as n → ∞, the generatedsurface will be perfectly smooth as shown in Figure 8.


Figure 7. Stacked Slabs

Figure 8. Smooth Surface

3. Parametrization of the Surface

Theorem 3.1. The curve of intersection of the surface and a vertical planethrough the origin is elliptic.

Proof. Without loss of generality, consider the curve obtained by intersectionof the surface and the vertical plane (passing through the diagonal of thesquare), as shown in Figure 9.

Figure 9. Intersection of the surface with the vertical plane


Since the surface has been obtained as n→∞ of the n stacked slabs, considerthe intersection of vertical plane passing through the diagonal of the squarewith the stacked slabs, along with the base of the (i+ 1)st slab and a pointQ(x, y, z),as shown in Figure 10.

Figure 10

To establish the relation among the coordinates of the point Q, the resultingintersection has been rotated about origin in xy-plane through the angle π

4 ,resulting the point Q(0,y,z), as shown in Figure 10. It can, thus, be writtenas

z =i

nR

y =

√(√

2R)2 − (i

n×√

2R)2 =√

2R

√1− (

i

n)2

y2 = 2R2(1− (i

n)2) = 2R2(1− (

z

R)2) = 2R2 − 2z2

y2 + 2z2 = 2R2

y2

(√

2R)2+

z2

(R)2= 1

Since the last equation represents the ellipse in the yz− plane, the arcs areelliptic. �

A square A1A2A3A4 has been considered in the xy-plane with side 2R. Thecenter of the square has been taken at the origin and its side A1A2 perpen-dicular to the x-axis, as shown in Figure 11.If (x, y, z) is any point on the quarter-circular arc CE, then the parametricequations of this arc are

x = R cos(t)y = 0z = R sin(t)

, t ∈[0,π

2

]Although the surface can be obtained by rotating the arc CED through anangle π, the suqare is divided into four triangles OA1A2, OA2A3, OA3A4 and


Figure 11

OA4A1. The purpose is to parametrize the surface. To generate the surfaceover the triangle OA1A2 (shown in Figure 12), a radial segment OP is sweptcounterclockwise from OA1 to OA2 such that the point P moves on thesegment A1A2. It follows that the length of OP depends on the angle r theOP makes with the positive x-axis.

Figure 12

To generate the surface over the triangle OA2A3, the radial segment OP willbe rotated from π

4 to 3π4 . However, the triangle OA2A3 is considered to be

rotated back to fit on the triangle OA1A2. The purpose is to use the cosineof the angle that OP makes with the positive x-axis. This actually gives theopportunity to control the parametrization of the whole surface in a single


formula. Similarly, the surfaces over the remaining triangles are obtained.

The parametrization of the square has been calculated as

x = Ra cos(r)

y = Ra sin(r)

, r ∈[−π4,

7π

4

]a = cos

(r − (i− 1)π2

), r ∈

[(−12 + (i− 1)

)π2 ,(−1

2 + i)π2

], i = 1, 2, 3, 4.

The above rotations can be performed using the following matrix transfor-mation. x

yz

=

cos(r) − sin(r) 0sin(r) cos(r) 0

0 0 1

Ra cos(t)

0R sin(t)

The required surface can, thus, be expressed in the following parametrization

x = Ra cos(r) cos(t)

y = Ra sin(r) cos(t)

z = R sin(t)

, r ∈[−π

4,

7π

4

], t ∈ [0,

π

2]

a = cos(r − (i− 1)π2

), r ∈

[(−12 + (i− 1)

)π2 ,(−1

2 + i)π2

], i = 1, 2, 3, 4.

4. Generalization

Extending the surface with square base to a surface with any regular n-gonalbase, the general result has been obtained as:

Theorem 4.1. The parametric representation of the surface with any regularn-gonal base is

x = Ra cos(r) cos(t)

y = Ra sin(r) cos(t)

z = R sin(t)

, r ∈[(−1

2

)2π

n,

(−1

2+ n

)2π

n

], t ∈

[0,π

2

],

a = cos(r − (i− 1) 2π

n

), r ∈

[(−12 + (i− 1)

)2πn ,(−1

2 + i)

2πn

], i = 1, 2, 3, . . . , n

Proof. The procedure is already explained in case of a squared base. However,in the generalized case quarter circular arc will be rotated on the completeboundary of the polygon, instead of semicircular arc rotated only on the halfboundary of the square. For instance, the surfaces for n = 5, 6 and 7 havebeen shown in Figure 13.

