Module M5.4 Applications of integration

Transcription

1 F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module M5.4 Opening items. Module introduction. Fst trck questions. Redy to study? Ares. Are under grph. Are between two grphs Solids of revolution. Volume of revolution. Surfce of revolution 4 Totls nd verges 4. The mss of n object of vrible density 4. Centre of mss 4. Moment of inerti 4.4 Function verges 4.5 Men vlue of distribution 5 Closing items 5. Module summry 5. Achievements 5. Exit test Exit module COPYRIGHT 998 S57 V.

2 Opening items. Module introduction You probbly know tht the volume of sphere of rdius r is 4 πr, but do you know how to prove tht this is the cse? One wy to proceed is to divide the sphere into set of thin discs, to find n pproximte expression for the volume of ech disc, then dd ll the pproximtions, nd so estimte the volume of the sphere. As we llow the discs to get thinner nd thinner, the ccurcy of the pproximtion improves nd pproches limiting vlue the limit of sum which is usully known s definite integrl. This module discusses severl physicl nd geometricl pplictions of integrtion, ll bsed on the fct tht definite integrl is the limit of n pproprite sum. This ide is probbly lredy fmilir to you, since you my well hve been introduced to definite integrls in the context of clculting res under grphs, where such n re is pproximted by set of thin rectngles. However, the module strts with review of the reltion between definite integrl nd n re, nd discusses cses where the re ctully corresponds to some physicl quntity. It goes on to show how definite integrls cn be used to find more complicted res those enclosed by two intersecting grphs. Then it discusses some exmples of solids (solids of revolution) whose volumes nd surfce res cn be written s definite integrls. (Here, you will find derivtion of the formul for the volume of sphere.) Finlly, it shows you how to express severl other quntities msses of objects whose density is not constnt, centres of mss nd moments of inerti of solid objects, verge vlues s definite integrls. COPYRIGHT 998 S57 V.

3 Study comment Hving red the introduction you my feel tht you re lredy fmilir with the mteril covered by this module nd tht you do not need to study it. If so, try the Fst trck questions given in Subsection.. If not, proceed directly to Redy to study? in Subsection.. COPYRIGHT 998 S57 V.

4 . Fst trck questions Study comment Cn you nswer the following Fst trck questions?. If you nswer the questions successfully you need only glnce through the module before looking t the Module summry (Subsection 5.) nd the Achievements listed in Subsection 5.. If you re sure tht you cn meet ech of these chievements, try the Exit test in Subsection 5.. If you hve difficulty with only one or two of the questions you should follow the guidnce given in the nswers nd red the relevnt prts of the module. However, if you hve difficulty with more thn two of the Exit questions you re strongly dvised to study the whole module. Question F Find the re of the region bounded by the grph of the function y = x + nd the line y = 5 x. Question F Given tht the integrl + u du =.4779, find, to two deciml plces, the re of the surfce of revolution generted by the grph of y = sin x s it is rotted bout the x-xis over the intervl x π/. COPYRIGHT 998 S57 V.

5 Question F A circulr disc hs rdius, mss M nd thickness t, nd its density t ny point is proportionl to the distnce of tht point from the xis of the disc (i.e. the line perpendiculr to the plne of the disc nd through its centre). Find the moment of inerti of the disc bout its xis. Express your nswer in terms of M nd. Study comment Hving seen the Fst trck questions you my feel tht it would be wiser to follow the norml route through the module nd to proceed directly to Redy to study? in Subsection.. Alterntively, you my still be sufficiently comfortble with the mteril covered by the module to proceed directly to the Closing items. COPYRIGHT 998 S57 V.

6 . Redy to study? Study comment In order to study this module, you will need to be fmilir with the following terms: centre of mss, definite integrl, improper integrl, integrnd, integrtion by prts, integrtion by substitution, limits of integrtion, modulus, moment of inerti, rnge of integrtion. If you re uncertin of ny of these terms, you cn review them now by referring to the Glossry which will indicte where in FLAP they re developed. In ddition, you will need to be fmilir with vrious trigonometric identities, nd you should know how to find stndrd integrls (such s the integrls of xn, or ex), nd to evlute definite integrls by the method of substitution, or by integrtion by prts. You will lso need to be ble to sketch grphs of stright lines, qudrtic nd cubic polynomils, reciprocl functions, circles nd ellipses; nd know how to find the points of intersection of two grphs. The following Redy to study questions will llow you to estblish whether you need to review some of these topics before embrking on this module. COPYRIGHT 998 S57 V.

7 Question R Evlute the definite integrls () 4 ( x 5 6 x ) dx ;4(b) 4 x dx ;4(c) / dx. x Question R Find the integrl R xe x dx, where nd R re positive constnts. Hence find the improper integrl xe x dx. Question R Sketch the grphs of () y = x 4x, (b) (x ) + (y + ) = 9. COPYRIGHT 998 S57 V.

8 Question R4 Find the points of intersection of the line y = x with the grph of y = x 5x + 4, nd sketch these two functions on the sme xes. Question R5 Two smll objects, of msses. kg nd. kg, re m prt. Find () the position of their centre of mss; (b) their moment of inerti bout n xis which psses through their centre of mss nd is perpendiculr to the line joining them. Question R6 In nswering this question, you should mke use only of trigonometric identities; you should not use your clcultor. () If cosθ =, wht re the possible vlues of sin θ? (b) If sinθ =, wht is the vlue of cos (θ)? COPYRIGHT 998 S57 V.

9 f(x) Ares. Are under grph b You lredy know tht definite integrl f (x) dx cn be relted to n A re under the grph of the function f(x). If f(x) is function tht is positive over the intervl x b, then the integrl b f (x) dx is equl to b x the mgnitude of the re enclosed by the grph of the function f(x), the verticl lines x = nd x = b, nd the x-xis (see Figure ). The re of this region is known s the re under the grph of f(x) over the intervl x b. It is worth reclling here the rgument tht reltes the b definite integrl f (x) dx to the re A shown in Figure, since we shll be using the sme line of resoning in mny different cses throughout this module. COPYRIGHT 998 S57 V. Figure Are under the grph of f(x) between x = nd x = b.

