AN ABSTRACT OF THE THESIS OF
Murari Prasad Ojha
May 27,
presented on
Mathematics
in
Doctor of Philosophy
for the degree of
1982
Smooth Approximations anduermassintegrals
Title:
Signature redacted for privacy.
Abstract Approved
Harry E. Goheen
Let [a,b] be a closed bounded interval of the real line, Eq
the q dimensional Euclidean space, and y
curve.
If to
c [a,b],
y(to)), and to
=
:
[a,b]
let to = sup {t E[a,b]
C [a,b]
inf ft
:
t
to
:
a continuous
E
t
to
,
y ([to,t]) =
, y([t,t0]) = y(t0)}.
Let P be a polygon inscribed in y corresponding to a partition
Q
:
a = tO < tl
.
< tn = b of [a,b]
.
Let 9(ti_iti,titi+1) =
the smallest non-negative angle between the directions
- y(ti_1) and -y(t1) - y(ti), with the proviso that
0(t1_1t1,titi+1) = 0 whenever y(ti_1) = y(ti) or y(ti) = y(ti+1).
Let K(P) = max {G(t1_1ti,titi+1), i = 1,2,...,n-1}
p(Q) = max {t;
- t1
,
1
= 1,2,...n}
.
,
and
A curve y can be smoothly approximated by inscribed polygons if
for every
c
such that
in the open interval (0,
p(Q) < d
K(P) <
implies
2
there exists a d > 0
)
c
.
The following approximation theorem is the main result in this
dissertation.
Approximation Theorem.
A curve is regular if and only if it
can be smoothly approximated by inscribed polygons.
Besides this approximation theorem, the dissertation also contains research work on integral geometry.
The
measure of r-planes that intersect a convex set C in Eq can be exis a constant depending on r,
pressed in the form x W (C) where
r r
Ar
and Wr(C) is the rth quermassintegral of C.
It is shown that it is
possible to extend this result to a non-convex set T of the type
T = U C.
,
where the C. are convex sets such that C.
n
1
> 1, provided W (C.) and W (C
r
1
n C14.1) are known.
cj
=0
for
By taking
r
C. as convex tubes, the calculations for the quermassintegrals are
1
carried out for the case q = 3.
SMOOTH APPROXIMATIONS AND QUERMASSINTEGRALS
by
Murari Prasad Ojha
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed May 27, 1982
Commencement June 1983
APPROVED:
Signature redacted for privacy.
Professor
'
Mathematics in charge of major
Signature redacted for privacy.
Head of Department of Mathematics
A
Signature redacted for privacy.
,PI ] ()
Date thesis is presented
11711
Typed by Bonnie Shula for
<g (41
)
Murari Prasad Ojha
ACKNOWLEDGEMENTS
I wish to thank Dr. Harry E. Goheen and Dr. J. Wolfgang
Smith for moral encouragement and valuable suggestions.
I am
especially grateful to Dr. Smith for his help with the proof in
Chapter II that smooth approximation implies regularity.
I also wish to thank Dr. Michael Papadopoulos for constant
encouragement, reassurance and moral support.
Finally, I owe my
eternal gratitude to my family for the love and patience they
expressed while I was working on this research.
Table of Contents
Introduction
1
Smooth Approximation of Space Curves by Inscribed
Polygons
4
Measure of Lines that Intersect a Non-Convex Set--17
Quermassintegrals of the Intersection of Two
Convex Tubes
32
Bibliography
41
LIST OF FIGURES
Page
Figure
A convex tube T(r,o).
17
A tube T around two intersecting line segments.
18
Number of (line) segments, x(AnT), in the intersection
of a line A with tube T (Figure 2).
19
A polygonal tube.
21
Intersection of convex tubes around two intersecting line
segments OA, OB.
26
of tube around OA, the plane V of inter-
Cylindrical surface
C1
section of the cylindrical regions of convex tubes T1, T2
respectively around line segments OA, OB, and the plane U
through 0 with OA perpendicular to U.
27
Plane U of Figure 6.
28
A convex set D
of intersection of two convex tubes.
Parallel convex set D + pB, where p >
33
0 and B is the
unit ball.
35
The region (D + pBAD below the plane V.
36
x)-plane where
Intersection of the region of Figure 10 with
(x3'
A is a line through the origin 0 on the plane U.
36
The planes of intersection (as in Figure 10) as A rotates
37
on the plane U.
Figure
Page
A plane as shown in Figure 12 (or, as in Figure 11).
37
Part of intersection of (D + pB)\D and the plane U.
37
The angle between a line x on U, and the line of
intersection of the plane V and the plane (x3, x).
39
SMOOTH APPROXIMATIONS AND QUERMASSINTEGRALS
CHAPTER I
Introduction
The fact that a continuous curve y : [a,b]
Eq
can be uniformly
approximated by inscribed polygons has played a basic role in
mathematics.
For example, it underlines the notion of rectifiability
and the definition of length.
Moreover, in the context of numerical
analysis a curve is generally approximated by a finite sequence of
Now the following question is
points, which is in effect a polygon.
a natural one:
if y is not only continuous, but also smooth in some
sense, and if one considers a sequence of inscribed polygons with mesh
tending to zero, how does the smoothness of y reflect itself in the
behavior of the polygonal sequence?
Is there some property of the
sequence iteself - which a computer can see - which is characteristic
of the fact that the limit curve is smooth?
The main result of this
dissertation constitutes an answer to this question.
smoothness to mean regularity, the desired property
mating polygons turns out to be the following:
given
If one takes
of the approxiE
> 0 there
exists a 6 > 0 such that, for every partition of [a,b] with mesh
less than 6, the angle between successive edges of the corresponding
inscribed polygon shall be less than
E
.
