CC BY
9ON THE NOTION OF SURFACE AREA FOR SOLID OF REVOLUTION (О понятии площади поверхности тел вращения)
N. Gorchev,
«St Cyril and St Methodius» Univverity oO Veliko Turnovo,
BULGARIA
Розроблений методично доцыьний i систематизований пгдхгд для введения понят-тя площ1 б1чно1 поверхт т1л обертання. Для ц1е1 мети використана м1ра Пеано-Жордано множини I принцип Дедектда. Мета розробки - допомогти студентам -майбуттм учителям опанувати цю тему.
Ключов1 слова: площа поверхт, цилтдр, конус, куля, опуклий многогранник.
1.Introduction. Rigorous study of sets of points, for which one can define axiomatically the notion of measure leads to the measurable, according to Peano-Jordan, sets of points. They are a minimal class of sets of points for which the measure is uniquely defined. In modern science Lebesgue measure is used more often than Peano-Jordan measure. However the later will not lose its methodological value because of its simplicity and naturality. This is why we will consider Peano-Jordan measure as a base of our study. In particular we will consider surface area of a cylinder, cone and sphere.
In the educational-methodological literature for secondary schools (not only in Bulgaria) two approaches for study of surface area are used:
"Unfolding" method;
Minkowski's method.
From methodological point of view these two methods are appropriate for high school students. However, when educating mathematics teachers, this notion should be defined with necessary rigor. They need to be aware of the eventual theoretical compromises connected with the use of these two methods. For example, not every surface admits "unfolding": From Differential Geometry is known that the surfaces that preserve their area after unfolding are the ones with zero Gaussian curvature. All of these surfaces are isometric to the plane. Cylinders and cones are examples of such surfaces, but the sphere is not. Suppose there was an isometric map j, from the sphere onto the plane. Then j would map a small spherical "hat" with center A onto a subset of the plane and every curve with length l would be mapped to a curve in the plane of the same length l. The boundary b of the "hat" would
be mapped to a circle with center j(A). The radius of that circle would be the length of the shortest arc, connecting A with b . But clearly the length of such an arc would be bigger than the length of b , divided by 2p , which is impossible.
Therefore one approach for motivating the first method consists of defining an isometric map from the surface into the plane.
The essence of Minkowski's method is to use the notion of derivative implicitly. One has to define e -neighborhood of a geometric body and a distance from a point to a geometric body. However the surface area, defined in this way, does not satisfy an important property that measure should have: additivity. To see this consider the following example: Let S be a bounded plane figure in a plane with Cartesian coordinate system Oxy, let Sj be the set of points in S with rational coordinates and S2 the ones with irrational coordinates. Then ju{S) = m(SJ ) = m(S2), where m is Minkowski's measure, and therefore
m(s = sj u s2) < m(S1)+m(s2), which
shows that m is not additive (Sj n S2 = 0).
2.Formulation of the Problem. The introduction of notion of surface area of a rotational surface in one of these two ways in the mathematical courses in our high schools is methodologically justified. But when educating students to become high school teachers another approach is needed for introduction of these notions with the necessary rigor.
Such approach is undertaken in the mathematical analysis courses.
In this note we use an approach based on Dedekind completeness principle [2] to study the surface area of cylinder, cone and sphere, suitable for high school.
©
Our argumentation will be based on polyhedrons (for which their surface area is known). A natural way to define surface area of any convex surface is by approximating it by surface areas of polyhedrons, inscribed in, or circumscribed around that surface. By an inscribed prism in a cylinder we mean a prism with all side edges, contained in the side boundary of the cylinder, and parallel to the axis of the cylinder, having bases, contained in the bases of the cylinder. By a circumscribed prism around a cylinder we mean a prism, each side face of which is tangent to the side boundary of the cone and parallel to the axis of the cylinder, having bases, containing the bases of the cylinder. By an inscribed pyramid in a cone we mean a pyramid with the same vertex as the cone, the base of which is a polygon inscribed in the base of the cone. By a circumscribed pyramid around a cone we mean a pyramid with the same vertex as the cone, the base of which is a polygon, circumscribed around the base of the cone.
Schwarz's example [3] shows that even for relatively simple surfaces such as the cylinder, the sequence of the surface areas of the inscribed polyhedrons is not convergent, although the right circular cylinder has a well defined surface area.
