The mean width of the oloid andintegral geometric applications of it

# The mean width of the oloid and integral geometric applications of it

Uwe Bäsel
###### Abstract

The oloid is the convex hull of two circles with equal radius in perpendicular planes so that the center of each circle lies on the other circle. We calculate the mean width of the oloid in two ways, first via the integral of mean curvature, and then directly. Using this result, the surface area and the volume of the parallel body are obtained. Furthermore, we derive the expectations of the mean width, the surface area and the volume of the intersections of a fixed oloid and a moving ball, as well as of a fixed and a moving oloid.
2010 Mathematics Subject Classification: 53A05, 52A15, 52A22, 60D05
Keywords: Oloid, convex hull, integral of mean curvature, mean width, Steiner formula, parallel body, intrinsic volumes, principal kinematic formula

## 1 Introduction

The oloid is the convex hull of two circles , with equal radius in perpendicular planes so that the center of each circle lies on the other circle (see Figures 2 and 2). Dirnböck & Stachel [4, p. 117] calculated the surface area and the volume of the oloid (see also [14], [15], and Equations (2), (6), (2) and (8) of the present paper). The surface is part of a developable surface [4], [2], [14], [15].

Finch [5] calculated surface areas, volumes and mean widths of the convex hulls of three different configurations of two orthogonal disks with equal radius. The mean width of every convex hull is determined twice: 1) using the integral of the mean curvature and the relation , 2) calculating directly.
According to [4, pp. 105-106], the circles with can be defined by

 kA:= {(x,y,z)∈R3:x2+(y+1/2)2=1∧z=0}, (1) kB:= {(x,y,z)∈R3:(y−1/2)2+z2=1∧x=0}

(see Fig. 1). A parametrization of the surface [2, p. 165, Eq. (2)] is

 (x,y,z)T=→ω(m,t)=(ω1(m,t),ω2(m,t),±ω3(m,t))T,0≤m≤1,−2π3≤t≤2π3, (2)

with

 ω1(m,t)=(1−m)sint,ω2(m,t)=2(m−1)cos2t+(2m−3)cost+2m−12(1+cost),ω3(m,t)=m√1+2cost1+cost.⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ (3)

To the authors knowledge, the mean width of the oloid is not aready known. In Section 3 we calculate the mean width of using the integral of mean curvature, and in Section 4 we calculate it directly. With the help of this result we derive the volume, the surface area and the mean width of the parallel body of in Section 5. Using the principal kinematic formula of integral geometry, the expectations of the mean width, the surface area and the volume of the intersections of a fixed oloid and a moving ball, as well as of a fixed and a moving oloid are calculated in Section 6.

## 2 Preliminaries

In the following, we work in the real vector space with its standard scalar product and its vector product for two vectors , . We denote the partial derivatives

 ∂→ω∂m,∂→ω∂t

of (see (2)) by , , and so on.
Using (3), for the coefficients , , of the first fundamental form (see e. g. [6, pp. 87-88], translation: p. 68) we find

 g11=⟨→ωm,→ωm⟩=3,g12=⟨→ωm,→ωt⟩=tan(t/2),g22=⟨→ωt,→ωt⟩=2(3m2−4m+1)cos2t−(4m−3)cost+1(1+cost)(1+2cost),g=det(gjk)=∣∣∣g11g12g21g22∣∣∣=g11g22−g212=2[(3m−2)cost−1]2(1+cost)(1+2cost).⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ (4)

Now, we able to calculate the surface area of the oloid :

 S(Ω1)= ∫∂Ω1dS=∫∂Ω1dS(m,t)=2∫2π/3t=−2π/3∫1m=0√g(m,t)dmdt = 4∫2π/3t=0∫1m=0√g(m,t)dmdt=4∫2π/3t=0∫1m=0√2[(2−3m)cost+1]√(1+cost)(1+2cost)dmdt = 2√2∫2π/302+cost√(1+cost)(1+2cost)dt. (5)

Mathematica evaluates this integral to

 S(Ω1)=2√2⋅√2π=4π. (6)

 V(Ω1)= 2∬zdxdy=2∫2π/3t=−2π/3∫1m=0ω3(m,t)∣∣∣∂(ω1(m,t),ω2(m,t))∂(m,t)∣∣∣dmdt = 4∫2π/3t=0∫1m=0ω3(m,t)∣∣∣∂(ω1(m,t),ω2(m,t))∂(m,t)∣∣∣dmdt.