�


Figure 13

5. Volume of the Solid

First the volume of solid with square base is calculated.Since the side of the cuboid is 2R with height R, the volume of the cuboidis Vc = (4R2)R = 4R3 and the volume of the pyramid is Vp = ( 1

3 )(Vc). Thus

the volume of the corresponding solid is V = Vc − Vp = ( 23 )(Vc) = (8

3 )R3

To determine the volume of solid corresponding to regular n-sided polygonalbase, each n-sided regular polygon has been partitioned into n equilateraltriangles, each with height R; an equilateral triangle is depicted in Figure 14,for n = 7.

Figure 14

The area of each equilateral triangle is determined as R2 tan(πn ), which shows

that the area of n-sided polygon will be nR2 tan(πn ).The volume VPrism of the corresponding prism with height R will beVPrism = AnR = nR2 tan(πn )R = nR3 tan(πn ) which shows that the volume

of corresponding pyramid will be VPyramid = 13VPrism

Thus the volume of solid can be calculated asV = VPrism − VPyramid = 2

3VPrism = 23nR

3 tan(πn )


6. GeoGebra Code

To implement the parametrization of the regular polygonal-based surface, aGeoGebra code is given in this section.

6.1. Generating the Polygon

The regular n-sided polygon is generated with the centreat the origin and one of the sides is perpendicular to thepositive x-axis:

• n is a slider taking discrete values from 3 to 10. (Prin-cipally, one can take n to be more than 10.)• k is also a slider taken to be the length of perpen-

dicular from the centre to a side of the polygon.

The vertices of the polygon are generated:

• A1 = (k/ cos(−1/2∗2π/n) cos(−1/2∗2π/n), k/ cos(−1/2∗2π/n) sin(−1/2∗2π/n))• A2 = (k/ cos((−1/2 + 1) ∗ 2π/n) cos((−1/2 + 1) ∗ 2π/n), k/ cos((−1/2 +

1) ∗ 2π/n) sin((−1/2 + 1) ∗ 2π/n))• A3 = (k/ cos((−1/2 + 1) ∗ 2π/n) cos((−1/2 + 2) ∗ 2π/n), k/ cos((−1/2 +







1) ∗ 2π/n) sin((−1/2 + 8) ∗ 2π/n))• A10 = (k/ cos((−1/2+1)∗2π/n) cos((−1/2+9)∗2π/n), k/ cos((−1/2+

1) ∗ 2π/n) sin((−1/2 + 9) ∗ 2π/n))• A11 = (k/ cos((−1/2+1)∗2π/n) cos((−1/2+10)∗2π/n), k/ cos((−1/2+

1) ∗ 2π/n) sin((−1/2 + 10) ∗ 2π/n))

Generating the sides of polygon :

• g = Segment(A1, A2)• h = Segment(A2, A3)• i = Segment(A3, A4)• j = Segment(A4, A5)• l = Segment(A5, A6)• m = Segment(A6, A7)• p = Segment(A7, A8)• q = Segment(A8, A9)


• r = Segment(A9, A10)• s = Segment(A10, A11)

6.2. Rotating Elliptic Arcs

r1 and t are taken to be the sliders with ranges a1 ≤ r1 ≤a11 and 0 ≤ t ≤ π

2 .

