18. Two different ways to find the volume of a cone

Transcription

1 . Two different was to find the volume of a cone Suppose ou didn t alread know that the volume of a solid cone of radius R and height H is 3 πr H. How would ou find out? One wa would be to chop up the cone into lots and lots of thin coaxial clindrical shells with sloping rooves, find the volume of each such shell and then sum the volumes to find the total. In Figure I have drawn onl eight such shells, and in Figure I have suggested onl twent, but I want ou to imagine that there are infinitel man of them; and because there are infinitel man of them, the thickness of each shell must be vanishingl small otherwise, ou couldn t possibl pack them all into the region occupied b the cone. Let V denote the total volume, i.e., the volume of the cone; and let δv denote the infinitesimal element of volume shown shaded in Figure e that is added to the part of the cone whose perpendicular distance from the axis of smmetr does not exceed t when t increases infinitesimall to t + δt (for < t < R). Observe that the direction in which t increases is perpendicular to the surface of the infinitesimal element of volume, and for that reason we refer to t as the transverse coordinate. So transverse equals radial.. Also observe that the INNER CIRCUMFERENCE of the infinitesimal volume element is πt, the OUTER CIRCUMFERENCE of the element is π(t + δt) and the thickness of the element is δt. Let h denote the INNER HEIGHT of the infinitesimal volume element, so that because it is infinitesimal its OUTER HEIGHT must be h + δh, where δh <. Then whatever the magnitude of δv, it must exceed the volume of a cuboid of thickness δt with length INNER CIRCUMFERENCE and breadth OUTER HEIGHT, but it cannot exceed the volume of a cuboid of thickness δt with length OUTER CIRCUMFERENCE and breadth INNER HEIGHT; that is, πt(h + δh)δt < δv < πh(t + δt)δt or πth δt + πt δh δt < δv < πth δt + πh δt () (see Figure f, where the bounding cuboids are sketched). But it is clear from Figures - that δh as δt in such a wa that δh = o(δt) () and (from the similar triangles in the vertical cross-section of Figure d) Thus, from ()-(3), with H R = h R t. (3) δv = πthδt + o(δt) () h = h(t) = H (R t). (5) R In this particular case, but not alwas see later In fact δh = h (t)δt = H R δt (because h is a linear function of t, the junk term o(δt) in δh = h (t)δt+o(δt) is precisel zero), from which it can be shown that () is satisfied with δv = πthδt+π(h H)δt πh 3R δt3 precisel. But this is far more than we need to know to find the volume of the cone: () is quite enough.

2 Figure : Using concentric clindrical shells with sloping rooves for the volume of a (right circular) cone. The solids in the last eight panels fit inside one another to ield the cone in the first.

3 a c e t b d f h t Figure : Using concentric clindrical shells with sloping rooves for the volume of a (right circular) cone. (a) The sloping roof of the cone, viewed from above with a tpical elementar volume shaded. (b) A vertical cross-section through the axis of smmetr, with a tpical elementar volume (whose cross section is a vertical strip) shaded. (c) Horizontal cross-section of the generic elementar volume; the transverse coordinate t increases perpendicularl to its surface (i.e., radiall). (d) Vertical cross-section of the generic elementar volume; the cone is traced out b rotating this triangle about the axis of smmetr through angle π or 36. (e) The generic elementar volume. (f) Rectangular slabs ielding upper (left) and lower (right) bounds on δv. From (), i.e., δv = πthδt + o(δt), we can now compute the volume as V = lim δv = lim {πth δt + o(δt)} δv δt t [,R] = lim δt = t [,R] t=r t= πth δt + lim δt πth dt + t [,R] t=r t= δt lim δt o(δt) δt dt = R πth dt (6) or, on using h = H (R t) from (5), R V = πh R R t(r t)dt = πh R R (Rt t )dt = πh R { Rt 3 t3} R = 3 πr H (7) 3

