Development of a Surface having Regular Polygonal Base and Elliptic Arcs
DDevelopment of a Surface having RegularPolygonal Base and Elliptic Arcs
Shahid Saeed Siddiqi and Abdul Rauf Nizami
Abstract.
This paper aims to develop the mathematical representationof a surface generated by elliptical arcs joining the sides of a regularpolygon to a point lying vertically upward on the central axis of thepolygon. The volume of the corresponding solid has also been deter-mined.
Subject Classification (2010)
Keywords
Regular polygon; Elliptic arcs; Parametric representation.
1. Introduction
To find the volume of 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.
Figure 1.
In this paper, the hemisphere has been developed using a different approach.The cylinder without the cone inside it has been sliced into n washers. Thesewashers, after converting to discs of volumes of corresponding washers, havebeen stacked as shown in Figure 2. a r X i v : . [ m a t h . HO ] A ug Shahid S. Siddiqi and A. R. Nizami
Figure 2.
The hemisphere is thus obtained as a limiting surface as n → ∞ . The ideaof obtaining hemisphere has been extended to generate a limiting surfaceobtained by stacking the slices of cuboid, excluding the pyramid inside it,after converting the slices into the slabs of volumes of corresponding slices.Lastly, the idea has been extended to obtain the surface with any regularpolygonal base.To develop the surface having a regular polygonal base along with the surfacegenerated by the elliptic arcs joining the sides of the polygon to the pointlying vertically upward on the central axis of the polygon, consider the solidcorresponding to square base, as shown in Figure 3. Figure 3.
Main SolidInitially, the geometrical development and mathematical representation of thecase with square base has been discussed in Sections 2 and 3, respectively.The case, with the regular polygonal base, has been developed in Section 4 . evelopment of a Surface having Regular Polygonal Base and Elliptic Arcs
2. Geometrical 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 CuboidThe cuboid with inserted pyramid has been sliced into n equal parts parallelto the xy -plane, as shown in Figure 5. Figure 5.
Sliced CuboidMoreover, 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 slabAlso, all the new slabs have been concentrically stacked with the same order,as shown in Figure 7.
Shahid S. Siddiqi and A. R. Nizami
Figure 7.
Stacked SlabsIt 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 8.
Smooth Surface
3. Mathematical Development 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. evelopment of a Surface having Regular Polygonal Base and Elliptic Arcs Figure 9.
Intersection of the surface with the vertical planeSince 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 point Q ( 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 ,resulting the point Q(0,y,z), as shown in Figure 10. It can, thus, be writtenas z = in Ry = (cid:114) ( √ R ) − ( in × √ R ) = √ R (cid:114) − ( in ) y = 2 R (1 − ( in ) ) = 2 R (1 − ( zR ) ) = 2 R − z y + z = 2 R y ( √ R ) + z ( R ) = 1Since the last equation represents the ellipse in the yz − plane, the arcs areelliptic. (cid:3) Shahid S. Siddiqi and A. R. Nizami
4. Development of the Surface
A square A A A A has been considered in the xy -plane with side 2 R . Thecenter of the square has been taken at the origin and its side A A perpen-dicular to the x -axis, as shown in Figure 11. 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 = 0 z = R sin( t ) , t ∈ (cid:104) , π (cid:105) Although the surface can be obtained by rotating the arc
CED through anangle π , the suqare is divided into four triangles OA A , OA A , OA A and OA A . The purpose is to parametrize the surface. To generate the surfaceover the triangle OA A (shown in Figure 12), a radial segment OP is sweptcounterclockwise from OA to OA such that the point P moves on thesegment A A . It follows that the length of OP depends on the angle r the OP makes with the positive x -axis.To generate the surface over the triangle OA A , the radial segment OP willbe rotated from π to π . However, the triangle OA A is considered to berotated back to fit on the triangle OA A . 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 singleformula. Similarly, the surfaces over the remaining triangles are obtained. evelopment of a Surface having Regular Polygonal Base and Elliptic Arcs Figure 12.
The parametrization of the square has been calculated as x = Ra cos( r ) y = Ra sin( r ) , r ∈ (cid:20) − π , π (cid:21) a = cos (cid:0) r − ( i − π (cid:1) , r ∈ (cid:2)(cid:0) − + ( i − (cid:1) π , (cid:0) − + i (cid:1) π (cid:3) , i = 1 , , , xyz = cos( r ) − sin( r ) 0sin( r ) cos( r ) 00 0 1 Ra cos( t )0 R 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 ∈ (cid:104) − π , π (cid:105) , t ∈ [0 , π a = cos (cid:0) r − ( i − π (cid:1) , r ∈ (cid:2)(cid:0) − + ( i − (cid:1) π , (cid:0) − + i (cid:1) π (cid:3) , i = 1 , , ,
5. Generalization
Extending the surface with square base to a surface with any regular n -gonalbase, the general result has been obtained as: Shahid S. Siddiqi and A. R. Nizami
Theorem 5.1.
The parametric representation of the surface with any regular n -gonal base is x = Ra cos( r ) cos( t ) y = Ra sin( r ) cos( t ) z = R sin( t ) , r ∈ (cid:20)(cid:18) − (cid:19) πn , (cid:18) −
12 + n (cid:19) πn (cid:21) , t ∈ (cid:104) , π (cid:105) ,a = cos (cid:0) r − ( i − πn (cid:1) , r ∈ (cid:2)(cid:0) − + ( i − (cid:1) πn , (cid:0) − + i (cid:1) πn (cid:3) , i = 1 , , , . . . , 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 , Figure 13. (cid:3)
6. Volume of the Solid
First the volume of solid with square base is calculated.Since the side of the cuboid is 2 R with height R , the volume of the cuboidis V c = (4 R ) R = 4 R and the volume of the pyramid is V p = ( )( V c ). Thusthe volume of the corresponding solid is V = V c − V p = ( )( V c ) = ( ) R 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.The area of each equilateral triangle is determined as R tan( πn ), which showsthat the area of n -sided polygon will be nR tan( πn ).The volume V P rism of the corresponding prism with height R will be V P rism = A n R = nR tan( πn ) R = nR tan( πn ) which shows that the volume evelopment of a Surface having Regular Polygonal Base and Elliptic Arcs Figure 14. of corresponding pyramid will be V P yramid = V P rism
Thus the volume of solid can be calculated as V = V P rism − V P yramid = V P rism = nR tan( πn )
7. Conclusion
In this article a surface has been geometrically developed and its parametriza-tion has also been given. A solid prism with a regular polygon base and hightequal to the radius of the in-circle of the base, contains a solid cone with baseas top of the prism and vertex at the center of the prism base. The cone hasbeen then pulled out of the cylinder, and the cylinder has been cut into n horizontal slices (washers). Corresponding to each washer, a disc with volumeequal to the volume of the washer has been considered. Then the discs havebeen stacked in the sequence of the corresponding washers. By indefinitelyincreasing the number of washers, the solid with smooth surface has beenobtained. Secondly, the curve of intersection of the surface and any verticalplane through the origin has been proved elliptic. Finally, parameterizationof both the regular polygonal base and the developed surface, have been de-termined. Moreover, following this technique the volume of the solid has beendetermined. 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. Shahid S. Siddiqi and A. R. Nizami [3] D. Roberts, Spheres and Hemispheres, https://mathbitsnotebook.com/Geometry/3DShapes/3DSpheres.htmlShahid Saeed SiddiqiFaculty of Information Technology, University of Central Punjab, Lahore-Pakistane-mail: [email protected]