Only one should add that
in as much as y need not be locally oneTto-one, the mesh of partitions
2
This, then, is the new concept
must be defined in an appropriate way.
to which one is led, the notion of "smooth approximation by inscribed
polygons."
Precise definitions and a proof of the Approximation
Theorem will be given in Chapter II.
Though this is undoubtedly our main result, it is not one which
we had envisaged from the start.
As it turns out, we began our
research by investigating two problems pertaining respectively to
differential and integral geometry of convex sets.
The first dealt
with the approximation of q-convex curves by inscribed polygons.
In particular, we were interested to find out whether q-convexity
was sufficient to guarantee certain good properties of the approximating polygons.
Eventually, however, it became clear that q-convex-
ity had little to do with the questions at hand, and this recognition
prompted us to consider the problem of which we have spoken before.
As concerns the integral geometry of convex sets, on the other
hand, our work has led to a number of results, which will be set forth
It is well-known that the measure of r-planes
in chapters III and IV.
intersecting a convex set C in Eq can be expressed in the form
where xis a constant depending on r and
Wr(C)
The question arises:
quermassintegral of C.
extended to the case of non-convex sets?
x
r Wr
(C)
is the so-called rth
can this formula be
We show that this can be
done for sets of the form T = U Ci, where the Ci are convex sets and
C.
n
c.j = 0
for
li-j1
>
1.
This in turn leads to the problem of
evaluating quermassintegrals for the intersection of convex sets,
3
which in general is a rather difficult matter.
We have
succeeded in carrying out the calculations for the cases
r = 1, 2 and q = 3 by taking C. as convex tubes.
1
CHAPTER II
SMOOTH APPROXIMATION OF SPACE CURVES BY INSCRIBED POLYGONS
Definitions
Let [a,b] be a closed bounded interval of the real line, E
the q dimensional Euclidean space, and Sq-1 the q-1 dimensional unit
sphere.
A curve is a continuous map y
A polygon P (in E
[a,b]
:
is a sequence
)
a0' a1 ' ...., an
and the line segment joining the points ai_1
inscribed in
P is
< tn= b such that
y
.
of
The points are called vertices
pointsinEqsuchthataa.1'
1+
of P.
E
ai is called an edge
,
if there exists a partition a = to< t1
=
ai
For every to
I
(t
X'
)
0
y(C).
=
0
[a,b], to #t
t
,
If xo
x'0
,
let u[t
0
Let t
[a,b],
E
,
t'] denote the unit
0
0
vector in the direction from xo to xO
let xo =
,
.
and define
0
= sup {
t6 = inf {
t E[a,b]
E[a,b]
t
Then a < t- < t
O.
(i)
< t
0
t]) = 1(t0) = xo 1
:
t
to
,
Y([to
:
t
to
,
Y([t, to])
,
=
1(t0) = x0}
b.
0
Suppose t
b.
A unit vector
ur
is a right tangent
if,for every sequence tn E[a,b] such that tn > to
lim
n
lim
= t+
t
n
0
'
.
n
u[t'
0
t]
n
= u
.
r
,
y(tn)
to
y at
xo
,
to
and
<
5
If t+
,
0
then y has no right tangent at to.
Suppose t- > a.
(b)
A unit vector u
is a left tangent to y at t
0
0
1
if, for every sequence
tn
lim tn =
t0
lim
u[t,tn] = - u1
'
n
c[a,b] such that tn < t
y (tn)
xo, and
.
n
If t- = a, then y has no left tangent at to.
0
Note:
If
(iii)
Suppose a<
ur or uI
exists, it is unique.
t0
A unit vector u is said to
< b.
t0 <
0
be a tangent to y at t if and only if
0
the right tangent ur exists at to
the left tangent ul exists at to
u = ur = ul
,
,
and
.
Suppose y has a unit tangent u(t) at every t
y
:
[a,b]
by setting
S
admits continuous tangents if
(D)
Let 0(t1t2
between u[ti
O(tit2
O(t/
,
,
,
We
= u(t).
We say
q-1
%
define a map y
E[a,b].
is continuous.
t3t4) = the smallest non-negative angle
,
t2] and u [t3
t3t4) = 0 if
t2t3) = g(t2t3
y
,
1(t1) =
,
t4]
y(t2) or
1(t3) =
y(t4),
t1) = the smallest non-negative angle
between u(ti) and u[t2, t3],
0(t1 ,t2t3)= 0(t2t3
9(t1
,
,
t1) = 0 if
y(t2) =
y(t3),
t2) = the smallest non-negative angle between u(ti) and u(t2).
6
If Q: a = t
< t <
0
= b is a partition of the interval
< t
1
[a,b], and P the corresponding polygon inscribed in y, let
K(P) = max
and
= max
p(Q)
0(ti_1ti, titi+1), i = 1,2,...,n-1 }
{
{t'i
-
,
i
= 1,2,...,n
.
A curve y can be smoothly approximated by inscribed polygons if
for every
E
in the open interval (0, -TT ) there exists a 6 > 0 such that
p(Q) < 6 implies
K(P)< E.
A rectifiable curve is regular if it is C1 with respect to
(E)
arc length parameterization.
Note:
Usually a regular curve is defined (for instance, see
[1], page 6) as follows:
y'(t)
0 for all t
A Cl curve y
E [a,b].
