1 Introduction

Hamnet Holditch, president of Caius College in Cambridge, published in [13] a remarkable theorem. Let \(C\;\) be a convex curve, and a chord h of length \(a+b\) be divided into parts of lengths a and b by a point A. Let \(C_{a,b}\) denote a curve traced out by the point A when the chord h slides around with both endpoints on C. Holditch proved that the area of a ring domain bounded by C and \(C_{a,b}\) is equal to \(\pi ab\), see Fig 1.

Arne Broman proved in [4] and [3] a much more general theorem and gave some kinematic applications. Further applications to mechanics were given in [11, 12], and [14]. Some additional remarks on Holditch’s theorem can be found also in [1, 8, 9], and [18], and recent related investigations are given in [15, 16], and [10].

In this paper we modify the Holditch construction in which one ring domain is considered. In our modification we deal with a family of pairs of ring domains and obtain a natural geometric generalization. As an application we derive some Crofton-type formula for a ring domain.

We denote by \({\mathcal {C}}^*\) the family of all closed strictly convex curves of class \(C^{1}\). Let \(C\in {\mathcal {C}}^*\) and let p denote a fixed support function of C. The parametric representation of the curve C has the form

$$\begin{aligned} z\left( t\right) =p\left( t\right) e^{it}+{\dot{p}} \left( t\right) ie^{it}\qquad \text { for }\quad t\in \left[ 0,2\pi \right] , \end{aligned}$$
(1.1)

where the dot denotes differentiation with respect to t, see [2] and [17]. We denote by \({\mathcal {C}}\) a subfamily of \({\mathcal {C}}^*\) defined as follows: a curve \(C\in {\mathcal {C}}^*\) belongs to \({\mathcal {C}}\) if and only if the function \(R=p+\ddot{p}\) satisfies the inequality

$$\begin{aligned} R=p+\ddot{p}>0. \end{aligned}$$
(1.2)

Note that the function R is the curvature radius of C if the curve is of class \(C^2\).

We fix \(\alpha \in \left( 0,\pi \right) \), and we denote by \(z_{\alpha }\left( t\right) \) the intersection point of the tangent lines at \(z\left( t\right) \) and \(z\left( t+\alpha \right) \). A curve \(C_\alpha :t\rightarrow z_{\alpha }\left( t\right) \) is called an \(\alpha \)-isoptic, see Fig. 2.

Fig. 1
figure 1

Illustration of Holditch’s theorem

Fig. 2
figure 2

An \(\alpha \)-isoptic of the curve C

We will use the notations introduced in [6] and [7], namely

$$\begin{aligned} z_{\alpha }\left( t\right) =z\left( t\right) +\lambda \left( t,\alpha \right) ie^{it}=z\left( t+\alpha \right) +\mu \left( t,\alpha \right) ie^{i\left( t+\alpha \right) }, \end{aligned}$$
(1.3)

where

$$\begin{aligned} \lambda \left( t,\alpha \right) \sin \alpha= & {} p\left( t+\alpha \right) -p\left( t\right) \cos \alpha -{\dot{p}}\left( t\right) \sin \alpha , \end{aligned}$$
(1.4)
$$\begin{aligned} \mu \left( t,\alpha \right) \sin \alpha= & {} p\left( t+\alpha \right) \cos \alpha -{\dot{p}}\left( t+\alpha \right) \sin \alpha -p\left( t\right) . \end{aligned}$$
(1.5)

Moreover, if

$$\begin{aligned} b\left( t,\alpha \right)= & {} p\left( t+\alpha \right) \sin \alpha +{\dot{p}} \left( t+\alpha \right) \cos \alpha -{\dot{p}}\left( t\right) , \end{aligned}$$
(1.6)
$$\begin{aligned} B\left( t,\alpha \right)= & {} p\left( t\right) -p\left( t+\alpha \right) \cos \alpha +{\dot{p}}\left( t+\alpha \right) \sin \alpha , \end{aligned}$$
(1.7)