10 Find the re under the grph of f(x) = x from x = to x =. (Note: In this subsection we ssume tht both x nd f( x) re dimensionless quntities.) The ide is tht we cn estimte vlue for A by dividing the re up into lrge number of thin rectngles. In Figure, the re under the grph of f(x) between x = nd x n+ = b hs been divided into n strips (lthough we only show six of them) which re then pproximted by rectngles: the first is of height f(x) nd width x, the second is of height f(x) nd width x, nd so on. The re of the ith rectngle, covering the intervl [xi, x i + x i ] is f(xi) x i nd the sum of the res of ll these rectngles provides good pproximtion to A, in other words, n A f (xi ) xi f(x) n strips A xi b x xi () i= As we llow the width of the rectngles to become smller nd smller (while, s consequence, n gets lrger) the sum on the right-hnd side of Eqution becomes n ever better pproximtion to A, nd in the limit, the sum is ctully equl to A. COPYRIGHT 998 S57 V. Figure An pproximtion to the re under the grph of f(x).

11 b This limit of sum is defined to be the definite integrl f (x) dx, so tht we hve b A= f (x) dx () The re under grph nd the mgnitude of the re bounded by grph In the bove discussion we ssumed tht the function f(x) is positive between nd b, nd we will need to mke minor djustment if f(x) is negtive or chnges sign in the intervl. In FLAP, we use the following definition: The re under the grph of function f( x) between x = nd x = b is equl to the definite integrl b f (x) dx. This hs the consequence tht in region where f(x) is lwys negtive, the re under the grph of f( x) is negtive quntity. We might, however, be interested insted in clculting the mgnitude of the re enclosed by the grph of f(x), the verticl lines x = nd x = b, nd the x-xis. Such mgnitude is, by definition, lwys positive. COPYRIGHT 998 S57 V.

12 You will find tht some uthors define re under grph in such wy tht it lwys gives the mgnitude of this re. However, we shll use the definition given bove, nd we will mke it very cler if we wnt you to clculte the mgnitude of n enclosed re. The essentil point for you to note is this: when clculting the re under grph, the res of the regions below the x-xis must be subtrcted from the res of the regions bove the x-xis. On the other hnd, when you re sked to find the mgnitude of the re bounded by the grph you must ensure tht ll the contributions to the re, from prts bove or below the x-xis, re positive. This mens tht you hve to consider seprtely the regions in which f(x) is positive nd those in which it is negtive, s in the following exmple. Exmple Find the sum of the mgnitudes of the res enclosed by the grph of f(x) = x x + nd the x-xis between x = nd x =. Solution As it is not immeditely obvious whether f(x) = x x + chnges sign between x = nd x =, we will strt by sketching the function. This qudrtic function fctorizes: x x + = (x )(x ). So the grph crosses the x-xis t x = nd t x =. When x =, f(x) =. COPYRIGHT 998 S57 V.

13 This gives us enough informtion to produce the sketch shown in Figure. We see tht the region of interest is divided into two prts: one lying bove the x-xis (lbelled A in Figure ) nd nother (A ) lying below. First we integrte f(x) between the limits x = nd x = to find n integrl I corresponding to the region A ( x I = f(x) [ x x + x ] = + =.8 x + ) dx = Now we integrte f( x) between x = nd x = nd find n integrl corresponding to the region A I = A ( x x + ) dx = [ x x + x ] A x = (8 ) ( 4 ) + ( ) =.67 Figure The function i.e. negtive nswer, becuse f(x) is lwys negtive in this region. f(x) = x x +. The mgnitude of the re required is therefore I + I = = =. 4 In question of this kind it is bsolutely essentil to be ble to determine where the function chnges sign; it will lso help if you re ble to sketch the grph of the function. COPYRIGHT 998 S57 V.

14 Question T Find the sum of the mgnitudes of the res enclosed by the grph of f(x) = ex, the x-xis, nd the lines x = nd x = loge. (You should strt by sketching the grph.)4 We cn summrize the previous discussion very netly in terms of the modulus of f(x): The mgnitude of the re bounded by the grph of y = f(x) between the points x = nd x = b is given by b the integrl f (x) dx. Since f(x) = f( x) when f(x) is positive, nd f(x) = f( x) when f(x) is negtive, the modulus sign tkes cre of b ny chnges in sign tht f(x) my undergo in the region of integrtion. The integrl f (x) dx b equl to f (x) dx ; the two re only equl if f(x) throughout the intervl x b. COPYRIGHT 998 S57 V. is not in generl

15 Although this description is quite net, in prctice we rrely try to integrte f(x) directly, nd usully we consider seprtely regions where the function is positive, nd regions where the function is negtive, s in Question T. The physicl significnce of the definite integrl So fr, we hve simply interpreted f(x) geometriclly, s the height of the grph y = f(x), in which cse b f (x) dx is indeed just the mgnitude of the re bounded by the grph of y = f( x), mesured in whtever scle units re used on the grph s xes. However, if f( x) represents some physicl quntity, then the definite integrl will of course hve different physicl significnce. Here re two exmples. COPYRIGHT 998 S57 V.

16 vx(t) Velocity time grphs Figure 4 shows grph of the velocity, vx(t) of n object moving long the x-xis, ginst time t; note tht v x (t) chnges sign twice. If we wnt to know the displcement sx of the object from its initil position (t t = ) fter given time T hs elpsed, we cn use the sme resoning tht led from Eqution to Eqution. n A f (xi ) xi (Eqn ) i= t/s b A= f (x) dx (Eqn ) Figure 4 Velocity time grph for n object moving long the x-xis. We divide the time T into mny short time intervls, ech of durtion t. During ny one of these time intervls, vx(t) is pproximtely constnt, so the corresponding displcement is pproximtely equl to v x(t) t; then the totl displcement is pproximtely equl to the sum vx(t) t. In the limit s t tends to zero, we find tht the T displcement is given by the integrl vx (t) dt, i.e. the re under the grph of vx(t) between t = nd t = T. COPYRIGHT 998 S57 V.

17 If, insted, we wnt to clculte the distnce trvelled by the object, we must recll tht distnce is the mgnitude of displcement. Thus the distnce trvelled in short time intervl t is equl to vx(t) t nd, T resoning s before, we find tht the distnce trvelled fter time T is given by the integrl vx (t) dt. So it is equl to the sum of the mgnitudes of the res enclosed by the grph of vx(t) nd the t-xis, between t = nd t = T. An object is moving long the x-xis so tht its velocity t time t is given by vx(t) = v sin(πt/t) m s, where v = 5ms nd T = s. Wht is its displcement fter s nd wht distnce does it trvel in the first seconds? Question T The grph shown in Figure 4 my be represented by the eqution v x = t + b t + c, where = 4ms 4, b = ms, c = 9ms. Clculte () the displcement of the object, (b) the distnce trvelled by the object between t = nd t = 4 s.4 COPYRIGHT 998 S57 V.

