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Calculus and Geometry GENERAL ¨ ARTICLE Calculus and Geometry Keerti Vardhan Madahar In this article we compute volume and surface area of the torus, volume of a slice of a 3-ball. Besides this we discuss some interesting facts like `derivative of the area of a circle with respect to its radius is its perimeter, derivative of the vol- The author earned his PhD ume of the sphere with respect to its radius is its degree in mathematics surface area and derivative of thevolumeof the (topology), 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 shapes like sphere, University.Hisareasof interest are combinatorial cone, cylinder andcircle. But anequally important geo- topology,algebraic metric object `torus' { a shape like a scooter tube or a topology, differential doughnut { is notdiscussed in school geometry. This is topology,braidgroups, knot perhaps due to the non availability of this shape atthe theory and graphtheory. time when Archimedes(287 BC{212 BC)was computing He also likes to play with mathematica software. volumes and surface areas 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 function 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 base 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 annulus of width h) =lim : h 0 h ! As h approaches zero, the outer boundary 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 ellipsoid is not the derivative of its volume and the perimeter of an ellipse 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-plane, 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 circles 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 surfaces 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 cylinders 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 curve C (plane equalto2¼2r2R: surface F) about an external line, 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- distance traveled by ders. The product lk is the sum of the lengths of these k its centroid 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 surface of revolution 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. RESONANCE ¨ July 2013 657 GENERAL ¨ ARTICLE 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<z < r,cuts the arcs ADC and ABC ¡ 2 2 2 2 atpointsx1 = R pr z and x2 = R + pr z respectively, so by¡using f¡ormula for thevolumeof¡the surface of revolution, we have 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 curves 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'.
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