Moreover consider a sequence of inscribed bodies in a right circular cone, each consisting of right circular cylinders with equal heights, and with heights going to zero. The limit of the surface areas of these bodies will be bigger than the surface area of the cone, since in a right angled triangle the sum of the catheti is bigger than the hypotenuse.
These examples show that the definition of surface area as a sequence of surface areas is not completely analogous to the corresponding definition of arc length of a curve.
3. Surface area of a right circular cylinder. We recall the following lemma due to Hadamar [1]:
Lemma: The surface area of a convex polyhedron is not bigger than the surface area of any convex polyhedron, encompassing the first. Equality holds if and only if both polyhe-dra coincide.
We say that a bounded convex surface P encompasses a bounded convex surface Q if Q is contained in the closure of the bounded subset of the space with boundary P . We will write P 3 Q.
Theorem 1: For each right circular cy-
linder there exists a unique number, bigger than the surface areas of the inscribed right prisms and less than the surface areas of the circumscribed right prisms.
Proof: Take a right circular cylinder C with bases k(O, R), k1 (O1, R) and heighth . We inscribe a right regular triangular prism P1 , double its edges of the base and obtain P2, double the edges of the base of P2, and obtain P3 and so on. In each step the vertices of the base of one prism are among the vertices of the subsequent ones. In this way we obtain a sequence P1,P2,k,Pn,... of inscribed right
regular prisms. Let m(P), mP),.■■, m(Pn),. be the sequence of their surface areas. We construct a sequence of circumscribed prisms Q1, Q2,..., Qn,. by taking the tangent planes at each edge with two vertices, one on the upper base and one on the lower base, for each prism from the sequence P1, P2,..., Pn,.... Let
m(Qi ),m(Q2 ),K,m(Qn),k be the sequence of the corresponding surface areas.
Since P c P+1 c Q1, Vi e N clearly the sequence {m(Pn)} is nondecreasing and bounded from above, therefore convergent. Let S be its limit: lim m(P„ )= S .
Let an be the length of the apothem of the Pn base. But the base of Pn and the base of Qn are similar polygons with similarity coefficient — and therefore m(Qn) = m(Pn )—
a„
a„
and lim m(Qn) = S, because of lim an = R.
Suppose that there exists a number W, bigger than the surface areas of all inscriber right prisms and smaller than the surface areas of all circumscribed right prisms. Since m(Pn)<W it followsS < W. Analogously from m(Qn)>W follows S >W, and therefore S = W.
It remains to show that if P and Q are any two right prisms (inscribed and circumscribed) for their surface areas m(P ) and m(Q) we havem(P)< S < m(Q).
From the fact thatm(Qn) > m(P) for all n we conclude that m(P) < S (true for each inscribed right prism). Equality is impossible because by doubling the faces of P we obtain a
(104>
prism P' that will have strictly bigger surface area (by Lemma), contradictingm(P') < S . We conclude that m(P)< S. Analogously m(Q) > S . This proves the theorem.
This Theorem motivates the following Definition 1: For each right circular cylinder, the number, greater than the surface areas of the inscribed right prisms and smaller than the surface areas of the circumscribed right prisms is called surface area of the cylinder.
Theorem 2: Let F1,F2,...,Fn,... be a sequence of right prisms, inscribed in a right circular cylinder, each containing the axis of the cylinder OO' and the lengths of the biggest base edges of which go to zero. Then the sequence of the surface areas of the prisms converges to the surface area S of the cylinder.
Proof: Since Fn is inscribed, from Theorem 1 follows m(Fn) < S . Let bn be the length of the longest edge of the base Bn of the prism Fn and let an be the distance from O to this edge. Homothety in the plane of k
R
with center O and coefficient — transforms
an
/
Bn into a polygon Bn , containing k (O, R). /
The right prism Fn with the same height as
/
the height of the cylinder and base Bn encompasses the cylinder.