From (3) it follows that

So we have

 V(Ω1)= 4∫2π/3t=0∫1m=0m√1+2cost1+cost1+(2−3m)cost1+costdmdt = 2∫2π/30√1+2cost(1+cost)2dt. (7)

Mathematica finds

 V(Ω1)=23[K(√3/2)+2E(√3/2)], (8)

where

 K(k)=F(π/2,k)=∫π/20dx√1−k2sin2x, (9)

is the complete elliptic integral of the first kind, and

 E(k)=E(π/2,k)=∫π/20√1−k2sin2xdx (10)

is the complete elliptic integral of the second kind. A numerical integration integration of (2) and evaluation of (8) with Mathematica yields the decimal expansion

 V(Ω1)≈3.05241846842437485669720053193

## 3 The integral of mean curvature

The surface of the oloid is piecewise continuously differentiable. We denote by the mean curvature in one point of . The circles and (see (1)) produce two edges and , respectively, that are smooth curves. Let denote the angle between the two unit normal vectors in every point of . Applying the general formula for the integral of the mean curvature (see [12, pp. 76-77, Eqs. (3.5), (3.7)]; cp. the formula for the mean width in [5, p. 3]) to gives

 M(Ω1)=∫∂Ω1HdS+122∑j=1∫εjαds=∫∂Ω1H(m,t)dS(m,t)+∫ε1α(t)dt (11)

For the unit normal vector one finds

 →n=(n1,n2,n3)T=→ωm×→ωt|→ωm×→ωt|=(sin(t/2),−cost2cos(t/2),√1+2cost2cos(t/2))\!\!T. (12)

Since is part of a developable surface and the line segments , are part of the generators of (see [4], [2]), it is not surprising that does not depend on . The mean curvature in a point of a surface is defined by

 H=12(κ1+κ2)=12g(g11b22−2g12b12+g22b11),

where , are the principal curvatures, are the coefficients of the second fundamental form (see e. g. [6, p. 99], translation: p. 79), and , are given by (4). In our case we have

 H(m,t)dS(m,t)=H(m,t)√g(m,t)dmdt=12√g(g11b22−2g12b12+g22b11)dmdt

and

 b11= L=⟨→ωmm,→n⟩=0,b12=M=⟨→ωmt,→n⟩=0, b22= N=⟨→ωtt,→n⟩=(3m−2)cost−1√2(1+2cost)√1+cost.

It follows that

 H(m,t)dS(m,t)= g11(m,t)b22(m,t)2√g(m,t)dmdt = 32⋅(3m−2)cost−1√2(1+2cost)√1+cost⋅√1+cost√1+2cost√2[(3m−2)cost−1]dmdt = 34√1+2costdmdt

and

 ∫∂Ω1HdS= 2∫1m=0∫2π/3t=−2π/3H(m,t)dS(m,t)=32∫10dm∫2π/3−2π/31√1+2costdt = 32∫2π/3−2π/31√1+2costdt=3∫2π/301√1+2costdt.

Mathematica finds

 ∫2π/30dt√1+2costdt=K(√3/2), (13)

where is the complete elliptic integral of the first kind (9), hence

 ∫∂Ω1HdS=3K(√3/2). (14)

A handmade proof for the relation in (13) may be found in Section 7.
Now we calculate the integral of mean curvature for the edges , (see (11)). Therefore, we consider . The first unit normal vector in a point , is given by (12), the second is . So one gets

 α(t)=arccos⟨→n(t),→n∗(t)⟩=arccos(−cost1+cost),

hence

 ∫ε1αdt=∫2π/3−2π/3α(t)dt=2∫2π/30α(t)dt=2∫2π/30arccos(−cost1+cost)dt.

We observe that the inverse function of the integrand is equal to the integrand, and hence the graph of the integrand symmetrical with respect to the line . As solution of

 f(t)=t=arccos(−cost1+cost)

we find , hence

 ∫ε1αdt= 4∫π/20[arccos(−cost1+cost)−t]dt=4∫π/20arccos(−cost1+cost)dt−4∫π/20tdt = 4∫π/20arccos(−cost1+cost)dt−π22=4∫π/20[π−arccoscost1+cost]dt−π22 = 4π∫π/20dt−4∫π/20arccoscost1+costdt−π22 = 3π22−4∫π/20arccoscost1+costdt. (15)

Mathematica and we, too, are not able to solve the last integral. It looks similar to Coxeter’s integral in [7, pp. 194-201]. The NIntegrate-function of Mathematica provides

 I:=∫π/20arccoscost1+costdt≈1.87738105428247449505835371657, (16)

hence

 ∫ε1αdt≈7.29488238450413994801832163353.

From (11), (14), (3), with (9) and (16), it follows that the integral of the mean curvature of is given by

 M(Ωr)=(3K(√3/2)+3π22−4I)r≈13.7644293270030696543343466299r.

For a convex body , the mean width is given by the relation (see [12, p. 78, Eq. (3.9)]). So we we have proved the following theorem.