18 Work Suppose tht n object is moving long the x-xis under the influence of constnt force F x in the x-direction. The work done by the force in moving the object from x = to x = b is W = Fxsx where sx is the displcement of the object, which in this cse is b. If the force, F x(x) sy, vries with x, then we cn divide the intervl x b into mny much smller subintervls, of width x, in ech of which Fx(x) is pproximtely constnt. The work done in moving the object through the smll subintervl between x nd x + x is pproximtely Fx(x) x. The totl work done is pproximtely given by dding ll these smll mounts of work, so tht totl work done W < Fx(x) x In the limit s x decreses towrds zero (nd the number of subintervls increses) this pproximtion to W becomes incresingly ccurte, nd b W= F(x) dx It follows tht we cn interpret W to be the re under the grph of Fx(x) between x = nd x = b. COPYRIGHT 998 S57 V.

19 Question T A positively chrged prticle is fixed t the origin nd second positive chrge moves wy from it long the xxis. The force cting on the second chrge is A x where the constnt A = 7. 6 N m. Clculte the work done on the chrge s it moves from x =. m to x =. m.4 COPYRIGHT 998 S57 V.

20 y. Are between two grphs Suppose tht we wnt to find the re A enclosed by the grphs of the two functions f(x) nd g(x) shown in Figure 5. Proceeding s before, we divide the region up into thin slices, ech of thickness x, then pproximte ech slice by thin rectngle. The height of the rectngle shown in Figure 5 is (f( x) g(x)), nd so its re is (f(x) g(x)) x. We now sum the res of ll the rectngles, nd let x tend to zero, so tht in the limit the sum becomes n integrl nd we hve A= y = g(x) A y = f(x) b [ f (x) g(x)] dx () x b x + x where nd b re the x-coordintes of the points of intersection of the two grphs. Notice tht f(x) g(x) in the intervl x b, which ensures tht f(x) g(x) is positive, nd therefore A is lso positive. x Figure 5 The re enclosed between two grphs. Suppose f(x) = x + 4x nd g(x) = x 4x + 4. The grphs of these functions intersect t x = nd x =. Use Eqution to find the re A enclosed by these grphs. COPYRIGHT 998 S57 V.

21 It is quite possible to find such n re without drwing the grphs (s the bove exercise shows), but if you re sked to clculte the re between the grphs of f(x) nd g(x), you will find it much esier if you begin by sketching the two grphs on the sme xes, nd, of course, finding the points of intersection (which you need in order to be ble to put in the correct limits in Eqution ). A= f(x) 5 4 y = 4 x + 5 b [ f (x) g(x)] dx (Eqn ) The following exmple shows you how to proceed. Exmple g(x) = x. Find the re between the line y = f(x) = 4x + 5 nd the grph of Solution The two grphs intersect where 4x + 5 = x. Rerrnging this eqution gives us qudrtic eqution to solve for x nmely 4x 5x + =, which fctorizes to give (x )(4x ) =. So we see tht the roots of this eqution re x = nd x = 4. COPYRIGHT 998 S57 V. y = /x.5 Figure 6 See Exmple. x

22 We cn now sketch the re enclosed by the two grphs, showing the points of intersection (see Figure 6). f(x) 5 We now evlute the re, using Eqution : A= 4 b [ f (x) g(x)] dx A= ( 4x + 5 x ) dx = [ x y = 4 x + 5 (Eqn ) + 5x log e x / 4 ] / 4 = ( 6 ) + 5( 4 ) log e + log e 4 = y = /x.5 Figure 6 See Exmple. Question T4 Find the re between the grphs of f (x) = 8 x nd g(x) = x. (Strt by sketching the two grphs.)4 COPYRIGHT 998 S57 V. x

23 In ll the exmples so fr, the functions f(x) nd g(x) hve been chosen so tht the grph of f(x) lies bove the grph of g(x). This ensures tht the integrnd in Eqution A= b [ f (x) g(x)] dx (Eqn ) is positive, so tht we obtin positive nswer for A. If we hd chosen our two grphs the other wy round, then the height of the rectngle in Figure 5 would hve been equl to (g(x) f( x)), nd it is this function tht we would hve integrted. Thus, wherever f(x) > g(x), we integrte (f(x) g(x)), to obtin the re; but in the cse tht g(x) > f(x), we integrte (g(x) f(x)). In both these cses, we cn write down the re of the region known s the re between the grphs of f(x) nd g(x) s b A = f (x) g(x) dx (4) It is importnt to use Eqution 4 (not Eqution ) if the two grphs intersect more thn twice, so tht (f(x) g(x)) is sometimes positive nd sometimes negtive, s in the next exmple. ) COPYRIGHT 998 S57 V.

24 f (x) = 4x 6x + y Exmple Figure 7 shows the grph of f(x) = 4x 6x + nd the line g(x) = x +. They hve three points of intersection, t x =, x = =. 9 nd x = + =. 77. Find the totl re enclosed by the two grphs. g (x) = x + Solution Over the region < x <.9, f(x) is greter thn g(x). Thus the re lbelled A in Figure 7 is given by integrting (f(x) g(x)) = 4x 6x x + from x = to x =.9. A =.9 ( 4 x 6 x.5 x + ) dx = [ x 4 x x + x ].5 A A.9 =.64 (.48) = x Over the region.9 < x <.77, f( x) is less thn g(x). Thus the re lbelled A in Figure 7 is given by integrting (f( x) g(x)) = g(x) f(x) = 4x + 6x + x from x =.9 to x =.77. COPYRIGHT 998 S57 V. Figure 7 See Exmple. Note tht the horizontl nd verticl scles on the grph hve been mde different for convenience of drwing.

25 A =.77 ( 4x + 6x + x ) dx = [ x 4 x x + x ].9.77 y.9 =.664 (.64) =.88 y = f(x) So the totl re is =.4.4 y = g(x) Figure 8 shows the grphs of two functions f( x) nd g(x); their points of intersection re given s x =., x =.5 nd x = 4.. Write down the two integrls whose sum gives the totl re enclosed by the grphs Figure 8 An re enclosed by two grphs between x =. nd x = 4.. COPYRIGHT 998 S57 V. x

26 The re of circle y Finlly in this subsection, it is worth noting tht we cn use Eqution 4 y = + (r x) b A = f (x) g(x) dx (Eqn 4) to find the res of shpes such s circles nd ellipses. For exmple, let us use it to prove tht the re of circle of rdius r is πr. The eqution of circle of rdius r centred on the origin is x + y = r; so, for given vlue of x (less thn r), y cn tke the two vlues + r x nd r x. Thus we cn regrd the region inside the circle s the re enclosed by the grphs of the two functions f(x) = + r x n d g(x) = r x (see Figure 9). The points of intersection of these two grphs re x = r nd x = + r. So Eqution 4 gives, for the re A of the circle, A= [ r r ( r x r x )] dx = ( r r COPYRIGHT 998 ) r x dx S57 V. ( r, ) (r, ) x y = (r x) Figure 9 The circle x + y = r.