4 as expected. Whenever we use integration to calculate a volume V, there is alwas a transverse coordinate t satisfing a t b () for suitable a, b and an element of volume of infinitesimal thickness δt whose (infinitesimal) volume δv is given with sufficient accurac b an equation of the form δv = f(t)δt + o(δt) (9) for suitable f, and V is alwas calculated as V = lim δv = lim {f(t)δt + o(δt)} = δv δt t [a,b] b a f(t)dt. () For example, with clindrical shells we have f(t) = πth(t), from (7). Nevertheless, there is considerable choice over what to use for a transverse coordinate and, correspondingl, how to chop up the volume V into suitable infinitesimal elements: we do not have to use thin clindrical shells. In particular, we can use thin circular disks with sloping rims instead. In Figure 3 I have drawn onl eight such disks, and in Figure I have suggested onl twent, but I want ou to imagine there are infinitel man of them; and because there are infinitel man of them, the thickness of each must be vanishingl small otherwise, ou couldn t possibl pack them all into the region occupied b the cone. Let δv now denote the infinitesimal element of volume that is added to the part of the cone whose perpendicular distance from the base does not exceed t when t increases infinitesimall to t + δt (for < t < H). The direction in which t increases is still (as alwas) perpendicular to the surface of the infinitesimal element of volume, but now transverse means axial not radial. Let r denote the base radius of the infinitesimal volume element (Figure d) so that because it is infinitesimal its roof radius must be r + δr, where δr < ; then the lower surface area of the volume element is πr, its upper surface area is π(r + δr) and its thickness is δt (as alwas). Whatever the magnitude of δv, it must exceed the volume of a clindrical disk (i.e., one having vertical rim) with thickness δt and cross-sectional area π(r + δr), but it cannot exceed the volume of a clindrical disk with thickness δt and cross-sectional area πr ; that is, π(r + δr) δt < δv < πr δt or πr δt + πr δr δt + πδr δt < δv < πr δt. () But it is clear from Figures 3- that δr as δt in such a wa that δr = o(δt) () and (from the similar triangles in the vertical cross-section of Figure d) Thus, from ()-(3), H R = H t. r (3) δv = πr δt + o(δt) ()

5 Figure 3: Using coaxial circular disks with sloping rims for the volume of a (right circular) cone. The solids in the last eight panels stack upon one another to produce the cone in the first panel. 5

6 a c R r b d t r Figure : Using coaxial circular disks with sloping rims for the volume of a (right circular) cone. (a) The sloping roof of the cone, viewed from above with a tpical elementar volume shaded. The view is identical to Figure a but the interpretation is completel different because the shading no longer represents the sloping roof of a clindrical shell instead it represents the sloping rim of a thin circular disk perpendicular to the cone s axis of smmetr. (b) A vertical cross-section through the axis of smmetr, with a tpical elementar volume (whose cross section is a horizontal strip) shaded. (c) Horizontal cross-section of the generic elementar volume, viewed from above; the transverse coordinate t increases perpendicularl to its surface, and hence to the page (i.e., axiall). (d) Vertical cross-section of the generic elementar volume; the cone is traced out b rotating this triangle about the axis of smmetr through angle π or 36. 6

7 with r = r(t) = R (H t). (5) H From () we now compute the volume as V = lim δv = lim {πr δt + o(δt)} δv δt or, on using (5), = lim δt = V = πr t [,H] t=h t= H H t [,H] πr δt + lim δt πr dt + t [,H] t=h t= δt lim δt o(δt) δt dt = (H t) dt = πr H { 3 (H t)3} H H πr dt (6) = 3 πr H (7) as expected. Both methods generalize to other volumes with an axis of circular smmetr (which need not be vertical); and with non-circular slices as elementar volumes, the second method generalizes to arbitrar volumes. As a general rule, however, we tr to avoid drawing three-dimensional diagrams like Figures and 3 because a cross-section through the axis of smmetr like those in Figures and usuall suffices. Indeed we can use the ver same diagram that we would use to calculate area: it is necessar onl to reinterpret it appropriatel. a b c d t x 3 3 x 3 x Figure 5: Using thin coaxial clindrical shells for a volume of revolution. The transverse coordinate is t = x; (b)-(d) show vertical cross-sections through the axis of smmetr. Consider, for example, the volume in Figure 5a, generated when the region shaded in Figure 5b (as opposed to the triangle in Figure d or Figure d) is rotated through 36 about the -axis (which is therefore in this case the axis of smmetr). If we use clindrical shells as elementar volumes, then the transverse coordinate is perpendicular In fact δr = r (t)δt = R H δt (again, because r is a linear function of t, the junk term o(δt) in δr = r (t)δt + o(δt) is precisel zero), from which it can be shown that () is satisfied with δv = πr δt + πrδrδt + πr 3H δt 3 precisel. But again this is far more than we need to know to find the volume of the cone: () suffices. 7