//
Construct the polygon Bn having sides /
parallel to those of Bn and tangent to k(O, R) // // . Let Fn be the prism with base Bn and the
same height as the height of the cylinder. // /
Clearly Fn is encompassed by in Fn and
/ //
therefore we have m(Fn ) > m(Fn ) > S from Lemma. Thus
m\Fn l MF
<
m(Fn ) m(Fn )
<
' R ^
and
v an j
a
S > j(Fn )> S —. From the condition R
lim bn = 0 follows lim an = R and after tak-
a
ing limit in the inequalities S > j(Fn ) > S —-
R
we obtain lim j(Fn ) = S .
We compute S.
Theorem 3: The surface area of a right circular cylinder with radius of the base R and heighth isS = 2pRh .
Proof: Let P1,P2,...,Pn,... be the sequence of Theorem 1. The sequence p1h,p2h,.,pnh,... of the surface areas of
P1,P2,k,Pn, where pn are the corresponding perimeters, is convergent by Theorem 2. But then S = lim pnh = 2pR, because
lim pnh = 2pR .
3. Surface area of a right circular cone.
The analogy between cones and cylinders allows us to just sketch the main points when introducing surface area of a cone.
Theorem 4: For each right circular cone there exists a unique number, greater than the surface areas of the inscribed right pyramids and smaller than the surface areas of the circumscribed right pyramids.
Proof: Take a right circular cone K with base k (O, R), vertex V and a side edge with length l . We inscribe a regular triangular pyramid P1, double its edges of the base and obtain P2 , double the edges of the base P2 , and obtain P3 and so on. In each step the vertices of the base of one pyramid are among the vertices of the subsequent ones and all pyramids have vertex V . In this way we obtain a sequence P1,P2,...,Pn,... of inscribed regular
pyramids. Let m(P), m(Pi),., m(Pn),. be the sequence of their surface areas. We construct a sequence of circumscribed pyramids Q1,Q2,k,Qn,... by taking the tangent planes at each side edge containing V and a vertex from the base, for each pyramid from the sequence P1, P2,..., Pn,.... Let
m(Qi ),m(Q2 ),.--,m(Qn),K be the sequence of the corresponding surface areas.
Since Pi c P+1 c Q1, Vn e N clearly the sequence {m(Pn)} is nondecreasing and bounded from above, therefore convergent. Let S be its limit: lim m(P„ )= S .
Let an be the length of the base edge of Pn, kn be the length of the apothem of the base, mn be the length of the apothem of the pyramid and bn is the length of the base edge of
//
/
Qn. Then = ^—. But then
m\Pn ) an mn
b R
— = —, lim kn = R and lim mn = l, thereat k„ n®¥ n®¥
nn
fore lim m(Qn) = lim m(p„) = S. The rest of
the proof is similar to the proof of Theorem 1. This Theorem motivates the following Definition 2: For each right circular cone, the number, greater than the surface areas of the inscribed pyramids and smaller than the surface areas of the circumscribed pyramids is called surface area of the cone. Theorem 5: Let F,F,•••,F ,... be a
1 5 2 5 5 n5
sequence of pyramids, inscribed in a right circular cone, each containing the axis of the cone OO' and the lengths of the biggest base edges of which go to zero. Then the sequence of the surface areas of the pyramids converges to the surface area S of the cone. Moreover any polyhedron, encompassing the cone has surface area bigger than S.
Proof: Since Fn is inscribed, from Theorem 4 follows ju(F„ ) < S . Let bn be the length of the longest edge of the base B„ of the pyramid Fn and let an be the distance from O to this edge. Homothety in the plane
R
of k(O, R) with center O and coefficient —
an
/
transforms B„ into a polygon B„ , containing
/
k(O, R) .The pyramid F„ with the same ver-
/
tex V as the cone and base B„ encompasses
//
the cone. Construct the polygon B„ having
/
sides parallel to those of B„ and tangent to //
k (O, R). Let Fn be the pyramid with base // // / B„ and vertex V. Clearly F„ e F„ and / //
therefore m(Fn ) > m(Fn ) > S from /
Lemma. Moreover F„ is encompassed by the pyramid, homothetic in space to the pyramid Fn with center of homothety O and coeffi-R
cient
and
therefore
a„
m Fn I
m(Fn )
m(Fn ) f a
s>m(Fn)>S an
<
f R >2
v an y
Thus
— I . From the condition
.R J
lim b = 0 follows lim a = R and after tak-
limit
in
the
ajL
R
we
inequalities obtain
ing
s >m(Fn )> S lirnm(F„ ) = s .