then we have

$$\begin{aligned} z\left( t\right) -z\left( t+\alpha \right)= & {} B\left( t,\alpha \right) e^{it}-b\left( t,\alpha \right) ie^{it}, \end{aligned}$$
(1.8)
$$\begin{aligned} \lambda \left( t,\alpha \right)= & {} b\left( t,\alpha \right) -B\left( t,\alpha \right) \cot \alpha , \end{aligned}$$
(1.9)
$$\begin{aligned} \mu \left( t,\alpha \right) \sin \alpha= & {} -B\left( t,\alpha \right) , \end{aligned}$$
(1.10)

see [7]. We denote by \(\xi _{\alpha }\left( t\right) \) the intersection point of the normal lines at \(z\left( t\right) \) and \(z\left( t+\alpha \right) \). We have

$$\begin{aligned} \xi _{\alpha }\left( t\right) =z\left( t\right) -f\left( t,\alpha \right) e^{it}=z\left( t+\alpha \right) -g\left( t,\alpha \right) e^{i\left( t+\alpha \right) }, \end{aligned}$$
(1.11)

see Fig. 3.

Fig. 3
figure 3

The functions f and g

Simple calculations lead us to the formulas

$$\begin{aligned} f\left( t,\alpha \right) \sin \alpha= & {} p\left( t\right) \sin \alpha +{\dot{p}} \left( t+\alpha \right) -{\dot{p}}\left( t\right) \cos \alpha , \end{aligned}$$
(1.12)
$$\begin{aligned} g\left( t,\alpha \right) \sin \alpha= & {} p\left( t+\alpha \right) \sin \alpha + {\dot{p}}\left( t+\alpha \right) \cos \alpha -{\dot{p}}\left( t\right) . \end{aligned}$$
(1.13)

Comparing (1.6) and (1.13), we get

$$\begin{aligned} g\left( t,\alpha \right) \sin \alpha =b\left( t,\alpha \right) . \end{aligned}$$
(1.14)

Moreover, it is easy to verify that

$$\begin{aligned} f\left( t,\alpha \right) =B\left( t,\alpha \right) +b\left( t,\alpha \right) \cot \alpha . \end{aligned}$$
(1.15)

2 Extremal chords

Let us fix a curve \(C\in {\mathcal {C}}\). We consider lengths of chords joining a fixed point \(z\left( t\right) \) and \(z\left( t+\alpha \right) \) for \(\alpha \in \left( 0, 2\pi \right) \). Let

$$\begin{aligned} H_{t}\left( \alpha \right) =\left| z\left( t\right) -z\left( t+\alpha \right) \right| \qquad \text {for }\quad \alpha \in \left( 0, 2\pi \right) \text {.} \end{aligned}$$
(2.1)

We denote by \(\left\langle -,-\right\rangle \) the Euclidean scalar product. Differentiating the function \(H_{t}\) given by the formula (2.1) and making use of (1.8) and (1.15), we obtain

$$\begin{aligned} H_{t}^{\prime }\left( \alpha \right) H_{t}\left( \alpha \right)= & {} \left\langle \frac{\partial }{\partial \alpha }(z\left( t\right) -z\left( t+\alpha \right) ) ,\;z\left( t\right) -z\left( t+\alpha \right) \right\rangle \\= & {} \left\langle -R\left( t+\alpha \right) ie^{i\left( t+\alpha \right) },\;B\left( t,\alpha \right) e^{it}-b\left( t,\alpha \right) ie^{it} \right\rangle \\= & {} R\left( t+\alpha \right) \left( B\left( t,\alpha \right) \sin \alpha +b\left( t,\alpha \right) \cos \alpha \right) \\= & {} R\left( t+\alpha \right) f\left( t,\alpha \right) \sin \alpha \end{aligned}$$

and

$$\begin{aligned} H_{t}^{\prime }\left( \alpha \right) =\frac{R(t+\alpha )}{H_{t} \left( \alpha \right) }f(t,\alpha )\sin \alpha . \end{aligned}$$
(2.2)

The above formula implies immediately the following statement.

If the chord joining a fixed point \(z\left( t\right) \)and a point \(z\left( t+\alpha \right) \)for some \(\alpha \in \left( 0, 2\pi \right) \)has maximal length, then the normal line at \(z\left( t+\alpha \right) \)intersects C at \(z\left( t\right) \).