8 to the axis of smmetr, hence in this case perpendicular to the -axis. Because of the circular smmetr, it does not matter in which direction we measure distance from the axis of smmetr; accordingl, we choose the direction parallel to the x-axis b setting t = x, x 3. () Then, because Figures 5b-5d are identical to Figure 5 of Lecture 7, we alread know from () of that lecture that the height of the generic element (Figure 5d) is x if x < 6 h(x) = 3 x + x 3 if 6 x < 7 (9) 3 x if 7 x 3. But we interpret h(x) as the height of a clindrical shell of radius x (as opposed to the height of a planar strip). Now, from () b analog with () and (6), we obtain δv = πxh(x)δx + o(δx) = V = Hence, from (9), { 6 V = π xh(x)dx + 7 where I, I and I 3 are defined b 6 xh(x)dx πxh(x) dx = π } xh(x) dx 3 xh(x) dx. () = π{i + I + I 3 } () I = I = I 3 = x x dx { x 3 x + x x 3x } dx x 3 xdx. () Each of these integrals is readil evaluated. For example, the substitution u = x = x = + u = dx = + u = du I = u= 6 u= ( + u )u dx du du = (u + u )du = ( 3 u3 + 5 u5) = 35 5 (3) and from Exercise we similarl find that I = 369 and I 3 =. So, from (), 5 V = π ( 35 + ) = 3773π. () Nevertheless, it would have been easier to use disks instead. Then the transverse coordinate is aligned with (as opposed to perpendicular to) the axis of smmetr; accordingl, we now choose t =, (5)

9 a b c d x 3 3 x x 3 Figure 6: Using thin coaxial circular disks for a volume of revolution. The transverse coordinate is t =, and (b)-(d) show vertical cross-sections through the axis of smmetr. The planar region (b) that generates the solid is the difference between planar region (d) and planar region (c). a b 3 x 3 x Figure 7: Using thin coaxial circular disks for a volume of revolution. The diagrams show a vertical cross section through the axis of smmetr of the two elementar volumes used. so that perpendicular distance from axis of smmetr is r = x. (6) Observe from Figure 6 that the planar region that generates the solid of revolution when rotated about the -axis (i.e., the region shaded in Figure 6b) is the difference between the region shaded in Figure 6d and the region shaded in Figure 6c. So we can use disks as elementar volumes to calculate the volumes generated b rotating the planar regions in Figure 6c and Figure 6d about the -axis, and then subtract the first from the second to deduce the volume generated b rotating the planar region in Figure 6b. From Lecture 7, the curved boundar of the region shaded in Figure 6c and sliced into disks in Figure 7a is given b x = L() = 3 +. (7) 9

10 Hence, for the volume of revolution it generates, we obtain δv = πr δt + o(δt) = πx δ + o(δ) = V = = π πr dt = πx d = π { }d = π { } {L()} d = π {3 + } d = π { } = 55π 5 () from (5)-(6) b analog with () and (6). Also from Lecture 7, the curved boundar of the region shaded in Figure 6d and sliced into disks in Figure 7b is given b Hence, for the volume of revolution it generates, we obtain x = R() = + 6. (9) δv = πr δt + o(δt) = πx δ + o(δ) = V = = π πr dt = πx d = π { }d = π { } {R()} d = π { + 6 } d = π { } = 696π 5 (3) again from (5)-(6) b analog with () and (6). Hence the volume of revolution generated b the planar region in Figure 6a is in agreement with (). 696π 5 55π 5 = 3773π (3)