If G is a polyhedron encompassing the cone, it is clear that, because G encompasses Fn for each n , we have /m(Fn) < m(G). Taking limit we get S < /(G).
We compute S.
Theorem 6: The surface area of a right circular cone with radius of the base R and lateral height l is S = pRl.
Proof: Let Px,P2,...,Pn,... be the sequence of Theorem 4. Let lx,l2,...,ln,... be the apothems of these pyramids and Pj, p2,..., pn, k be the semiperimeters of the bases. The sequence pjlj, p 2l2,., pnln,. of the surface areas of Pj,P2,...,Pnis convergent by Theorem 5. But then S = lim pnln = pRl, because lim pn = pR
and lim ln = l.
Corollary: Let P be a polyhedron, encompassed by a truncated cone K and Q be a polyhedron, encompassing K . Then for the surface areas of those we have
/(p )</(k )</(q ).
Proof: Follows from Theorem 5, the Lemma and the fact that a truncated cone, as a set, is a difference of two untruncated cones.
5. Surface area of the sphere. Consider a sphere T with center O and radius R . A convex polyhedron P we will call inscribed in T if all the vertices of P lie on the sphere. The convex polyhedron Q we call circumscribed around T if all faces of Q are tangent to T.
If we draw some longitudinal and latitudinal lines on the sphere, the intersection points will be vertices of an inscribed convex polyhedron. The faces of this polyhedron will be triangles (around the poles) and the rest trapezo-ids. A polyhedron of this type we will call a
//
/
2
pyramidal body. In the case when the longitudinal planes split the equator plane into equal sectors we call it a regular pyramidal body. In analogous way we define circumscribed pyramidal bodies.
Theorem 7: For each sphere there exists a unique number, greater than the surface areas of the inscribed pyramidal bodies and smaller than the surface areas of the circum-scribedpyramidal bodies.
Proof: First we construct a sequence {Pn} of regular inscribed pyramidal bodies with surface areas m(Pn) for which we construct the sequence of pyramiddal bodies {Q'n}, homo-thetic to {Pn} and encompassing the sphere, with surface areas m(Q'n), and for which the sequences {m(Pn)} and {m{Q'n)} have same limits.
Let e > 0 . Consider the homotheties in space j and j2 with center O and coefficient R + e and R -e respectively. Let j (T) = T1 and j2 (T )= T2. We take a grid on T1 in an appropriate way by longitudinal planes of equal angles and latitudinal planes, such that the diameter of any face of the inscribed regular pyramidal body is less than e. Let the regular pyramidal body thus obtained
be P'. Let j-1(P') = P which is a regular pyramidal body, that is clearly inscribed in T . Let j2 _1(P') = Q which is a regular pyramidal body, encompassing T. Then
m( P) = j2
mQ) IR + e) .
Let e1,e2,. ,en,. be a monotonously decreasing sequence of positive numbers, going to zero. For each en, as in the previous paragraph, we define a regular inscribed pyramidal body Pn (and its corresponding body /
Qn ). Thus we obtain sequences {Pn} and
{Qn} satisfying lirnm(Pn ) = limmjiQn J =
l
for some l .
We construct a sequence of regular circumscribed pyramidal bodies {Qn} by taking appropriate grids on T and taking the planes, tangent to T at the centers of each «spherical face». The grids are chosen so that the diameter of each face of Qn is less than en. Using
homotheties we construct a corresponding sequence {Pn} of bodies, encompassed by the sphere T and satisfying
lim ml Pn 1 = lim m(Qn ) = m for some m .
n®¥ ^ J n®¥
We use the Lemma to derive that m = l: / //
Qn encompasses Pk for all k and n , therefore l > m ; Qn encompasses Pk for all k and n , therefore l < m .
Denote this limit by S :
lim m(Pn ) = lim mj Q'n \ =
= limmk ) = limm(Qn ) = S.
The rest of the proof is similar to the end of the proof of Theorem 1.
Definition 3: For each sphere, the number, greater than the surface areas of the inscribed pyramidal bodies and smaller than the circumscribed pyramidal bodies is called surface area of the sphere.