Let

$$\begin{aligned} ww(C)=\min \limits _{t\in [0,2\pi ]}\max \limits _{\alpha \in [0,2\pi ]}H(t,\alpha ). \end{aligned}$$
(2.3)

Now we consider the particular but important case of ellipses.

Proposition 2.1

Let us fix an ellipse E, \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where \(a>b>0\). Then the maximal length of chords is given by the formula

$$\begin{aligned} {\left\{ \begin{array}{ll} 3^{\frac{3}{2}}\dfrac{a^2b^2}{(a^2+b^2)^{\frac{3}{2}}}, &{} \text {if } a>b\sqrt{2},\\ 2b, &{} \text {if } a\le b\sqrt{2}. \end{array}\right. } \end{aligned}$$
(2.4)

Proof

For a given point P(rs) of E we consider the normal line to E at P. This normal line intersects E at the second point \({\tilde{P}}({\tilde{r}}, {\tilde{s}})\), where

$$\begin{aligned} ({\tilde{r}}, {\tilde{s}})=\left( r\frac{a^6s^2-b^6r^2-2a^4b^2s^2}{b^6r^2+a^6s^2}, s\frac{b^6r^2-2a^2b^4r^2-a^6s^2}{b^6r^2+a^6s^2}\right) . \end{aligned}$$
(2.5)

Hence the distance between the points P and \({\tilde{P}}\) is \(d(P,{\tilde{P}})=2 (b^4r^2+a^4s^2)^\frac{3}{2}(b^6r^2+a^6s^2)^{-1}.\) Since \(a^2s^2=a^2b^2-b^2r^2\), it suffices to find the minimum of the function

$$\begin{aligned} d(r)=2b\frac{(a^4-(a^2-b^2)r^2)^\frac{3}{2}}{a^6-(a^4-b^4)r^2} \qquad \text { for } \quad r\in (-a,a). \end{aligned}$$

\(\square \)

Example

Let us fix an ellipse \(E, \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Each chord of the length m which slides around the curve E with endpoints on E determines some curve \(E_m\). We find the equation of the curve \(E_m\). For this aim we solve the system of equations

$$\begin{aligned} {\left\{ \begin{array}{ll} b^2x^2+a^2y^2-a^2b^2=0,\\ b^2ux+a^2vy-a^2b^2=0, \end{array}\right. } \end{aligned}$$

where the second equation represents the line dual to an exterior point (uv) with respect to E. Then its intersection points with E are

$$\begin{aligned}&(r, s)=\left( a^2\frac{b^2u+v\sqrt{a^2v^2+b^2u^2-a^2b^2}}{a^2v^2+b^2u^2}, b^2\frac{a^2v-u\sqrt{a^2v^2+b^2u^2-a^2b^2}}{a^2v^2+b^2u^2}\right) ,\\&({\tilde{r}}, {\tilde{s}})=\left( a^2\frac{b^2u-v\sqrt{a^2v^2+b^2u^2-a^2b^2}}{ a^2v^2+b^2u^2 },b^2\frac{a^2v+u\sqrt{a^2v^2+b^2u^2-a^2b^2}}{a^2v^2+b^2u^2}\right) . \end{aligned}$$

Since \(m^2=(r-{\tilde{r}})^2+(s-{\tilde{s}})^2\), we get

$$\begin{aligned} 4(b^2x^2+a^2y^2-a^2b^2)(b^4x^2+a^4y^2)-m^2(b^2x^2+a^2y^2)^2=0, \end{aligned}$$
(2.6)

where we substituted xy instead of uv, respectively.