11 Exercises. For the integrals defined b (): (a) Use the substitution u = 3 x to show that I 3 = 5. (b) Show that I = The region R is bounded above b the line = x +, to the right b the parabola x = 3 + and below b the line =. (a) Use integration with respect to x to find the volume generated b rotating R about the x-axis. (b) Use integration with respect to to find the volume generated b rotating R about the x-axis. (c) Use integration with respect to x to find the volume generated b rotating R about x =. (d) Use integration with respect to to find the volume generated b rotating R about x =. Suitable problems from standard calculus texts Stewart (3): pp. 5-53, ## -36 and -9; pp. 5-53, ## -6 and Reference Stewart, J. 3 Calculus: earl transcendentals. Belmont, California: Brooks/Cole, 5th edn. Solutions or hints for selected exercises. (a) Here the transverse coordinate t = x is aligned with the axis of smmetr. We must therefore use disks. To find the volume V generated b rotating R, i.e., the darker shaded region in the diagram below, we subtract the volume V generated b rotating the lighter shaded region from the volume V of the cone generated b rotating both regions together. This cone has base radius 3 and height 6. Hence V = 3 π3 6 = π. The element of volume for V is δv = πr δt + o(δt) = πr δx+o(δx), where r is the distance from the axis of smmetr to the circumference of the disk, and hence to the curve x = 3 + or = 3(x ). Hence r = =

12 3(x ) = r = 3(x ). Now we have V = lim δv δv = x= x= = lim πr dx = 3π δt t [LOWEST,HIGHEST] {πr δt + o(δt)} = (x )dx = 3π (x ) or V = 7π. Thus V = V V = π 7π = 9π. 3 t=highest t=lowest πr dt = 3π { ( ) } 3 x (b) The transverse coordinate t = is now perpendicular to the axis of smmetr. We must therefore use clindrical shells. The generic shell is obtained b rotating the generic strip, which is aligned with the axis of smmetr and therefore stretches from the point (x,) to the point (x,), where (x,) lies on the line = x + or x = ( ), and (x,) lies on the parabola x = 3 +. In other words, x = ( ) and x = 3 +. So the height of the strip and hence of the generic shell is h = h() = x x = x x = 3 + ( ) = (note that x > x = x x = x x ). The volume element is δv = πrhδt + o(δt) = πrh()δ + o(δ), where r is the distance from the axis of smmetr to the clindrical wall of the shell. Hence r =. Now we have V = = lim δv = lim δv =3 = π h() d = π δt t [LOWEST,HIGHEST] 3 {πr h δt + o(δt)} = = π { } = 9π as before. t=highest t=lowest πr h dt { } d = π { } 3 (c) The transverse coordinate t = x is again perpendicular to the axis of smmetr (which is now vertical), and so again we must use clindrical shells. To find the new volume V generated b rotating R about this new axis of smmetr, we subtract the volume V 3 generated b rotating the lighter shaded region about x =

13 from the volume V of the cone generated b rotating both regions together. The new cone has base radius 6 and height 3. Hence V = 3 π6 3 = 36π. The generic (lighter shaded) shell is obtained b rotating the generic (lighter shaded) strip, which is aligned with the axis of smmetr and stretches from the x-axis to the parabola = 3(x ). In other words, the height of the strip and hence of the generic shell is h = h(x) = 3(x ). Thus the element of volume for V 3 is δv 3 = πrhδt + o(δt) = πrh(x)δx + o(δx), where r is the distance from the axis of smmetr to the clindrical wall of the shell. Hence r = x. Now we have V 3 = lim δv3 δv 3 = x= x= = lim δt t [LOWEST,HIGHEST] π( x)h(x)dx = π {πr h δt + o(δt)} = ( x) 3(x ) dx. t=highest t=lowest πr h dt This integral is best evaluated b means of the substitution u = 3(x ) or x = 3 u +, so that dx = u and du 3 V 3 = π = π x= x= u=3 u= ( x) u= 3( ) 3(x ) dx = π u= 3( ) ( { 3 u + } ) u 3 udu = 3 π 3 3 = π ( 3 3u u) du = π ( u 3 u5) ( x) 3(x ) dx du du ( 3 3 u) u du = 3 π ( ) = 7π 5. Thus V = V V 3 = 36π 7π 5 = π 5. (d) The transverse coordinate t = is aligned with the axis of smmetr. We must therefore use disks as in (a), and so the element of volume for V 3 is δv 3 = πr δt + o(δt) = πr δ + o(δ), where r is the distance from the axis of smmetr x = to the curve x = 3 +. Hence r = x = 3 3, and V 3 = lim δv3 δv 3 = =3 = = lim πr d = π = π ( ) 3 δt t [LOWEST,HIGHEST] 3 {πr δt + o(δt)} = ( 3 3 ) d = π 3 So again we have V = V V 3 = 36π 7π = π 5 5 t=highest t=lowest ( ) d = π ( ) π = 7π 5. πr dt 3