[a,b]
:
Eq is regular if
Our definition concurs with this
definition because any C1 curve y
:
Eq with y'(t)
[a,b]
rectifiable, and C1 with respect to arc length.
0 is
However, unlike the
usual definition, our definition is independent of parameterization.
We give an example of a curve which is regular, but not regular with
respect to a given parameterization.
Example
Define y: [-2, 1]
E2 by
y(t) = (- t2, t3), -2
=
(t2, t3), 0
0,
t
t
1.
Here y'(0) = 0, and so y is not regular in the usual sense.
s(t) =
ftlly1(T)11 dT. f
-2
/4T2 + 9i4 dT.
-2
Solving this equation for t in terms of s, and setting
y(t(s)) = ; (s), we find that
Let
7
((4Q)3/2
((40) 3/2 - 27s)213 -4
27s)213 -4
-
3/2
-
-;,(s)= (
9
9
(40)3/2-8
0s
27
(27s - (40) 3/2 + 16) 2/3 - 4
(
(
(27s - (40)3/2 + 16)2/3 -4
9
9
(40)3/2 -8
(13)3/2 + (40)3/2 - 16
s
27
.
y is C
1
3/2
27
with respect to s, and so y is regular.
Approximation Theorem.
A curve is regular if and only if it can
be smoothly approximated by inscribed polygons.
I.
First we show that if a curve is regular then it can be smoothly
approximated by inscribed polygons.
Let yr
:
[a,b]
Eq be a regular curve, and let y
denote the arc length parameterization of yr.
:
[0,L]
It suffices to show that
y can be smoothly approximated by inscribed polygons. For suppose that for
every
E
in the open interval (0,
there exists a 6 > 0 such that if
2
p(Q) < 6 then
K(P) < E
,
where Q is a partition of [0,L], and P
the corresponding polygon inscribed in y
continuous function f: [a,b]
[0,L]
.
There exists a monotone
such that yr = y o f.
Moreover, by uniform continuity of f, there exists a
5
0 such that,
for every partition Qr of [a,b] with p(Qr) '< 6r, one has p(Q) <
Let Pr be the polygon inscribed in yr corresponding to Q.
6.
8
So if p(Qr) < Sr then
Let y
length s.
:
Eq be a regular curve parameterized by arc
[0,L]
E [0,L]
For every s, s*
y(S) - y(S*) -
h
y'(5*)
,
s*
s
let o(h) =
,
, where h = s - s*.
Given
Lemma 2.1
K(p) < 6
(Pr) =
> 0 there exists a 6 > 0
E
110(h)11
then
such that if 0 <lhl < 6
<
6
.
Ihl
Proof of lemma 2.1
(yi(S), y2(5),
,
Let
(5) =
yq(s)). Then
Y(S)
11o(h)11
0 and y
>
e
Yl(s*)11
Y(54`)
=
1h1
/
[
yl
Yl(s*)
(s) -
Since
given
E
72
(s*)
yi
y E Cl [0,L]
If we choose 6 =
6i
min(61
+
C1
> 0 there exists 6.> 0
such that if 0
(S) -Yq (S*)
*
Y.(s)
1
y(s) -
So
yi(s*)
then
,
y!(s*)
1
h
11(S*)]
[0,L] for each i = 1,2,...,q.
62
then for each i = 1,2,...,q, if 0 <1h1
YiSs) -
2
<
C
FIT
<
6 then
So given
110(h)11
E
> 0 there exists
6 > 0 such that if 0 <1h1
<
6
then
<
Ihl
62
E
9
Now to show that y can be smoothly approximated by inscribed
polygons we need to show that given
6 > 0 such that for every so, sl, s2
<
12.
1s/
- s21
= g(si
,
<
in (0,
there exists
6 [0,L]
with so <,s1 < s2
E
6 then 9(s0s1, s1s2) < E.
Let
,
if 0
a = O(sis2, s1), and
soy.
By the triangle inequality on Sq-1
, we have
a+
g(sosi, s1s2)
(2.1)
We have
cos a = u(s1).u[sl, s2]
Y(Si)
Y(S2)
=
(Si).yI
Y(S1)11
IlY(S2)
hy1(s1) + o(h)
2 + 2hyl(s
)
.o(h) +
llo(h)112
1
where h = s2 -
s
1
h + yl(s1).o(h)
h\,/
1
+ 2 yi(s1).
(1J0(h)11
0(h)
h
)
o(h)
1
+
(s 1 )
(2.2)
+ 2 y'(s
).
1
o(h)
110(h)II
(
)2
10
Let h =
.
By lemma 2.1, given
110(h)11
h
El
> 0 so that we have in (2.2)
E1
2
1
<
h
> 0, we can choose
E
> 0 such
< 61.
ly.(so. o(h)
Given
6
1
that if 0 < h < 6 then 11)(h)11
Also
> 0 there exists
E
s2 - s1
-
1.
cos a
<
2
Given
E
in (0__)
0 so that a <
, we can choose 62 >
2
That is, given
there exists 6 > 0 such that
in (0,
E
2
if 0 <
Is, -
s21
a<
< 6 then
(2.3)
Similarly
if 0 <
(so
-
then
13
<
(2.4)
2
Substituting (2.3) and (2.4) in (2.1), it follows that given
in (0,
)
there exists 6 > 0 such that if 0
0
<
1s1
-
<
2
s11
<
E
6
Is0
- s21 < 6 then O(sosi
,
s1s2) < E.
Hence a regular curve can
be smoothly approximated by inscribed polygons.
II.
Now assume that y can be smoothly approximated by inscribed
polygons.
Let to
We will show that y is regular.
E [a,b].