27 The definite integrl cn be evluted by mking the substitution x = rsin u. Then r x = r ( sin u ) = rcos u, nd dx = rcos u du. When x = r, sin(u) =, so u = π, nd when x = r, u = π ; thus the new limits of integrtion re π nd π. So the integrl becomes A = r π/ cos u du π/ We cn evlute this integrl by mens of the trigonometric identity cosu = + cos (u); substituting this into the integrl, we find A = r π/ π/ π/ π/ { + cos(u)} du = r u + sin(u) π/ π/ cos u du = r π π = r + s i nπ + sin ( π ) = π r COPYRIGHT 998 S57 V.

28 Question T5 Find the re enclosed by the ellipse x 6 + y 9 =.4 COPYRIGHT 998 S57 V.

29 y Solids of revolution P (h, r). Volume of revolution A Suppose tht we wnt to find the volume of cone, of height h nd bse r rdius r. Such cone is shown in Figure ; we hve drwn it so tht its vertex is t the origin nd its xis of symmetry lies long the x-xis. D O To find the volume we cn pproximte the cone by set of thin discs, Q h x ech of thickness x. One of these discs, D, is shown in Figure. The centre of the disc D hs coordintes (x,). We clculte the volume of ech disc, nd then obtin n estimte for the volume of the cone by dding these volumes. We then llow x to tend to zero, nd in the limit x the sum becomes definite integrl between the limits nd h, which x gives the exct vlue of the volume of the cone. In order to be ble to evlute this integrl, we must, of course, obtin n expression for the Figure A cone of height h nd volume of typicl disc in terms of x. bse rdius r. To clculte the volume of typicl disc D, we need first to know its rdius. This is equl to the y-coordinte of the point A, nd to express this in terms of x, we need to find the eqution of the line OP in Figure. This line psses through the origin nd the point (h,r); its grdient is r h nd its intercept is zero. So the eqution of the line is y = rx h. Consequently, the rdius of the disc D is rx h. COPYRIGHT 998 S57 V.

30 Hence r cross-sectionl re of D = π x h r volume of D = re thickness = π x x h nd Hence the volume V of the cone is given by the integrl r V = π h h h r x = πr h x dx = π h n expression tht you hve probbly seen before. We will show shortly tht we cn generlize this strtegy to find the volume of ny solid which cn be regrded, to good pproximtion, s series of discs ll centred on the x-xis. But we first introduce some terminology. COPYRIGHT 998 S57 V.

31 Look t Figure ; imgine rotting the tringle OPQ (which lies in the (x,y) plne) bout the x-xis. As you do so, the line OP will sweep out the surfce of the cone. We cn sy, therefore, tht the cone is generted by rotting the re OPQ bout the x-xis. This is n exmple of solid of revolution solid which cn be obtined by rotting the re under grph (or prt of grph) bout some xis. The volume of solid of revolution is known s volume of revolution. y P (h, r) A r D O Q h Describe the solid of revolution obtined by rotting the re under the line y = between x = nd x = 6 bout the x-xis. x x Figure A cone of height h nd bse rdius r. COPYRIGHT 998 S57 V. x

32 Figure shows n rbitrry solid of revolution, obtined by rotting the re under the grph of the function f(x) between the points x = nd x = b bout the x-xis. As we did with the cone, we cn pproximte this volume by set of thin discs of thickness x; we show one such disc, D, in Figure. The rdius of D is equl to the y-coordinte of the point A; tht is, it is equl to f(x). So the cross-sectionl re of D is π[f(x)], nd its volume is π[f (x)] x. Summing over the volumes of ll the discs, nd llowing x to tend to zero, we obtin definite integrl for the volume V of the solid of revolution: b V = π [ f (x)] dx solid disc D y A b x Exmple 4 The solid of revolution obtined by rotting the semicircle f(x) = r x bout the x-xis is sphere of rdius r. Use Eqution 5 COPYRIGHT 998 x (5) to show tht the volume of sphere of rdius r is y = f(x) 4 πr. S57 V. Figure Solid of revolution obtined by rotting the re under the grph of f(x) bout the x-xis.

33 Solution We first sketch the semicircle (see Figure ); this shows us tht the limits of integrtion re x = r nd x = + r. Then we substitute these limits nd f(x) = r x into Eqution 5, b V = π [ f (x)] dx y (r x) (Eqn 5) r to obtin r x r V = π ( r x ) dx Figure r Evluting the integrl gives us V = π[ r x x ] r = π r { r } ( r ) = 4 See Exmple 4. πr 4 Question T6 Find the volume of revolution obtined by rotting the re under the grph of f (x) = x between x = nd x = bout the x-xis.4 COPYRIGHT 998 S57 V.

34 y. Surfce of revolution P (h, r) Suppose now tht we wnt to find, not the volume but the surfce re of the cone shown in Figure. By now, you my be thinking tht you know wht is coming. Perhps wht we must do is to pproximte the cone by set of thin discs (just s we did to clculte its volume), work out the surfce re of typicl disc, dd up ll these surfce res, nd let the thickness of the discs tend to zero, to obtin n integrl giving the surfce re of the cone. This process does NOT work, nd n exmple should convince you tht this is the cse. The disc D in Figure hs rdius rx h, its circumference is π rx h, nd so its surfce re (= circumference thickness) is π rx h x. Thus the integrl giving the totl surfce re of the discs s their thickness tends to zero is π r h h x dx = π h r x = πrh h (6) However, Eqution 6 does not give the correct nswer for the surfce re of the cone s we cn esily verify! COPYRIGHT 998 S57 V. A r D O Q h x x Figure A cone of height h nd bse rdius r. x