###### Theorem 1.

The mean width of the oloid is

 ¯b(Ωr)=(32πK(√3/2)+3π4−2πI)r≈2.19067696623158876633263049436r,

where is the complete elliptic integral of the first kind (9), and

 I=∫π/20arccoscosx1+cosxdx.

## 4 Direct calculation of the mean width

Let

 P={(x,y,z)∈R3:ax+by+cz=d}

be a supporting plane of given in the Hesse normal form. So with , is the normal unit vector of and is the distance of from the origin. intersects the plane in the line

 Lxy={(x,y)∈R2:ax+by=d},

and the plane in the line

 Lyz={(y,z)∈R2:by+cz=d}.

The equation of in Hesse normal form is

 ax√a2+b2+by√a2+b2=d√a2+b2,

therefore, the distance of from the center of is

 d1=∣∣∣a√a2+b2⋅0+b√a2+b2⋅(−12)−d√a2+b2∣∣∣=b/2+d√a2+b2

(see e. g. [3, p. 172]). Since is tangent to , we have

 b/2+d√a2+b2=1⟹d=√a2+b2−b2.

Analogously one finds that the distance of from the center of is

 d2=b/2−d√a2+b2=1,

hence

 d=√a2+b2+b2.

It follows that the distance between the support plane and the origin is

 p=max{√a2+b2−b2,√a2+b2+b2}.

Now we use spherical coordinates and as coordinates of the unit normal vector :

 a=cosφsinϑ,b=sinφsinϑ,c=cosϑ.

So we have

 √a2+b2−b2= (1−12sinφ)sinϑ, √a2+b2+b2= 12sinφsinϑ+√sin2φsin2ϑ+cos2ϑ,

and can write as

 p(φ,ϑ)=max{(1−12sinφ)sinϑ,12sinφsinϑ+√sin2φsin2ϑ+cos2ϑ}.

Clearly, is the support function of in the direction . Hence the width of in this direction is given by

 b(φ,ϑ)=p(φ,ϑ)+p(π+φ,π−ϑ).

In order to calculate the mean width of we have to integrate over all directions, hence over the unit hemisphere. Let denote the surface element of the unit sphere, we have

 ¯b(Ω1)= ∫π0∫π0b(φ,ϑ)dS(φ,ϑ)∫π0∫π0dS(φ,ϑ)=∫π0∫π0[p(φ,ϑ)+p(π+φ,π−ϑ)]dS(φ,ϑ)∫π0∫π0dS(φ,ϑ) = ∫π0∫π0p(φ,ϑ)dS(φ,ϑ)+∫π0∫π0p(π+φ,π−ϑ)dS(φ,ϑ)∫π0∫π0dS(φ,ϑ)=2∫π0∫π0p(φ,ϑ)dS(φ,ϑ)∫π0∫π0dS(φ,ϑ)

(cp. [12, p. 78, Eq. (3.9)]), where the last equal sign follows from the fact that there are two congruent portions of in the half spaces and . Due to the symmetry of with respect to the planes and we can restrict the spherical coordiates to the intervals and , respectively, hence

 ¯b(Ω1)= 2∫π/20∫π/20p(φ,ϑ)dS(φ,ϑ)∫π/20∫π/20dS(φ,ϑ)=2∫π/20∫π/20p(φ,ϑ)sinϑdϑdφ∫π/20∫π/20sinϑdϑdφ = 4π∫π/20∫π/20p(φ,ϑ)sinϑdϑdφ = 4π[∫π/6φ=0∫ξ(φ)ϑ=0(12sinφsinϑ+√sin2φsin2ϑ+cos2ϑ)sinϑdϑdφ +∫π/6φ=0∫π/2ϑ=ξ(φ)(1−12sinφ)sin2ϑdϑdφ (17)

where

 ξ(φ)=arccos√−1+2sinφ√−2+2sinφ

is the solution of the equation

 (1−12sinφ)sinϑ=12sinφsinϑ+√sin2φsin2ϑ+cos2ϑ

for . Numerical integration of (4) with Mathematica gives

 ¯b(Ω1)≈2.19067696623.