We first show that y has a tangent at
suffices to consider the case where y([a,t0])
y(b).
It
to.
y(a) and y([to, b])
11
Consider any sequence
y(t
)
and
lim
sequence u[t
= t
t
n
n ÷
0
E[a,b] such that t, > t+
tn
y(t
,
0
H
)
11
By compactness of 5q-1, the
.
0
t ] has a limit point
,
ur.
Similarly if
y(tI)
tr'il
and
y(to)
m
is any sequence in [a,b1 such that
lim
= t- then the sequence u[t
t'
m
0
<
,
0
t-6
ti] has a
m
limit point - ul.
In order to show that y has a tangent at t
,
it suffices to show
0
that ur = ul.
Suppose ur
u/ and u
C
Let ip be the smallest non-negative angle between
Let 0 = 0(t'tt t ).
Then 4, # 0.
.
in (0,
ul.
)
ff2
m 0'
so that E <
.
0 n
We may choose
By the definition of smooth
approximation, there exists a 6 > 0 such that if 0 <It -tI< 6,
n
0 <
itm - tol
<
6 then 0
<
E
.
0
But lim 0 = 4), which is a contradiction.
We now show that y admits continuous tangents.
It suffices to
7
2 ) there exists 6 > 0 such that if 0
show that given E in (0,
<
It -
t01
<
Lemma 2.2
t, to
< E.
6 then 9(t,t0)
Given E in (0,
)
there is a 6 > 0 so that for every
< 6 then 0(t,t0t) < E
[a,b], if 0 < t - t
and 0(t o, tot) < E.
0
Proof of Lemma 2.2
Let 2 > 0 and determine the corresponding
6 > 0 from the definition of smooth approximation.
Let 0 < t - to
< 6.
12
If y (t) = y (t
y(t).
)
then g(t,tot) = g(to, tot) = O.
So assume
y(to)
Then
t-
= inffTE [a,b]
t, y([T, t]) = y
T
:
(0} >
to
and there is a sequence tn such that to < t< t-,
y(t), and
y(tn)
t t-) <
,
g(t,ut n
nllm.
27-.
By our choice of 6,
tn = t-.
Now let n
Then g(t t,
co.
n
c
2
Likewise, if
< E.
t+
= SupfTE[a,b]
0
:
T
,
t0
lim
0
g(t+tm'
tm
=
1(t0)} < t,
tm < t,
,
t
0
co
liM
M
As before g(t
tm = tl".
T])
Y([t0
there is a sequence tm such that t
m
= 0(tot-, t-)
0
=
0
g(t+0
t+
0
'
t) =
< E.
2
co
Lemma
2.3
77
Given E in (0,
)
there exists 6 > 0 such that for
2
every t, to E[a,b], if It - tol < 6 then g(t, to) <
Proof of lemma
lemma
y (to), and
y (tm)
2.2.
Let It -
2.3
toI
Let
E
2
E.
as in
> 0 and choose 6 > 0
< 6.
If t >
to
g(t,to)
If t <
t0
g(t,to
+ g(t t,t )
0
0
<
2
2
.
13
g(t,tto) + g(tt0,t0) <
g(t,t0)
2
We now show that y is rectifiable.
Given
Lemma 2.4
E
in (0,
) there exists 6 > 0 such that for
[a,b], if 0 < It -
every t, tl, t2, t
< 6
and t < t1 < t2 < t'
c.
then 0(tti,tit2) <
Proof of lemma 2.4
Given
> 0 choose
E
6 > 0, the smaller of
3
the delta in the definition of smooth approximation and in lemma 2.3.
Then
g(tv,t) + o(t,ttl) + e(tt1,t1t2) <
e(tti,t1t2)
1r-+
E
3
Given
Lemma 2.5
every to, tn
E [a,b]
) there exists 6 > 0 such that for
in (0,
E
if
A
Sn
< sec
0 < tn - to < 6 then
E
, where Axn = y(tn) -
11Axnll
y(to), and ASn = length of the inscribed polygon with vertices at
y(t ), y(t1),..., y(tn), where to < t1 <
< tn.
0
Let H. be the hyperplane through y(t)
Proof of lemma 2.5
such that
13i
is normal to Hi.
AXn
Let ai = IlY(ti) - Y(ti_011
= distance between the hyperplanes H1..1 and Hi
9.
=g(t t
1
0 n'
t.
1-1
t
Ito -
and
)
i
By lemma 2.4, given
if 0 <
,
E
in (0,
tn1 < 6 then Oi
< E,
)
i =
there exists 6 > 0 such that
Hence
=
14
aj
= sec gi
< sec
Axn
Let u =
(2.5)
.
Then
.
IlAxn11
l[y(t.) - y(t.
=
0
1- 1
<E<
O.
)].ul = [y(t) - y(t.
1-
)].0 because
Consequently,
.
2
Ax
=
n
[1(t.) - (t.
E
)]
,
and
1-1
i=1
E[1()-
y(t
t.1
)]
.
i-1
i=1
u =
E
1lAxnII =
Ax
u =
13.
i=1
?
Thus,
A Sn
FlAx
< sec E
2
1
+
+
fi
2
1
,
by (2.5).
+
fin
y is rectifiable.
Lemma 2.6
Proof of lemma 2.6
corresponding polygon.
Let Q be any partition of [a,b] and P its
There is a fixed partition Ql such that
successive partition points of Q/ differ by less than the d
corresponding to E =
1
in lemma 2.5.
P* denote the corresponding polygon.