Theorem 8: The surface area of sphere with radius R is S = 4pR2.
Proof: We construct a sequence of inscribed regular pyramidal bodies P1,P2,...,Pn,... for a sphere T of radius R in the following way: For each n we draw n equally spaced meridians and 2n latitudinal lines that split the "north" and "south" part into n sectors, such that the angles ZAiOAi+1 between two consecutive latitudinal planes, inter-
p
secting a meridian, are all equal to — (see
2n
figure).
In each "truncated sphere" sector we inscribe a truncated cone K nm (the north- and
nm \
south- pole caps will be untruncated cones). We denote Kn = U Knm - an inscribed con-
m
ical body. For every e > 0 from Theorem 7 we can construct a circumscribed pyramidal body Q for T such that
m(Q) > m(T) > m(Q)-e. Let Qm be the part of the polyhedron Q situated between the two latitudinal planes ofKnm . From the Corollary and the Lemma we have m(Knm) < m(Qm). Taking sum and union we conclude m(Pn )<m(Kn )<m(Q ). it follows m(Kn) < m(T)+ e and since e > 0 was arbi-
-(i07)
trary we obtain m(Kn) £ mT) for all n . We conclude m(Pn) < m(Kn) £ mT) and taking limit it follows lim j(Kn) = j(T) = S by the
proof of Theorem 7.
We consider the northern hemisphere T.
S
— = S1 + S2 + ■■■ + Sn-1 + Sn , Where Si iS
the surface area of the truncated cone defined by the points Ai-1, Bi-1, Bi, Ai and Sn is the
cone defined by the points An-1, Bn-1, C (see figure). To compute Si we consider similar triangles Ai-1 AiD, where D is the foot of the height of At to Ai-1 Bi-1 and triangle OMiE, where E is the midpoint of the line segment
P_ Pt. But then
S,.
-i A
A. B. + A.B.
2kMxO .Pt_1Pt.
2—MnOPn_xC and
Also Sn S
^ = 2—(M1O .op + MOO .Pf2 +... +
+Mn Op_ 2Pn, + MOPn CC),
n_1 n_2 n_1 n n_1
MO = R cos| — |. Then ' I 4n J
S P
- = 2pR cos — (OP1 + P1P2 + ... + Pn-C ) =
= 2pR cos—.
4n
Taking limit we get S = 4pR2. This proofs the Theorem.
Mn c
A = А»
О - Pn
В - Hi,
Remark: The convex polyhedron method that we used can be employed for more general convex surfaces (as done by Hadamard [1]). However from methodological point of view it has some advantage to consider "nicef polyhedrons for defining the surface areas of rotational surfaces.
1Адамар Ж. Элементарная геометрия. Ч ii. Стереометрия / Ж.Адамар. - М. : Учпедгиз, 1961.
2. Фихтенгольц Г.М.. Курс дифференциального и интегрального исчисления. Т. 1 / Г.М.Фихтенгольц. -М.: Наука, 1968.
3.Фихтенгольц Г.М.. Курс дифференциального и интегрального исчисления. Т. 3 / Г.М.Фихтенгольц. -М.:Наука, 1966.
2
Резюме. Горчев Н. О ПОНЯТИИ ПЛОЩАДИ ПОВЕРХНОСТИ ТЕЛ ВРАЩЕНИЯ.
Разработан методически целесообразный и систематизированный подход для введения понятия площади боковой поверхности тел вращения. Для этой цели использована мера Пеано-Жордано множества и принцип Дедекинда. Цель разработки - помочь студентам - будущим учителям в овладении этой темой..
Ключевые слова: площадь поверхности, цилиндр, конус, шар, выпуклый многогранник.
Abstract. Gorchev N. ON THE NOTION OF SURFACE AREA FOR SOLID OF REVOLUTION. We develop a methodologically comprehensive and systematical approach for the introduction of the notion of surface area of rotational surfaces (cylinder, cone and sphere). We use Peano-Jordan measure, based of the continuity principle of Dedekind Our approach is a simplified version of an approach by Hadamard and is suited for university students.
Key words: surface area, cylinder, cone, sphere, convex polyhedron.
Стаття представлена професором В.Б.Мтушевим.
На^йшла доредакцп 13.03.2013р.