## 5 The parallel body

For a convex body and , the set (Minkowski sum)

 K+Bnϱ={x∈Rn:d(x,K)≤ϱ}

is the parallel body of at distance , where is the -ball of radius ,

 Bnϱ={x∈Rn:∥x∥≤ϱ},

and is the distance between the point and . The volume of the parallel body is given by the Steiner formula

 Vn(K+Bnϱ)=n∑j=0ϱn−jκn−jVj(K), (18)

where

 κk=πk/2Γ(1+k/2) (19)

is the volume of the -dimensional unit ball , and are the intrinsic volumes of . [11, p. 2, pp. 12-13, p. 600]
Using the relations in [10, p. 301], where denotes the Euler characteristic, the intrinsic volumes of are

 (20)

with (see (10)), (see (9)), and (see (16)). Let denote the parallel body of at distance . Due to (18), its volume is

 V(Ωr,ϱ)=V3(Ωr,ϱ)= κ0V3(Ωr)+κ1V2(Ωr)ϱ+κ2V1(Ωr)ϱ2+κ3V0(Ωr)ϱ3 = V3(Ωr)+2V2(Ωr)ϱ+πV1(Ωr)ϱ2+4π3V0(Ωr)ϱ3 = V(Ωr)+S(Ωr)ϱ+M(Ωr)ϱ2+4π3ϱ3 = V(Ω1)r3+S(Ω1)r2ϱ+M(Ω1)rϱ2+4π3ϱ3.

Applying [12, p. 82, (3.17)] allows to calculate the surface area of :

 S(Ωr,ϱ)=S(Ωr)+2M(Ωr)ϱ+4πϱ2=S(Ω1)r2+2M(Ω1)rϱ+4πϱ2.

Clearly, the mean width of is equal to , hence

 M(Ωr,ϱ)=2π[¯b(Ωr)+2ϱ]=2π¯b(Ωr)+4πϱ=M(Ωr)+4πϱ=M(Ω1)r+4πϱ

###### Theorem 2.

The integral of mean curvature, the surface area and the volume of the parallel body are given by

 M(Ωr,ϱ)= M(Ω1)r+4πϱ,S(Ωr,ϱ)=4πr2+2M(Ω1)rϱ+4πϱ2, V(Ωr,ϱ)= 23[2E(√3/2)+K(√3/2)]r3+4πr2ϱ+M(Ω1)rϱ2+4π3ϱ3

with

 M(Ω1)=3K(√3/2)+3π22−4∫π/20arccoscosx1+cosxdx.

## 6 Intersections with an oloid

Now, we apply our results and the principal kinematic formula to derive some expectations for the intersections of the oloid and the three-dimensional ball of radius , and of two oloids .
The principal kinematic formula (see [10, p. 301]) for a fixed convex body and a moving convex body is for given by

 Ij(K,M):=∫SOn∫RnVj(K∩(ϑM+→x))dλ(→x)dν(ϑ)=n∑k=jαnjkVk(K)Vn+j−k(M) (21)

with the notation

group of proper (orientation-preserving) rotations [11, p. 13],
proper rotation, ,
translation vector,
Lebesgue measure on ,
unique Haar measure on with [11, p. 584],

and

 αnjk=k!κk(n+j−k)!κn+j−kj!κjn!κn,αnjk=αnj(n+j−k),αnjj=αnjn=1,

where is the volume of the unit -ball (see (19)). Since the intersection of two convex sets is a convex set, we have

 I0(K,M)= ∫SOn∫Rnχ(K∩(ϑM+→x))dλ(→x)dν(ϑ) = ∫SOn∫Rn\mathbbm1K∩(ϑM+→x)≠∅dλ(→x)dν(ϑ), (22)

where is the indicator function of the event . So we see that is the measure of the set of rigid motions bringing into a hitting position with (see [11, p. 175], [8, p. 262, p. 267]).
For , (21) gives

 I0(K,M)=V0(K)V3(M)+12V1(K)V2(M)+12V2(K)V1(M)+V3(K)V0(M),I1(K,M)=V1(K)V3(M)+π4V2(K)V2(M)+V3(K)V1(M),I2(K,M)=V2(K)V3(M)+V3(K)V2(M),I3(K,M)=V3(K)V3(M).⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ (23)

From (6) it follows that

 E[V(K∩M)]= E[V3(K∩M)]=I3(K,M)I0(K,M). (24)

is the expected volume of . Analogously, we get the expected mean width and the expected surface area:

 E[¯b(K∩M)]= 12E[V1(K∩M)]=I1(K,M)2I0(K,M), (25) E[S(K∩M)]= 2E[V2(K∩M)]=2I2(K,M)I0(K,M). (26)

Clearly, it is possible to reverse the roles of the fixed body and the moving body.

###### Example 1.

As an example we calculate the expected values (25), (26) and (24) for and , or, equivalently, for and . For the ball one easily gets

 V0(Br)=χ(Br)=1,V1(Br)=2¯b(Br)=4r,V2(Br)=12S(Br)=2πr2,V3(Br)=V(Br)=4πr33.⎫⎪⎬⎪⎭ (27)

Note that these terms also follow from the general formula

 Vk(Br)=Vk(B3r)=Vk(B31)rk=(3k)κ3κ3−krk<