L(P)
L(P*)
Let Q* = Q U Ql
,
and let
Then
L(Pi) sec 1
by lemma 2.5. where L(P) is the length of P.
Thus, y is rectifiable.
Finally we show that y is Cl with respect to arc length.
Let a < t
< b.
0
E[a,b] so that
Consider any sequence
tn
tn > t0'
15
y(t )
y(t ), and
0
lim
Let Axn = y(t
tn = to.
)
- y(to), and
n
ASn = arc length measured from 1(t0) to y(tn).
Since y admits
continuous tangents by lemma 2.3, it suffices to show that
Axn
lim
where
-
u(to)
u(to),
n
is the unit tangent at
,
so
to
ASn
that y is C1 with respect to arc length.
By lemma 2.5, given
E
in (0,
)
2
0 <
Ito - tni
there exists 6 > 0 such that if
< 6 then
A
IlAxn1I
1
AS
11
>
ASn
where
ASn
cos E.
ASn
is the length of any inscribed polygon beginning at y(t
)
and ending at y(tn).
Since y is rectifiable by lemma 2.6, there exists a partition
Q of [to, tn] so that if ASn is the length of the corresponding inscribed
polygon, then
AS
n
>
1
-
.
ASn
Hence if 0 <Ito - ti < 6 then
IlAxnll
1
>(1
>.
-
E) cos
ASn
lim
n
11AXnII
co AS
= 1.
(2.6)
16
We have
111(tn) - y(to)
Axn
ASn
u [to, tn]
ASn
IIIAXn11
_
u[t0 ,t ]
ASn
Axn
lim
lim
=
-9-
fl
u [to, tn] = u(to), by (2.6).
n
ASn
A non-regular C1 curve.
An example:
Let y
:
[0,1]
E2 be the
curve defined by
y(t) = (t3, t3 sin
= (0,0)
), 0 < t
t = 0.
,
Let n be some positive integer, and
We show that y is not regular.
Q
:
1
0 <
2
<
4n + 1
< ... <
4n + 1
be a partition of the interval
,
1
yt
)
-
Y(0)
[0,1].
4n +
1
4n +
1
We have
1
,° )
(
4n+ 1
,
(4n + 1)3
2
Y(
1
1
)
y(
)
4n + 1
4n + 1
=
( 78 )
(4n+1)3
'
(4n + 1)3
7
If 0 is the angle between these two vectors, then cos 0 -,
113
So 0 7.'-±
0 as n
.
That is, y cannot be smoothly
approximated by inscribed polygons.
Hence y is not regular.
17
CHAPTER III
Measure of Lines That Intersect a Non-Convex Set
Measure of Lines that Intersect a Convex Set in E3.
3.1
Let A be a line in E3, and let m(A:An C#(p) be the measure of all
lines
A that intersect a convex set C in E3.
Then (see [2], [3], or
[ 4 ])
M(A:An C)
=2
IdA
S,
(3.1)
Af1C#(;)
Where S is the surface area of C.
Consider a line segment OA of length a in E3.
Let B(x,r) be a ball
B(x,r) is a convex tube.
U
xcOA
The surface area of
us denote this tube by T(r,o), or simply by T.
of radius r, centered at x.
The union
Let
a
T(r, a)
Fifyire
this tube is
27 ra + 47 r2
1
The measure of all lines A that intersect
.
the tube T is given by
m( A : AfIT#
(p)
= f
dA
=
(27ra +
4Tr
r
(3.2)
AnT# (1)
3.2. Tube Around Two Intersecting Segments
Consider two line segments OA, OB of lengths 01,G2 respectively in E3.
Let the angle between them be Y) ,
0 <
< ff.
18
Let Ti(r, al) =
B(x,r), and T2(r, 02)
U
xE0A
=
U
B(x,r).
xEOB
Let T = TIU T2.
Let
A be a line in E3, and let x(A n-r) = number of segments in All T.
Then x(All T) = 0, 1, or 2.
Figure
x(A
T)
(3.3)
0, if ART =
=
2
if A meets T in one segment (that is, if
meets T1 fl T2, or,
A
Ti but not T2, or, T2
but not TO
if A meets T in two segments (that is, if A
meets both T1 and T2, but not T1
Figures (a), (b)
,
x(A 11T) = 1, and
and (c)
rl T2).
respectively show the cases where x(AR T) = 0,
x(A fl T) = 2.
(See following page for figures).
Theorem 3.1.
f X(AR T) dA
A
=
f dA
+
T
dA
An T2
ARTi
-
I
dA
A n (Ti fl
T2)
(3.4)
Proof:
Case (i):
x
(ART) = 0
If ART = 4), then All
.
, An
T2, and An (T1 n T2) are all empty.
both sides of the equation (3.4) are zero, and the equation is true.
Hence
19
Figure
20
x( MIT) =
Case (ii):
Let AR (T1 A T2)
is empty.
#
1
of
Then neither
(1).
the sets
nn Ti,
An T2
So, the lines that intersect T are counted once on the right
(Because the lines that intersect T are
hand side of equation (3.4).
counted in each of the integrals on the right hand side of (3.4), and
hence are counted 1 + 1
Let ANT,
n
T2) =
-
=
1
So, the equation (3.4) holds.
time).
1
Then one of the sets An Ti ,AnT2 is empty and
(p.
So equation (3.4) holds in this case.
the other is not empty and equals ART.
Case (iii):
Here An
(ART) = 2.
(T1 n T2) =
and A nT2
cp,
A (1T1
(1)
The lines A that intersect T are counted 1 + 1 - 0 = 2 times on the
So the equation holds in this case
right hand side of equation (3.4).
too.
