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Home , Cylinder, Surface (topology), Torus

GENERAL ¨ ARTICLE and

Keerti Vardhan Madahar

In this article we compute and of the , volume of a slice of a 3-ball. Besides this we discuss some interesting facts like `derivative of the area of a with respect to its is its perimeter, derivative of the vol- The author earned his PhD ume of the with respect to its radius is its degree in and derivative of thevolumeof the (), in 2000, from torus with respect to the radius of its meridian Panjab University, circle is its surface area'. Chandigarh and since then he has been teaching 1. Introduction analysis, algebra, calculus and discrete mathematics at In school one learns how to compute area,volume,sur- DES-MDRC, Panjab face area and perimeter of various like sphere, University.Hisareasof interest are combinatorial , andcircle. But anequally important geo- topology,algebraic metric object `torus' { a like a scooter tube or a topology, differential doughnut { is notdiscussed in school geometry. This is topology,braidgroups, perhaps due to the non availability of this shape atthe theory and graphtheory. time when (287 BC{212 BC)was computing He also likes to play with mathematica software. and surface of sphere, cone and cylinder. Hadthisshape been available atthat time, Archimedes would have computed its volume and surface area and consequently today's school children might have been using these formulas along with the formulasof sphere, cone and cylinder. These days torus-likeshapes are fre- quently seen atmany places, so one would be interested to learn about their volumes and surface areas. 2. Derivatives of Areas and Volumes of Some Geometric Shapes Keywords The area, A(r)=¼r2,of a circle of radius r is a poly- Sphere–Cylinder Theorem, vol- umeandsurfaceareaofthe nomial of a single variable r and its derivative torus, volume and surface area with respect to r is 2¼r, which is its perimeter. The 4 3 of a slice of a solid sphere. volume, V (r)=3 ¼r ,of a sphere of radius r is a also a

654 RESONANCE ¨July 2013 GENERAL ¨ ARTICLE polynomial function of a single variable r and itsderiva- tive 4¼r2 is its surface area.Thevolume,V (r; s) = ¼r2s, of a cylinder of radius r and height s ispolynomial function of two variables. Its derivative with respect to r is its curved surface area and its derivative with respect to s is its or lid area.Finally the volume, V (r; R)=2¼2r2R,of a torus is a polynomial function of two variables and its derivative with respecttor is its surface area.Thecommonreason for all these facts Figure 1. is given below by considering the example of a circle (Figure 1). By de¯nition of the derivative, wehave dA A(r + h) A(r) =lim ¡ dr h 0 h ! (area of an of width h) =lim : h 0 h ! As h approaches zero, the outer of the annulus approaches its inner boundary. And for small values of h,thearea of the annulus is equaltotheproduct of the length of its inner boundary (which will benearly the same as the length of its outer boundary) and itswidth. To see this, divide the annulus into a large number (say n)of rectangularparts, of width h,bydissecting it with n radii of the outer boundary. The area of each rectan- gularpart is the product of its length (which lies along the inner circle) and height h.Sothesumof the ar- easof all these rectanglesisthesumof allthe lengths multiplied by their common height h.Thesumof all these lengths is the perimeter of the inner circle of the annulus, so the last limit becomes (Perimeter of the inner circle h) lim £ h 0 h ! = Perimeter of the inner circle: Exactly in the same manner, we canshow that the derivative of the volume of a sphere of radius r with respect to r is its surface area and derivative of the vol- ume of a cylinder with respect to its radius is itscurved surface area etc.

RESONANCE ¨ July 2013 655 GENERAL ¨ ARTICLE

Remark.Fromthisdiscussion one mayconcludethat the derivative of the volumeof any geometricalshape with respect to a certain variable is its surface area and derivative of the area is its perimeter. But it is not always true asthesurface area of an is not the derivative of its volume and the perimeter of an is not the derivative of its area. 3. Torus We now compute volume andsurface area of the torus and observe that the derivative of its volume is its sur- face area.We¯rstde¯ne the torus precisely. When a circle in the xz-, say(x R)2 + z2 = r2, revolves about the z-axis and completes¡ one revolution, then the surface tracedinR3 is called a torus (a portion of the torus is shown in Figure 2). Thecirclesdrawn on this torus, which are parallel to the z-axis are called meridians and parallel to the xy-plane are called longitudes of the torus. In order to get thevolume and surface area of the torus we cut the torus into slices along the meridian circles. If the number of slices is su±ciently large then each slice

z

C

D r P R B x

R - A

R +

y Figure 2.