35 We hve nother wy of finding this surfce re: we cn imgine cutting the cone long stright line running from its bse to its vertex nd spreding the cone out flt. This gives us the sector of circle shown in Figure 4; the rdius of the circle is equl to the slnt height l of the cone (i.e. the length of the line OP in Figures nd 4), nd the length of the rc PR is equl to the circumference of the bse of the cone, πr. (If you do not believe tht this is the figure obtined, try doing the O experiment in reverse. Cut out sector of l θ circle; you will find tht you cn roll it up into cone.) P The surfce re of the cone is equl to the re of the sector shown in Figure 4. This sector is frction θ π of the complete circle of rdius l, where θ is the ngle PO R (mesured in rdins). P (h, r) A r D O Q h x R x Figure 4 An unrolled cone. Figure A cone of height h nd bse rdius r. So its re must be equl to (θ π) the re of this circle (πl), i.e. to COPYRIGHT 998 y S57 V. θl. x

36 The ngle θ is equl to the length of the rc PR divided by the rdius l of the circle, θ = πr l, so we find πr l = πrl surfce re S of cone = l (7) NOT the sme s the result obtined in Eqution 6. π h O θ P l R Figure 4 An unrolled cone. h r r x dx = π x = πrh h h (Eqn 6) The reson why Eqution 6 gives the wrong nswer is simply tht, while the volume of thin slice of the cone is well pproximted by the volume of thin disc of the sme thickness nd with rdius equl to the verge rdius of the slice, the surfce re of such slice is not equl, even pproximtely, to the surfce re of the corresponding disc. However, we cn rrive t n integrl giving the surfce re of the cone if we consider crefully wht the surfce re of thin slice of it ctully is. COPYRIGHT 998 S57 V.

37 Figure 5 shows such slice Ω, whose thickness (equl to the length AB) we will cll s. We cn think of this slice s being generted by rotting the line segment AB bout the x-xis; in which cse we see tht the point B trvels totl distnce equl to π BC. So if we imgine cutting the slice long AB, nd lying it flt, we will obtin the shpe shown in Figure 5b prt of ring. The re of this region is, to good pproximtion, given by the product of its thickness s nd the length of the outer rc bounding it, π BC: re of slice Ω = π BC s y A Ω B C x () (8) Before we cn use Eqution 8 to give us n integrl equl to the surfce re of the cone, we must express ll quntities in it in terms of x. The length BC is equl to the y-coordinte of the point B, which is equl to rx h. We now need to relte s to x, nd this cn be done using Pythgors s theorem. Ω Figure 5 () A slice of cone. (b) The unwrpped slice Ω. A (b) COPYRIGHT 998 S57 V. B

38 B s From Figure 5c, we see tht ( s) = ( x) + ( y) = ( x) ( + ( y) ( x) ) i.e. s = x + ( y) ( x) The rtio y x is simply equl to the grdient of the line OP, which is r h. So we hve s = x + A x Figure 5 (c) The length s in terms of x nd y. r h nd substituting this, nd BC = rx h in Eqution 8, re of slice Ω = π BC s (Eqn 8) we find re of the slice Ω = π (c) y r r x + x h h 444 BC 4 s COPYRIGHT 998 S57 V.

39 We now dd up ll such res, llow the thickness of the slices to tend to zero, nd so obtin definite integrl for the surfce re S of the cone: h h r r r r S = π x + dx = π + x dx h h 44 h h everything here is constnt Evlute the bove integrl, to get n expression for S. COPYRIGHT 998 S57 V.

40 The nswer you hve found does not yet look quite the sme s the nswer obtined in Eqution 7, which ws πrl. However, it is not hrd to show tht it is in fct the sme. Figure 6 shows tht the slnt height l, the height h nd the rdius r of the cone re relted by Pythgors s theorem: l = h + r which cn be rerrnged to give l = h + r h. So the right-hnd sides of Eqution 9 S = π r r + h h h x = πrh + r h πr l = πrl l l r h x (Eqn 9) nd Eqution 7 surfce re S of cone = y Figure 6 Reltion between slnt height l, height h nd rdius r of cone. (Eqn 7) re identicl. Now tht we hve seen how to write the surfce re of cone s n integrl, we cn generlize the method to find the re of ny surfce of revolution tht is, ny surfce produced by rotting grph bout n xis. COPYRIGHT 998 S57 V.

41 Figure 7 shows the grph of the function f(x) nd its surfce of revolution between the points x = nd x = b. To find the surfce of revolution S, we pproximte the grph by set of line segments of equl length s, s shown in Figure 7b. When one of these is rotted bout the x-xis, it genertes surfce Ω whose surfce re is pproximtely equl to tht of slice of the solid of revolution. The re of the surfce Ω is pproximtely equl to its thickness s multiplied by the distnce trvelled by one end of the line segment s it is rotted bout the x-xis, πf(x). So we hve re of Ω πf(x) s () the surfce Ω of the edge of the disc y y y = f(x) Figure 7 () The surfce of revolution produced by the grph of f(x) between the points x = nd x = b. (b) An pproximtion to the grph of f(x). s b () COPYRIGHT 998 x x (b) S57 V.

42 Figure 7c shows tht in this generl cse, we hve the pproximte reltion s ( x) + ( y) = x + ( y x) () This pproximtion will become better s x nd s get smller nd smller. We cn lso write dy y x i. e. y f (x) x 4 () dx nd gin, this pproximtion will improve s x tends to zero. Substituting Equtions nd into Eqution re of Ω πf(x) s (Eqn ) gives re of the thin strip Ω π f (x) + [ f (x)] x length of the strip Ω s width of the strip Ω COPYRIGHT 998 S57 V. s (c) y x Figure 7 (c) Approximtion for the length s in terms of x nd y.

43 nd when we dd up the re of ll the slices, nd llow x to tend to zero, we obtin n integrl for the surfce S of revolution: b S = π f (x) + [ f (x)] dx () Exmple 5 The solid of revolution obtined by rotting the semicircle y = r x bout the x-xis is sphere of rdius r. Use Eqution to show tht the surfce re of sphere of rdius r is 4πr. Solution With f (x) = r x, f (x) = So + [ f (x)] = + r x. r x x r = nd + [ f (x)] = x r x COPYRIGHT 998 S57 V. r r. x

44 y Thus the integrnd in Eqution b S = π f (x) + [ f (x)] dx (r x) (Eqn ) is simply equl to r. The limits of integrtion re r nd r (s in Exmple 4; see Figure ). r r x So the surfce re is Figure See Exmple 4. r S = πr dx = πr [ x ]r r = 4 πr 4 r Question T7 Find the re of the surfce of revolution obtined by rotting the grph of f(x) = x between x = nd x = bout the x-xis.4 COPYRIGHT 998 S57 V.