• a1 = −1/2 ∗ 2π/n• a2 = (−1/2 + 1) ∗ 2π/n• a3 = (−1/2 + 2) ∗ 2π/n• a4 = (−1/2 + 3) ∗ 2π/n• a5 = (−1/2 + 4) ∗ 2π/n• a6 = (−1/2 + 5) ∗ 2π/n• a7 = (−1/2 + 6) ∗ 2π/n• a8 = (−1/2 + 7) ∗ 2π/n• a9 = (−1/2 + 8) ∗ 2π/n• a10 = (−1/2 + 9) ∗ 2π/n• a11 = (−1/2 + 10) ∗ 2π/n

• d = If(a1 ≤ r1 ≤ a2, k/ cos(r1), a2 ≤ r1 ≤ a3, k/ cos(r1 − 2π/n), a3 ≤r1 ≤ a4, k/ cos(r1− 2 ∗ 2π/n), a4 ≤ r1 ≤ a5, k/ cos(r1− 3 ∗ 2π/n), a5 ≤r1 ≤ a6, k/ cos(r1− 4 ∗ 2π/n), a6 ≤ r1 ≤ a7, k/ cos(r1− 5 ∗ 2π/n), a7 ≤r1 ≤ a8, k/ cos(r1− 6 ∗ 2π/n), a8 ≤ r1 ≤ a9, k/ cos(r1− 7 ∗ 2π/n), a9 ≤r1 ≤ a10, k/ cos(r1− 8 ∗ 2π/n), a10 ≤ r1 ≤ a11, k/ cos(r1− 9 ∗ 2π/n))

Moving the point P along the boundary of the polygon:

• P = (d cos(r1), d sin(r1))

The starting point of the elliptic arc in 3D:

• Pn = (d cos(r1), 0, k sin(t))

Matrix of rotation of elliptic arc:

• M1 = {{cos(r1),− sin(r1), 0}, {sin(r1), cos(r1), 0}, {0, 0, 1}}• Q = (M1Pn)• ellipticArc = Locus(Q , t)

6.3. Development of the Surface

• Surface(If(a1 ≤ r1 ≤ a2, k/ cos(r1), a2 ≤ r1 ≤ a3, k/ cos(r1−2π/n), a3 ≤r1 ≤ a4, k/ cos(r1− 2 ∗ 2π/n), a4 ≤ r1 ≤ a5, k/ cos(r1− 3 ∗ 2π/n), a5 ≤r1 ≤ a6, k/ cos(r1− 4 ∗ 2π/n), a6 ≤ r1 ≤ a7, k/ cos(r1− 5 ∗ 2π/n), a7 ≤r1 ≤ a8, k/ cos(r1− 6 ∗ 2π/n), a8 ≤ r1 ≤ a9, k/ cos(r1− 7 ∗ 2π/n), a9 ≤r1 ≤ a10, k/ cos(r1−8∗2π/n), a10 ≤ r1 ≤ a11, k/ cos(r1−9∗2π/n)) cos(t) cos(r1),If(a1 ≤ r1 ≤ a2, k/ cos(r1), a2 ≤ r1 ≤ a3, k/ cos(r1− 2π/n), a3 ≤ r1 ≤a4, k/ cos(r1− 2 ∗ 2π/n), a4 ≤ r1 ≤ a5, k/ cos(r1− 3 ∗ 2π/n), a5 ≤ r1 ≤a6, k/ cos(r1− 4 ∗ 2π/n), a6 ≤ r1 ≤ a7, k/ cos(r1− 5 ∗ 2π/n), a7 ≤ r1 ≤a8, k/ cos(r1− 6 ∗ 2π/n), a8 ≤ r1 ≤ a9, k/ cos(r1− 7 ∗ 2π/n), a9 ≤ r1 ≤a10, k/ cos(r1−8∗2π/n), a10 ≤ r1 ≤ a11, k/ cos(r1−9∗2π/n)) cos(t) sin(r1),k sin(t), t, 0, pi/2, r1, a1, a11)


References

[1] H. Anton, IRL C. Bivens and S.L. Davis, Claculus, 10th Edition, John Wileyand Sons Inc., USA, 2012.

[2] G. B. Thomas, M. D. Weir and J. R. Hass, Claculus, 13th Edition, Pearson,2014.

[3] D. Roberts, Spheres and Hemispheres, http-s://mathbitsnotebook.com/Geometry/3DShapes/3DSpheres.html

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