Hence, we have:
I
(4
X( A 11T) dA =
Tr
r2+
27T r02) -
2ff
r
al)
(4r2 +
S(Ti n T2),
Tr2
Therefore,
where S(1-1 A T2) is the surface area of T1 n T2.
x (ART) dA
1.3.
=
2
(01
2) +4Tr2 r2
2
s(Tin
T2).
(3.5)
Tube Around a Polygon.
Consider a polygon
lengths uk, k = 1,2,....,n.
P in E3, consisting of the edges ek of
Let T(r,
a,)
=
U
xeek
B(x,r).
Assume that
21
=
(/)
for
li-j1
li-j1
>
1.
To see that this can be done with r small enough,
ei Aej
=
(1)
for
>
Let r be such that T(r,
1.
Let min
we choose r as follows.
Ix
=ij
xj
> 0.
n
T(r,
Let min
li-j1
x.ce.
G.)
=
>
1
x.Ee
Choose r so that 2r <
cS
Then
.
(3.6)
T(r,
oi)
A T(r,
=
(1)
for
li-j1
>
1.
dge ek of length ak
Figure 4
22
Call P closed if
el
n
en
is a point such that the last vertex of P
in en is also the first vertex of P in e, and ei
n ej
=0
for 1
<
0
Ii
<
-
ji
i
-
< n - 1.
If P is closed we can choose r similarly.
n - 1, let
Ix. -x
min
xi
E
1
j
= 6
ij
> 0.
ei
For 1
Let
min
1
<
li-j1
< n-1
6
xi
= 6.
Now choose r so that 2r < 6.
T(r, P) =T(r,o1) U
and
ei
Let
(3.7)
U T(r, 0.11),
(All T(r, P)) = number of segments in
An T(r, P).
(3.8)
Theorem 3.2
I x(A n T(r, P)) dA
=
E
k=1
dA
An 1-(r, 0k)
f
-
E
I dA
k=1 An [(T(r, 0k) n T(r, uk4.1)] (3.9)
where in the second integral, en+1 stands for el if P is closed,
and
=
en+1
(f)
if P is not closed.
23
Proof:
Let x(A
T(r
,
P)) =
n).
The lines A that intersect T(r, P) are counted t times on the left hand
side of equation (3.9).
By reasoning similar to that in Theorem 3.1,
the lines A that intersect T(r, P) are counted
9,
times on the right hand
side of equation (3.9).
Hence we have
I x(AnT(r, P)) dA = AI - A2, where
n
2
A1 =
iT
(2 7
rak
471.
r2)
Tr
(3 .10)
2 r LipN + 2
7 2 r2n,
k=1
where L(P) =
E
GI,
is the length of P, and
k=1
A2 =1L
2
YS[T(r,
k=1
ak
)
n
T(r, akil)]
7
=
2
2
k=1
S(Dk),
(3.11)
24
Where
001),
Dk = T(r, ok) A T(r,
and S(Dk)
(3.13)
is the surface area of Dk.
First let P be closed.
Then
n
2E S(Dk)
A2=
(3.14)
k=1
SOU +
Tr
E
2
2
[S(Dk)
k=1
k=1
- S(Br)
]
1,
Where S(Br) is the surface area of the ball of radius r, in E3.
Therefore,
4r2
A2 -
[S(Dk) - S(B )]
(3.15)
k=1
and so,
fX(A
72 r L(P)
=
T(r,P))dA
4 2 [s(1),) - so )]
(3.16)
k=1
Next, suppose P is not closed.
Then
n-1
A2 -
Tr
(21Tr2 + 27r2) +
2
IT2
ok) n
E
T(r,
k=S[T(r,
ak+i)]
II
(the surface area of the two half balls at the ends of P)
It
n-1
.
2
7.
2
4nr2 +
4ffr2
+
2
7
2
k=1
S(Dk)
n-1
{
k=1
n-1
s(Br )
+
k=,
[S(Dk)
- S(Br)]
1
25
7
2
IT
ffr2 +
fx(A
[S(D) -
'IT
+
E
2
n T(r, P))d
E
2
n-1
2nr2n
=
n-1
7
4 Trr2 (n-1) +
S(Dk) - S(Br)]
k=1
S(Br)], and so
k=1
A
=
n-1
7
Tr2r L(p)
E
2
[S(Dk) - S(Br)]
(3.18)
k=1
we therefore have,
1
fx
2
7r
(A nT(r, P)dA
-
L(P)
S(D) - S(Br)
2-7IT
n-1
E=
where
k
E
if P is closed, and E =
if P is not closed.
k=1
k=1
Theorem 3.3.
S(Dk) - S(Br)
0
asTk
7
,
wherek is the angle between
the edges ek and ek+, of P.
Proof.
To show this, let us consider two line segments OA, OB of lengths
ol
and a2 respectively.
Let Y) be the angle AOB.
segments draw balls of radius r.
T2(r,
With centers on these
Then we get two tubes Ti(r, al) and
02).
Let CI and
respectively.
C2
be the cylindrical surfaces of the tubes T1 and T2
(That is, CI is the surface of the tube Tl with half
balls removed on the two ends of the segment OA).
of the surfaces C1 and
C2
The intersection
is a plane curve, which is in fact an ellipse.
26
,u
'V
I
Figure 5
27
Let V be the plane of intersection CI 0 C2.
angle between OA and OB.
The plane V bisects the
In other words, if W is the plane containing
the segments OA and OB, the line of intersection of the planes V and
W, makes an angle
with OA, and an angle +. with OB.