45 4 Totls nd verges A 4. The mss of n object of vrible density x Suppose tht we wnt to find the mss of rod of length L nd uniform L x cross-sectionl re A, whose density ρ(x) vries with distnce x from one end of the rod (see Figure 8). We proceed by dividing the rod into thin slices of thickness x. The volume of one of these slices is A x, nd since Figure 8 A rod of vrying density. the density within this slice is pproximtely constnt, nd equl to ρ(x), the mss of the rod between x nd x + x is pproximtely ρ (x)a x. The totl mss M of the rod is pproximtely given by the sum of the msses of ll the thin slices. In the limit s x tends to zero, this sum becomes definite integrl, exctly equl to the mss of the rod: L M = A ρ (x) dx (4) Find the mss M of rod of length L nd cross-sectionl re A whose density is given by ρ (x) = where ρ is constnt. COPYRIGHT 998 S57 V. ρ L + x

46 The method used to derive Eqution 4 R L M = A ρ (x) dx (Eqn 4) h cn be generlized to find the mss of some solids which re not in the shpe of rods. For exmple, suppose we wnt to find the totl mss of disc, of rdius R nd depth h whose density ρ(r) vries with the distnce r from the xis of the disc (see Figure 9). Since the density depends only on r, nd on neither the ngulr displcement round the disc nor the depth, we cn esily write down n pproximte expression for the mss of thin ring-shped portion of the disc (shown in Figure 9b) if we know the volume of such portion. We therefore divide the disc into thin concentric rings of rdius r nd thickness r, clculte the mss of one such ring, nd dd up the msses of ll the rings; then, s usul, s we let r tend to zero, the resulting integrl gives us the mss of the disc. () r r h (b) Figure 9 () A disc of vrying density. (b) A thin ring-shped portion of the disc. COPYRIGHT 998 S57 V.

47 To find the mss of the ring, note tht its volume is pproximtely equl to the product of its inner circumference πr, its thickness r nd its depth h. So tht mss of the ring < volume of ring πr h r ρ{ (r) { { { circumference depth thickness density of ring of ring of ring nd the mss M of the whole disc is given by: R M = πh rρ (r) dr (5) Question T8 Use Eqution 5 to find the mss of disc of rdius R = 4 cm nd height h = cm, whose density ρ (r) = ρ + r R where ρ = kg m.4 COPYRIGHT 998 S57 V.

48 The sme strtegy cn be used to find the mss of sphere of rdius R whose density ρ(r) depends only on the distnce r from the centre of the sphere. Here, we divide the sphere into thin sphericl shells of thickness r, concentric with the sphere. The volume of such shell is pproximtely equl to the surfce re of its inner surfce, 4πr, multiplied by its thickness r. Write down n pproximte expression for the mss of this sphericl shell. Question T9 Due to the effects of grvity, the density ρ (r) of str vries with distnce r from the centre of the str. Assuming tht the density of sphericl str of rdius R is given by ρ (r) = ρ ( r R ) where ρ is constnt, use Eqution 6 R M = 4π r ρ (r) dr 4 (Eqn 6) to clculte the mss of the str.4 COPYRIGHT 998 S57 V.

49 4. Centre of mss If we hve set of prticles of msses m i distributed t positions xi long the x-xis, the position x c of the centre of mss of this set of prticles is given by xc = mi xi i (7) mi i Suppose for exmple tht we hve three smll spheres of led of mss. kg,.5 kg nd. kg, ttched to thin rod, mde of luminium, t distnces 5 cm, cm nd 75 cm, respectively, from one end of the rod. The centre of mss is the point bout which the rod will blnce, ) nd ccording to Eqution 7 this point will be pproximtely cm < 7.78 cm from one end. We re ignoring the mss of the luminium rod in this clcultion since it is presumed to be smll in comprison to the other msses. COPYRIGHT 998 S57 V.

50 If we remove the led weights then we cn no longer ignore the mss of the rod, in which cse we re deling with mss distributed uniformly insted of number of discrete msses. In the cse of uniform rod, mde of luminium sy, it is cler tht it will blnce bout its centre point, but generlly we my need to use integrtion to find the position of centre of mss. Suppose, for exmple, tht we wnt to find the centre of mss of the nona uniform rod shown in Figure 8, of length L, cross-sectionl re A nd density ρ(x) (which vries long the length of the rod). We divide the rod x into slices of thickness x, nd now we my tret it s number of discrete L x msses nd pply Eqution 7 mi xi Figure 8 A rod of vrying density. xc = i (Eqn 7) mi i to obtin n pproximte expression for x c. COPYRIGHT 998 S57 V.

51 The mss of the slice between x nd x + x is pproximtely ρ(x)a x; nd now we multiply the mss of this slice by its x-coordinte, sum over ll slices, nd divide this sum by the sum of the msses of the slices: xc x ρ (x) A x ρ (x) A x As x tends to zero, this rtio of sums becomes rtio of integrls, nd we obtin n exct vlue: L xc = A x ρ (x) dx L (8) A ρ (x) dx The integrl in the denomintor here is (from Eqution 4) equl to the mss M of the rod; so n lterntive wy to write Eqution 8 is xc = L A xρ (x) dx M COPYRIGHT 998 (8b) S57 V.

52 L xc = A x ρ (x) dx L (8) A ρ (x) dx Use Eqution 8 to find the position of the centre of mss of rod with uniform cross section nd of length L, whose density ρ(x) t point distnce x from one end is given by ρ(x) = C(x + L), where C is constnt. COPYRIGHT 998 S57 V.

53 4. Moment of inerti When you try to push hevy object, the difficulty increses with the object s mss. On the other hnd, if you try to rotte n object bout n xis, the difficulty increses with quntity known s its moment of inerti bout tht xis. You my hve seen pictures of someone hving difficulty opening the mssive doors of bnk vult; this is not usully becuse there is resistnce in the hinges, but becuse the doors hve lrge moment of inerti bout the xis of the hinges (nd once you get the door strted it is just s difficult to stop it). The mss of n object is simply one of its intrinsic properties; but while the moment of inerti is relted to n object s mss, it lso depends crucilly on the choice of xis. The moment of inerti of telegrph pole bout the xis of circulr symmetry of the pole is reltively smll, but its moment of inerti bout n xis through one end of the pole, nd perpendiculr to the pole, is quite considerble. The moment of inerti I of set of point msses mi bout given xis is defined s I = N mi ri (9) i = where ri is the perpendiculr distnce of mss mi from the xis nd N is the totl number of msses. Some objects re designed so s to hve very lrge moments of inerti. For exmple flywheel is constructed to be s mssive nd with s lrge dimeter s is convenient, with most of its mss s fr from the xis s possible. We will see why such design is sensible shortly. COPYRIGHT 998 S57 V.