656 RESONANCE ¨July 2013 GENERAL ¨ ARTICLE will look like a right circular cylinder. We know how Torus is a surface of to compute the volume and surface area of the cylinder. revolution and for By adding the volumes and curved surface areas of these such volume slices we will get volume and surface area respectively and surface area can of the torus. be computed by Volume of the Torus. Suppose we divide the torus using a well-known into k each of length l;thenthevolume of Theorem of Pappus, each cylinder will be ¼r2l.Thevolumeof the torus which states that the will be the sum of the volumes of these k cylinders, i.e., surface area (volume) ¼r2lk. The product lk is the sum of the lengths of these of a surface of k cylinders, so it must be equalto2¼R,i.e., the length of revolution, generated the middle longitudinalcircleof the torus. Substituting by rotating a plane this value of lk we get the volume of the torus to be C (plane equalto2¼2r2R: surface F) about an external , is equal Surface Area of the Torus. Again considerthedi- to the product of the vision of the torus into k cylinders each of length l.The curved surface area of each such cylinder will be 2¼rl; length of the curve C consequently the surface area of the torus will be 2¼rlk, (area of F) and the i.e., sum of the curved surface areasof these k cylin- traveled by ders. The product lk is the sum of the lengths of these k its during one cylinders, so it must be equalto2¼R, i.e., the length of revolution. For the middle longitudinalcircleof the torus. Substituting example:- this value of lk we get the surface area of the torus as Surface Area of the 4¼2rR. torus = (2S r).(2S R). Remark. Although we have obtained correct values of the volume and surface area of the torus, weverify it by using calculus because in the proof givenhere, some approximations have been made thatmay lead towrong conclusions. 4. Computations Using Calculus Since a torus is a and for such sur- faces there are well-de¯ned techniques in calculus, to compute volume and surface area,weuse these tech- niques as follows.

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Volume of the Torus. Notice thatvolumeof the torus is the di®erenceof the volumes of the surfaces of revolution obtainedbyrevolving the arcs ABC and ADC (see Figure 2),of the circle (x R)2 + z2 = r2, about the z-axis.Sinceanyhorizont¡allineinthexz- plane, for r

r 2 V = ¼ R + pr2 z2 dz r ¢ ¡ Z¡ ³ r ´ 2 ¼ R pr2 z2 dz ¡ r ¢ ¡ ¡ Z¡ r ³2 ´ = ¼ R + pr2 z2 ¢ r ¡ ¡ Z¡ ½ ¾ ³ ´ 2 R pr2 z2 dz ¡ ¡ ½³ ´ ¾ r =4¼R pr2 z2 dz ¢ r ¡ Z¡ =2¼2r2R :

Surface Area of the Torus. This canbeobtained by adding the surface areas of the surfaces of revolution obtained by revolving the arcs ABC and ADC about the z-axis (see Figure 2).Weknow thatif a curve x = g(z) is rotated about the z-axis, then the surface area of the surface of revolution thus obtained, can be computed from the following formula:

b 2 2¼ g(z) 1+ g0(z) dz : a ¢ ¢ f g Z h p i By applying this formula on the ABC and ADC we get

658 RESONANCE ¨July 2013 GENERAL ¨ ARTICLE

Surface area r z2 p 2 2 = 2¼ R + r z 1+ 2 2 + r ¢ " ¡ ¢ r z # Z¡ n o r ¡ z2 p 2 2 R r z 1+ 2 2 dz " ¡ ¡ ¢ r z # n o r ¡ r r = 2¼ 2R dz 2 2 r ¢ pr z ¢ Z¡ ¡ ¸ r 1 =4¼rR dz 2 2 ¢ r pr z Z¡ ¡ =4¼2rR:

5. A Deduction from the Archimedes' Sphere{ Cylinder Theorem Archimedes showed thatthesurface area of a sphere of radius r is equal to the curved surface area of a circum- scribing cylinder of radius r and height 2r.Infact he proved the following more generalresult:`If a cylinder { circumscribing a sphere { is cut by a plane parallel to its base then the slice of the cylinder and the sphere un- derneath this plane will have same surface areas'. The result is known as Archimedes' Sphere{Cylinder The- orem. Archimedes considered this result ashismost beautiful and important achievement. He desired that his favourite picture of sphere and cylinder be carved on his grave after his death and his desire was ful¯lled.

By using Archimedes' result we shall compute the vol- Figure 3. ume of the slice of the sphere cut out by a plane at a height h from the ground asshowninFigure 3. Assume that the boundary circle of the slice of the sphere of radius r,hasradius a and also assume that m trian- gles (m is su±ciently large) have been drawn on this slice. If we join the vertices of these triangles with the center of the sphere, we get m , each having a tri- angularbase and height r.Noticethatthesumof the

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Figure 4.

volumes of these m cones minus the volume of a cone with base radius a and height r h gives the volume of the required slice (see Figure 4).¡ Required volume =

m 1 1 ( area of ith triangle r) ¼a2(r h) 3 £ ¡ 3 ¡ i=1 X 1 = (surface area of the sphere's slice) r ¼a2(r h) 3f £ ¡ ¡ g 1 = (surface area of the cylinder's slice) r ¼a2(r h) 3f £ ¡ ¡ g (by using Sphere{Cylinder Theorem) 1 = (2¼rh) r ¼a2(r h) 3f £ ¡ ¡ g 1 = 2¼r2h ¼a2(r h) : (1) 3f ¡ ¡ g From Figure 4, it follows that a2 +(r h)2 = r2 ¡ a2 = r2 (r h)2 = h(2r h): (2) ) ¡ ¡ ¡ From (1) and (2) we get

660 RESONANCE ¨July 2013 GENERAL ¨ ARTICLE

Volume of the sphere's slice 1 = 2¼r2h ¼h(2r h)(r h) 3f ¡ ¡ ¡ g Address for Correspondence 1 Keerti Vardhan Madahar = ¼ 2r2h +2rh2 h3 2r2h + rh2) 3 f ¡ ¡ g Assistant Professor in Mathematics (DES) 1 2 = ¼h (3r h)(3)Panjab University 3 ¡ Chandigarh, India Email: Corollary. Substituting h =2r in (3) proves [email protected] 4 3 thatthevolumeof thesphereis 3 ¼r .

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