3 Sliding a chord around a curve

We denote by \({\mathcal {N}}\) a family of functions \(\nu :[0, +\infty )\rightarrow {\mathbb {R}}\) of the class \(C^1(0, +\infty )\) satisfying the following conditions:

$$\begin{aligned}&{\dot{\nu }} >0 \quad \text { for } \quad t\in (0, +\infty ), \end{aligned}$$
(3.1)
$$\begin{aligned}&t<\nu (t)<t+\pi \quad \text { for }\quad t\in [0, +\infty ),\end{aligned}$$
(3.2)
$$\begin{aligned}&\nu (t+2\pi )=\nu (t)+2\pi \quad \text{ for } \quad t\in [0, +\infty ). \end{aligned}$$
(3.3)

Let C be a curve \(t\rightarrow z(t)\) given by (1.1), and \(\nu \in {\mathcal {N}}\) be a function. We associate with C a vector field Q along the curve C, defined as follows:

$$\begin{aligned} Q(t)=z(t)-z(\nu (t)). \end{aligned}$$
(3.4)

In view of (1.8) we have

$$\begin{aligned} Q(t)=B(t, \nu (t)-t)e^{it}-b(t,\nu (t)-t)ie^{it}. \end{aligned}$$
(3.5)

Differentiating (3.5) and using the formulas

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial b}{\partial \alpha } = R(t +\alpha )\cos \alpha , \\ \\ \frac{\partial B}{\partial \alpha } = R(t + \alpha )\sin \alpha , \end{array} \right. \qquad \begin{array}{l} \frac{\partial b}{\partial t} = B(t, \alpha ) + R(t + \alpha )\cos \alpha - R(t), \\ \\ \frac{\partial B}{\partial t} = - b(t, \alpha ) + R(t + \alpha )\sin \alpha \end{array} \end{aligned}$$
(3.6)

given in [7], we obtain

$$\begin{aligned} \dot{Q}={\dot{\nu }} R(\nu )\sin (\nu -t)e^{it}+(R-{\dot{\nu }} R(\nu )\cos (\nu -t))ie^{it}. \end{aligned}$$
(3.7)

We note that

$$\begin{aligned} \langle Q,{\dot{Q}}\rangle&=\dot{\nu }BR(\nu ) \cdot \sin (\nu -t) -\left( R-\dot{\nu }R(\nu ) \cdot \cos (\nu -t) \right) b \\&=\dot{\nu }R(\nu ) \left( B\sin (\nu -t) +b\cos (\nu -t) \right) -bR\\&=\left( \dot{\nu }R(\nu ) \cdot f(t, \nu -t)-Rg(t, \nu -t)\right) \sin (\nu -t). \end{aligned}$$

Hence we have equivalence of the following conditions:

$$\begin{aligned}&\langle Q,{\dot{Q}}\rangle =0, \end{aligned}$$
(3.8)
$$\begin{aligned}&{\dot{\nu }}=\frac{R(t)\cdot g(t, \nu -t)}{R(\nu )\cdot f(t,\nu -t)}. \end{aligned}$$
(3.9)

With respect to (3.9) we consider the implicit equation

$$\begin{aligned} p\left( t\right) \sin \Gamma +{\dot{p}} \left( t+\Gamma \right) -{\dot{p}}\left( t\right) \cos \Gamma =0. \end{aligned}$$

Differentiating the above equation and using the formulas (1.2) and (1.4), we get

$$\begin{aligned} (R(t+\Gamma )-\lambda \sin \Gamma ){\dot{\Gamma }}=\lambda \sin \Gamma -R(t+\Gamma )+R\cos \Gamma , \end{aligned}$$

and therefore

$$\begin{aligned} {\dot{\Gamma }}=\frac{R(t)\cos \Gamma }{R(t+\Gamma )-\lambda (t,\Gamma ) \sin \Gamma }-1. \end{aligned}$$
(3.10)

If the maximal width is attained at \(t=t_0\), then \(\Gamma (t_0)=\pi \). We note that \(\Gamma (t)>\frac{\pi }{2}\), since for the orthoptic curve we have \(f=-\mu \not =0\) and the considered function \(\Gamma \) is differentiable.