54 Two equl point msses of mgnitude 5 kg re fstened to the ends of light metre rule. Wht is the pproximte moment of inerti bout n xis perpendiculr to the rule () through its centre (b) through one end? Finding the moment of inerti of number of point msses is reltively esy. When the mss is distributed throughout n object we generlly need to employ integrtion, but not in the following cse. A flywheel is designed so tht most of its mss M is distributed round the rim of the wheel, of rdius R sy. If you re designing flywheel to hve the gretest possible moment of inerti, is it better to double the mss nd keep the rdius fixed or to double the rdius nd keep the mss fixed? The following exmple is bsed on similr ide. COPYRIGHT 998 S57 V.

55 xis of circulr symmetry Exmple 6 Find the moment of inerti of thin-wlled hollow cylinder of rdius R nd mss M bout its xis of circulr symmetry (see Figure ). Solution We divide the cylinder up into thin verticl slices ech of mss m. All points on such slice hve perpendiculr distnce r from the xis so, from Eqution 9, I = R N mi ri (Eqn 9) i = the moment of inerti I = ( m)r. Since R is constnt, it cn be tken outside the summtion sign, so tht I = R m. But to the totl mss M of the cylinder, so m is simply equl The moment of inerti of thin hollow cylinder bout the xis of the cylinder I = MR () slice of mss m Note tht this result does not depend on the height of the cylinder.4 Figure COPYRIGHT 998 S57 V. See Exmple 6.

56 The previous cses were esy becuse the object could be divided into number of smll msses which were (pproximtely) the sme distnce from the xis; when the distnce from the xis vries we will need to employ more sophisticted summtion process, which will led to definite integrl. Here re some exmples: Thin rods Exmple 7 Find the moment of inerti of thin uniform rod (i.e. of constnt cross-sectionl re nd uniform density), bout n xis PQ perpendiculr to the rod nd pssing through one end, in terms of its mss M nd its length L (see Figure ). xis P L x Solution We divide the rod up into smll slices of length x. (Note tht x we hve chosen the xis to be situted t the end x = of the rod, which will Q mke it esy to write down the distnce of ech slice from the xis.) As the rod is of uniform density nd cross section, we cn sy tht the mss per Figure See Exmple 7. unit length of the rod is M L, so tht ech thin slice hs mss M M = x L Since the slice is presumed to be very thin, the perpendiculr distnce of ll points within the slice from the xis is pproximtely x. COPYRIGHT 998 S57 V.

57 We cn now set up the integrl we wnt to evlute. The moment of inerti of ech slice is pproximtely M x 4 x4 L distnce from xis squred mss of slice M x x. L As x tends to zero, this sum becomes n integrl giving the moment of inerti I of the rod bout the xis PQ: so the totl moment of inerti of the rod is pproximtely I = M L L x dx Evluting this integrl, we find Moment of inerti of rod bout n xis through one end I= M L L x = ML 4 COPYRIGHT 998 S57 V.

58 Question T Find the moment of inerti of thin uniform rod, of mss M nd length L, bout n xis perpendiculr to the rod nd pssing through the centre of the rod. (Hint: Tke the origin of coordintes to be t the centre of the rod.)4 This pproch cn esily be dpted to the cse where the density of the thin rod is not constnt, but vries long its length. Suppose tht the rod in Exmple 7 hs density ρ(x) t distnce x from one end, nd constnt cross-sectionl re A. xis P Wht is the moment of inerti of thin slice of the rod of thickness x bout the xis PQ shown in Figure? L x x Q Figure See Exmple 7. Now write down n integrl giving the totl moment of inerti of the rod bout the xis PQ. COPYRIGHT 998 S57 V.

59 Question T A thin rod of length cm nd constnt cross-sectionl re. mm hs density ρ(x ) = B + Cx, where B = 5 kg m nd C = kg m 4. Clculte its moment of inerti bout n xis perpendiculr to the rod, nd pssing through one end.4 COPYRIGHT 998 S57 V.

60 r Solid cylinders nd discs Eqution I = MR (Eqn ) r gives us the moment of inerti of hollow cylinder bout its xis of circulr symmetry. We cn use this result to clculte the moment of inerti of solid cylinder of uniform density bout its xis of circulr symmetry (see Figure ). We simply divide the cylinder into lrge number of concentric thin-wlled h hollow cylinders, use Eqution to write down the moment of inerti of typicl one of these, dd up ll such moments of inerti, nd so rrive t n integrl. We first need to clculte the mss of typicl cylindricl shell of thickness r, nd, since we re ssuming tht the density of the cylinder is uniform, this just mens finding the volume of the shell. Its volume is pproximtely equl to the product of its circumference (πr), its thickness ( r) nd its height (h); so if the cylinder hs density ρ, Figure A solid cylinder of rdius R nd height h. COPYRIGHT 998 S57 V. R xis of circulr symmetry

61 constnt height M = the mss of typicl cylindricl shell = density } } πr r h { ρ circumference thickness Its rdius is r; so (replcing M by M nd R by r in Eqution ) I = MR (Eqn ) we find tht the moment of inerti of this hollow cylinder is πrhρ r r = πr ρh r The totl moment of inerti of the cylinder is therefore pproximtely given by πr ρh r. As r tends to zero, the sum becomes n integrl, giving the moment of inerti I exctly: R I = πρh r dr We evlute the integrl, to find I = πρh[ 4 r 4 ] = R πρhr4 COPYRIGHT 998 () S57 V.

62 It is usully more convenient to hve n expression for I in terms of the mss M of the cylinder, rther thn the density ρ. We know tht the volume of the cylinder is πrh so ρ = M πr h. Substituting for ρ in Eqution I = πρh[ 4 r 4 ] = R πρhr4 (Eqn ) gives M I = πhr 4 = πr h 44 MR density ρ The moment of inerti of uniform solid cylinder bout its xis I= MR () Agin, this result does not depend on the height of the cylinder (s in Eqution ), I = MR (Eqn ) nd becuse of this, Eqution pplies eqully well to thin disc s to long cylinder. COPYRIGHT 998 S57 V.