Let 0 be the origin, and the axis of C1 (that is, the direction
along OA) be the x3-axis, and the plane U through 0 containing
end cross section of C1 be the (xl, x2)-plane.
Let the line of intersection of the plane W (containing the line
segments OA, OB) and the plane U be the x2-axis.
Then the line of
intersection of the planes U and V will be the x1-axis.
Since x3 is perpendicular to the plane U, and OA makes an angle
(D
F
2
with the line of intersection of the planes V and W, it is clear
that the angle between the planes U and V is
7
2
cl
Figure
6
2
28
Let us now compute the surface area of the cylinder C1 below
If xl is the distance from cl n C2 to the plane (xl, x2),
the plane V.
then
7
x3 = tan
(
2
cot
x2
)
2
2
x2 - y cot
(P
2
sin
, where
11)
Figure 7
Cross-section of
(r, ip)
CI
is the polar coordinate corresponding to the point (x1
in the plane U.
T
,
x2)
Therefore, the surface area of C1 below the plane V
7
=
2
f
x3(10ds(fl
Since the arc length
(1p)
= rip, the surface area of the cylinder below
the plane V
7
=
2
r2 cot
f
y-sintp dip
2
=
2r2cot
2
Similarly, the surface area of the cylinder C2 above the plane V =
2r2cot
(19
.
2
Hence S(D),
the total surface area of
T1 n
T2,
-
2Tr
= 4r2cot
+
2
= 4r2cot
S(D)
S(D)
2
2r2(7 +LP).
+
2r2(ir +)
+
= 4r2[cot
+
4r[cot
+
- S(Br)
(7O2
'29
-P)
21T
- S(Br) = 4r2cot
Let F(V) = cot
(TT
41rr2
2
- 4nr2
2
TT
+
(P2
2
1-cosec2
P(v) F(y)
4-
F(R) = 0 +
But
(
)
2
0
2
as Y" t from 0 to
=0
Tr2
0 as
IT
->
TT
Let us now consider a regular space curve y
ized by arc length s.
Let Q
:
0 = so < s1 <
:
[0,L]
E3, parameter-
< sn = L be a par-
tition of [0,L] and P be the corresponding polygon inscribed in
Let
ek
and s
in (0,
y.
be the kth edge of P obtained by joining the vertices
sk-1
By the Approximation Theorem in Chapter IL given
.
)
E
there exists 6 > 0 such that if p(Q) < 6 then lyk -
where Yk is the angle between the edges ek and ek+1.
Let r <
1
min {ak, k = 1,2,...,n}, where
Gk is the length of ek.
Trl
< 6,
30
Theorem 3 .4.
S(Dk)
E
- 5(Br)
4- 0 as
4- 0.
1_1(Q)
Proof.
We have, for g2
'
-22
<
4
tan 0 = 0 +
Where
Bn
1787
285
+
+
15
315
+
3
22n(22n - 1) Bn
+
82n-1
4.....
(2n)!
represents the nth Bernoulli number.
7
= tan(
cot
.
03
2
2
-
)
lp
+
.*. F(Y) = cot
Tr+2
(
'
3
+
(
-Y')5
240
24
(7
2
'
)3
7
2
2
2
2
3
cI
"s3
for sufficiently small
Tr-
Co.
4
in (0,
Since y is regular, given
E
if P(Q) < 6
< E.
.
then
- 71
S(Dk) - S(Br) = 4r2
[S(Dk) - S(Br)]
Choose r so that r <
P('P)
there exists 6 > 0 such that
Jr -Y013
<
r2E3
< nr2c3.
ok
min
< r
)
2
S
< min
-
5k-1
,n
31
Then
sk - Ski
nr < n min
z
S(Dk) -
2
2
S(B
)
L
2
63
± 0 as
p(Q)
0.
32
CHAPTER IV
Quermassintegrals of the Intersection of Two
Convex Tubes
Quermassinte2rals of a Convex Set:
convex set, and B be the unit ball, in Eq.
B(x,p).
X
Definition.
For p
Let C be a
0, let C + pB
The set C + pB is called "the parallel set in the
C
distance p of the convex set C".
If V(C + pB) is the volume of C + pB,
then we have the formula (see [3] pp. 220-221):
q
V(C + pB)
=
p=0
nW (C) pP
P
P
This is called Steiner's formula for parallel convex sets.
0, 1, 2, ...
,
q, is called the pth quermassintegral of C.
W (C), p =
The constant
term W 0(C) gives the volume of C.
The intersection of the convex tubes T1(r') and -12(r a2 ) is a
convex set D.
In this chapter we show how to compute the
quermassintegrals of D.
33
TT +
Figure 8
To find the quermassintegrals of D, we compute the volume V(D + pB)
We haveD+ pB =DU
UB(x,p).
xe)D
Let us first compute the volume of D.
Let x be a ray on the (x1,x2)
- plane with xl- axis as the polar axis, 0 as the pole, and let X
with the x1-axis.
make an angle lp
4 Trr 3
V(D) 3
.
Then
7
2
ifixdxdkix3
Tr
34
We have
x
=
X2
3
cot
sin
-
tp
cot
2
2
Tr
V(D) =
7
1T+
4Tr r3
2
4r3
3
(
it
7r
cot
xdxedx3
4)=0 A=o x3=0
+
2
xsi n
f
+
3
r
)
2
4r3 cm3
--"
v,
2
To compute the volume V(D + pB), we di vide the region D + pB into four
regions: D, I, II, III so that
V(D + pB) = V(D) + V1 + V2+ V3,
where V1, V2, V3 are the volumes of the regions I, II, and III,
respecti vely.