We associate with the curve C and \(\nu \in {\mathcal {N}}\) a curve \(C_\nu \) defined by

$$\begin{aligned} t\rightarrow w_\nu \left( t\right) =z\left( t\right) +\lambda \left( t,\nu \left( t\right) -t\right) ie^{it}\qquad \text { for }\quad t\in \left[ 0,2\pi \right] . \end{aligned}$$
(3.11)

Theorem 3.1

The integral formula

$$\begin{aligned} \int _{0}^{2\pi }\left[ w_\nu \left( t\right) ,\;{\dot{Q}}\left( t\right) \right] dt=0 \end{aligned}$$
(3.12)

holds, where \([a+bi, c+di]=ad-bc.\)

Proof

Let \(\alpha \left( t\right) =\nu \left( t\right) -t\). Using (1.10), (1.7), and (1.8), we obtain

$$\begin{aligned}&\int _{0}^{2\pi }\left[ w_\nu ,\;{\dot{Q}}\right] dt\\&\quad =\int _{0}^{2\pi }\left[ pe^{it}+\left( {\dot{p}}+\lambda \right) ie^{it},\dot{\nu }R(\nu )\cdot \sin \alpha e^{it}+\left( R-\dot{\nu }R(\nu )\cdot \cos \alpha \right) ie^{it}\right] dt \\&\quad =\int _{0}^{2\pi }\left( p\left( R-\dot{\nu }R(\nu )\cdot \cos \alpha \right) - \frac{p(\nu ) -p\cos \alpha }{\sin \alpha }\dot{\nu }R(\nu )\cdot \sin \alpha \right) dt \\&\quad =\int _{0}^{2\pi }\left( pR-\dot{\nu }R(\nu )\cdot p(\nu ) \right) dt=0 \text {.} \end{aligned}$$

\(\square \)

4 The main theorem

Now we assume that a chord of constant length m slides around with both endpoints on C which is given by formula (1.1). The endpoints of the sliding chord determine an increasing function \(\nu \in {\mathcal {N}}\). We assume that \( |z(0)-z(t_0)|=m \) for some \(t_0\in (0,2\pi ).\) Thus the function \(\nu \) satisfies the differential equation (3.9) with the initial condition \(\nu (0)=t_0\). For a fixed \(\xi \in [0,1]\) we consider a curve \(C(m,\xi )\) given by the formula

$$\begin{aligned} t\rightarrow v_\xi (t)=z(t)-\xi Q(t) \quad \text { for } \quad t\in [0,2\pi ]. \end{aligned}$$
(4.1)

Obviously, we have \(|Q|\equiv m\). We note that

$$\begin{aligned} \left[ v_\xi , \dot{v}_\xi \right] =[z, \dot{z}]-2\xi [z, \dot{Q}]+\xi [z,Q]^. +\xi ^2[Q, \dot{Q}]. \end{aligned}$$

Hence we get immediately

$$\begin{aligned} {\text {Area}}C- {\text {Area}}C(m,\xi ) = -\xi \int \limits _0^{2\pi }[{\dot{z}}, Q]dt-\pi \xi ^2m^2. \end{aligned}$$
(4.2)

We note that the graphs of the curves C(m, 0) and C(m, 1) coincide with the graph of C. Thus for \(\xi =1\) from (4.2) it follows immediately that \(\int \limits _0^{2\pi }[{\dot{z}}, Q]dt=-\pi m^2.\) Now formula (4.2) can be rewritten in the form

$$\begin{aligned} {\text {Area}}C- {\text {Area}}C(m,\xi )= \pi m^2\xi (1-\xi ). \end{aligned}$$
(4.3)

Letting \(m=a+b\) and \(\xi =\frac{a}{a+b}\), we get the well-known Holditch formula

$$\begin{aligned} {\text {Area}}C- {\text {Area}}C\left( a+b,\frac{a}{a+b}\right) =\pi ab. \end{aligned}$$
Fig. 4
figure 4

The curves C, \(C_{\nu ,\xi }\), \(C_{\nu }\), \(D_{\nu ,\gamma }\)

Now we associate with C and \(C_{\nu }\) a certain curve \(D_{\nu ,\gamma }\) defined as follows:

$$\begin{aligned} t\rightarrow v\left( t\right) =w_\nu \left( t\right) +\gamma Q\left( t\right) \text { for }t\in \left[ 0,2\pi \right] , \end{aligned}$$
(4.4)

where \(\gamma \) is a nonnegative constant.