63 If the density ρ (r) of the cylinder vries with distnce r from its xis, the moment of inerti bout the xis cn lso esily be written down s definite integrl. It is: R I = πh r ρ (r) dr (4) where R nd h re the rdius nd height of the cylinder. Question T Derive Eqution 4.4 COPYRIGHT 998 S57 V.

64 Solids of revolution solid disc D y A Since the height of the cylinder does not pper in Eqution, I= MR y = f(x) (Eqn ) tht formul pplies eqully well to thin disc s to long cylinder. This mens tht we my use it to find the moment of inerti, bout the x-xis, of ny solid of revolution. We simply pproximte the solid of revolution by set of thin discs (s we did to find its volume), clculte the moment of inerti of one such disc using Eqution, nd, in the usul wy, rrive t n integrl. Figure shows n rbitrry solid of revolution, produced by rotting the re under the grph y = f(x) over the intervl x b bout the x-xis. If the solid hs uniform density ρ, find n expression for the moment of inerti of the disc D bout the x-xis. b x Figure Solid of revolution obtined by rotting the re under the grph of f(x) bout the x-xis. Write down n integrl giving the moment of inerti I bout the x-xis of the solid of revolution in Figure. COPYRIGHT 998 S57 V. x

65 Exmple 8 Find the moment of inerti of uniform sphere of rdius R, density ρ nd mss M bout dimeter. (Give the nswer in terms of M nd R.) Solution We recll first tht sphere of rdius R is the solid of revolution obtined by rotting the semicircle y = R x bout the x-xis (compre Exmple 4, nd Figure ). Then the x-xis is dimeter of the sphere; so we my use Eqution 5, b I = πρ [ f (x)] dx 4 4 y (Eqn 5) (r x) with f (x) = R x, nd limits of integrtion R nd R. r This gives I = πρ R R x ) dx = πρ ( R (R4 4 R [ [ f ( x )] 4 = πρ R 4 x R x + 5 x 5 4 ) R x + x 4 dx R ] R R = 8 5 πρr 5 COPYRIGHT 998 S57 V. Figure r See Exmple 4. x

66 To express the nswer in terms of R nd M, we write ρ = so tht M I = 4 58 πr 5 = πr 44 5 mss = volume 4 M, πr MR.4 density ρ Question T Find the moment of inerti bout the xis of symmetry of cone of rdius r, uniform density ρ, height h nd mss M. Give the nswer in terms of M. (Hint: You my like to refer bck to the clcultion of the volume of cone in Subsection..)4 COPYRIGHT 998 S57 V.

67 4.4 Function verges In everydy lnguge the word verge is much bused term. Of course, we know roughly wht we men by sying n verge mn or n verge dy for the time of yer. We men tht the mn or dy is in some wy good representtive of ll men or dys. In this subsection nd the next we will discuss two forms of verge tht re quite distinct, the point being tht our choice of mening for the word verge depends on the context. Our first illustrtion concerns verge velocity. You re probbly fmilir with the definition of verge velocity between two times t nd t s totl displcement between t = t nd t = t totl time (t t ) In the cse of n object trvelling long the x-xis, we sw in Subsection. tht if we know the velocity vx(t) s t function of time, we cn find the totl displcement between two times s n integrl, vx (t) dt. t So, if we introduce the nottion vv for verge velocity, then t vv = vx (t) dt t t t COPYRIGHT 998 S57 V.

68 The velocity vx(t) of n object t time t is given by v x(t) = t. Wht is the verge vlue of the velocity from t = to t = T? We cn use the sme method to clculte verge vlues of other functions. For exmple, consider the electricl power used by domestic pplince. The power P supplied to prticulr pplince by the mins in the United Kingdom is designed so tht it vries with time ccording to the formul P(t) = P sin (ω t) P(t) P where P is constnt depending on the power rting of the prticulr pplince. Figure π/ω t Sketch of P(t) = Psin (ωt). A sketch of P ginst t is shown in Figure. The power rting quoted for ny domestic pplince is ctully defined to be the verge power consumed. We cn find n expression for this verge power Pv in terms of P by clculting the verge over just one cycle, since the symmetry of the grph mens tht the verge over mny cycles is the sme s the verge over one cycle. COPYRIGHT 998 S57 V.

69 So we will clculte the verge power consumed between t = nd t = π ω, which is given by the integrl of P(t) between these times, divided by the time intervl: π/ ω Pv = P(t) dt π ω = ω π π/ ω P sin (ω t) dt To evlute the integrl, we use the trigonometric identity sin (ω t) = Pv = ω P π [ cos(ω t)], so tht π/ ω [ cos(ω t)] dt = ω P π π/ ω t sin( ω t) ω nd since sin π = sin =, we finlly find ω P π P Pv = = π ω So the verge power is hlf the pek power P. COPYRIGHT 998 S57 V.

70 h So fr in this subsection we hve tken the integrtion vrible to be time t, so tht the verges in question were time verges. However, the notion of the verge vlue of function cn be defined quite generlly: for ny function f(x), the verge vlue fv over the intervl x b is defined s b f v = f (x) dx b (6) h Question T4 Figure 4 shows the cross section of wter surfce between two glss pltes. The height h of the wter surfce t position x is given by h(x) = h + bx for x. Find the verge height of the surfce (i.e. the verge vlue of h(x) over the intervl x ).4 O Figure 4 COPYRIGHT 998 S57 V. x See Question T4.

71 4.5 Men vlue of distribution The word verge is often used in completely different sense to the one introduced in Subsection 4.4. For exmple, if you st three exms nd scored 8% on the first, 5% on the second nd 74% on the third, you 8% + 5% + 74% might sy tht your verge score on ll three ws = 69%. But it would be more correct to cll this your men score. Generlly speking, the men of N numbers is defined s the sum of the numbers, divided by N. There re severl different wys of writing n expression for men vlue. To illustrte the point, let us suppose tht we hve severl, sy N, prticles moving with different constnt speeds nd we wnt to know the men vlue v of their speeds. We mesure the speed of ech one, nd find tht N of them hve speed v, N hve speed v, nd so on up to Nn hving speed vn (so tht N + N + + Nn = N). Insted of dding up the mesured vlues one by one to obtin the men vlue of ll our mesurements, we cn multiply ech of the n vlues obtined by the number of times it occurs, dd the results, nd divide by N. Wht we obtin is the men vlue of the distribution: v = N n Nivi (7) i= COPYRIGHT 998 S57 V.