Region I
=
The region between the balls of radii r and r + p,
as shown,
Region II = The region between the cylindrical surface C1 of Ti ( r, Gi
and the cylindrical surface of T1 ( r + poi)
up to the height C1 n C2, and the similar region
between the cylindrical surface C2 of T2 ( r,
the cyl inarica I
surface of T2 ( r +
CY 2 )
p, 02), and,
Region III = the crescent-like region, above the region II,
obtained by adding the balls on C1 n C2.
and
35
Figure
9:
D + pB
/
plane
x3
= x2 cot
2
/'
This portion is extra (that is does not belong
to D +
pB).
36
The portion of II and III below the plane x
x3
= x2 cot
line of intersection of
the plane x3 = x2 cot
and plane (x3,
2
x),
a, p sin a)
(p cos
tip L
=
tan
a=
r sin
cot
Figure 11
=
sin
ty
cot
-
Je=
-
r)L
37
Figure 12
x2cot
x3
2
Figure 14
Figure 13
iT
V2
=41
(I xclxclx3
V3 = 4 f
dip
)
(I AcIxdx3
)
P2 = ((r + p) cos ty, (r + p) sin ty
P3 =
+ p) cos tp, (r + p ) sin ty
P4 = (r cos ty, r sin
P5 = ((r + p COS
)
chp
P3PL,P5
PiPzP3PI,
P1 = (r cos q,, r sin
)
)
,
r sin to cot
/
r sin tp cot
cosg, ,
(r + p cos a ) sin
tP,
r sin
4,
cot4-+ p sin a)
38
4 Tr
(r + p)3
=
.
2
(I
0
r
dx3
= 4
xdx
)
cot-T+ (A-r) tan a
r + pCOSa
I [ f
r sin
o
r sin 4) cot
+p
)AdA +
dx3
(f
p2-
+
cot
r sin IP
+ p cos a
IP
f
r
where tan a = sin
d
)
o
Tr
V3
Tr
2
r sin ip.cot
r+ p
= 4i
2
3
Tr
Tr
V2
r3
4 rr
+
Tr
3
(A.
-
r) 2
dx3
)
Adx] d
,
cot
r sin IP
cot -T
IV
It
r+P
V2 =4 I
r sin IP
I
cot
2
XdAd
r
Tr
=4
2
f
ke
cot-T[
r sinIP
(r + p) 2
r2
2
] 05
2
0
=
Now
(2rp
2r cot
+ p 2)
+ (x-r) tan a
dx3), AdA
+ p cos a
r sin ip cot
=
I
r + p cos a
- AO tan ad?,
(2
r sin 4) cot
rp 2 cos a sin a
03 COS2 a sin a
and
2
3
+ p
r sin
+ p cos a
dx3
I
r sin q) cot
(
tt,
cot
2_
+
)
AdA
(x_r)
2
r+p
=I
/p2
r+p COSa
_
(A-0 2
Ack
39
=
f
p2 _ u2 (r
u) du
p COS a
=
1/2 a - 1/4 sin 2 a) + 1/3 p3 sin 3a
rp 2 (
p3 cOS2a
V3
sin a
3
--
I
4p3
2
3
sin a dip
+
f
2rP2
2
I/
0
4133
3
rp 2
p3 Sirl3aa
2
3
adip
sin ip cot
1
92
+ 2rp2
dip
sin2
+ cot2i.)
2
Figure 15
sin ip cot
1
_
4P3
3
(
4- )
2rP2
where
TI
2
G(tP) =
)
o
o
'if
dip
IT
"if
=
)
I arc tan (sin ip cot
Y.'
2
)
dip
2
f
arc tan(sinipcot(l°2 )clip
40
V1 =
2/3
V2 = 2r cot
V3 -
(3r2
(21
(71-
(2rp
2122--
4p3
3
3rP 2 +
P
+ p 2),
and
2 G('1,
-T-) + 2rp
-
P3)
where
Tf
2
GO1
ic
arc tan (sin ip cot
V(D + pB) = V(D) +
[2r2
(
)
dip
71-1-) + 4r2cot
+ [2r( Tr+ Y)) + 2r cot
+ 44,o
[
4_1-
rr
-2- +
p
+ 2r G(Y')]
- -2-
p2
P3
= Wo(D) + 3 WI(D) + 3 W2(D) + W3(D).
W (D) = V(D) -
4r3
4 r3
Tr
k-2-
3
W1(D) = 1/3 [2r2(TT+
-2-J
+ 4r
-T- cot 2.-
2 Cot 2-1
= 1/3 S(D),
where S(D) is the surface area of D
W 2(D) = 1/3 [2r( fl+
+ 2r cot -2 + 2r GM], where
7T
2
GCP) = I
o
W3(D) -
.
3
arc tan (sin ip cot -\c-
)
dip
41
BIBLIOGRAPHY
do Carmo, M., Differential Geometry of Curves and Surfaces,
Prentice-Hall, Inc., New Jersey, 1976, pp. 2-22.
Kendall, M.G., and Moran, P.A.P., Geometrical Probability,
Charles Griffin and Company, Ltd., London, 1963,
pp. 16-20, 73-74.
Santalo, L.A., Integral Geometry and Geometric Probability,
Addison-Wesley Publishing Company, Massachusetts,
1976, pp. 220-227, p. 233.
Santa16, L.A., Integral Geometry, Studies in Global Geometry
and Analysis.
(S.S. Chern, Ed.), Math. Assoc. Amer.,
1967, pp. 147-193.