Let us fix \(\xi \in (0,1)\). We consider a curve \(C_{\nu , \xi }\), \(t\rightarrow v_{\nu , \xi }(t)=(1-\xi )z(t)+\xi z(\nu (t))\) for \( t\in [0, 2\pi ]\), see Fig. 4.

Theorem 4.1

If \(C\in {\mathcal {C}}\), \(\nu \in {\mathcal {N}}\), \(\xi \in (0,1)\) and \(\gamma \not =0\), then the following formula holds:

$$\begin{aligned} \frac{{\text {Area}}D_{\nu ,\gamma }- {\text {Area}}C_{\nu }}{{\text {Area}}C-{\text {Area}}C_{\nu ,\xi }}=\frac{\gamma ^2}{ \xi (1-\xi )}. \end{aligned}$$
(4.5)

Proof

We have

$$\begin{aligned} \left[ v,{\dot{v}}\right] =\left[ w_\nu ,\dot{w_\nu }\right] -2\gamma \left[ w_\nu ,{\dot{Q}}\right] +\gamma \left[ w_\nu ,Q\right] ^{\cdot }+\gamma ^{2}\left[ Q,{\dot{Q}}\right] \text {.} \end{aligned}$$

Applying Theorem 3.1, we obtain

$$\begin{aligned} 2{\text {Area}}D_{\nu ,\gamma }-2{\text {Area}}C_{\nu }=\gamma ^2\int _{0}^{2\pi }\left[ Q\left( t \right) ,\;{\dot{Q}}\left( t\right) \right] dt. \end{aligned}$$
(4.6)

It was proved in [5] that

$$\begin{aligned} 2{\text {Area}}C-2{\text {Area}}C_{\nu ,\xi }=\xi (1-\xi ) \int _{0}^{2\pi }\left[ Q\left( t \right) ,\;{\dot{Q}}\left( t\right) \right] dt. \end{aligned}$$
(4.7)

Comparing the formulas (4.6) and (4.7), we get (4.5). \(\square \)

As corollaries of Theorem 4.1 we have the following Holditch-type formulas.

Corollary 4.2

If \(a+b<ww(C)\), then

$$\begin{aligned} {\text {Area}} D_{\frac{b}{a+b}}-{\text {Area}} C_{a+b}=\pi b^{2}. \end{aligned}$$
(4.8)
Fig. 5
figure 5

Curves C, \(C_m\), \(D_{\frac{1}{m} }\)

Corollary 4.3

If \(m<ww\left( C\right) \), then

$$\begin{aligned} {\text {Area}}D_{\frac{1}{m}}-{\text {Area}}C_{m}=\pi \text {.} \end{aligned}$$
(4.9)

Figure 5 illustrates Corollary 4.2

5 Crofton-type integral formulas

In this section we provide an interesting application of the developed theory and derive a new, geometrically justified Crofton-type formula.

Let us fix \(C\in {\mathcal {C}}\) and \(r<ww\left( C\right) \). The function \(\nu \left( t,m\right) \) determined by \(m\in \left( 0,r\right) \) satisfies the condition (3.4). Differentiating (3.4) with respect to m, we obtain

$$\begin{aligned} m=\left\langle Q,\frac{\partial Q}{\partial m}\right\rangle =-R(\nu ) \cdot \frac{\partial \nu }{\partial m}\left\langle Q,ie^{i\nu }\right\rangle =R(\nu ) \cdot \frac{\partial \nu }{\partial m}\left[ Q,e^{i\nu }\right] . \end{aligned}$$

Let \(\alpha \left( t,m\right) =\nu \left( t,m\right) -t.\) With respect to (3.5) and (1.15) we have

$$\begin{aligned} \left[ Q,e^{i\nu }\right] =\left[ Be^{it}-bie^{it},\cos \alpha \cdot e^{it}+\sin \alpha \cdot ie^{it}\right] =B\sin \alpha +b\cos \alpha =F\sin \alpha . \end{aligned}$$

The above calculations imply that

$$\begin{aligned} \frac{\partial \nu }{\partial m}=\frac{m}{R(\nu ) \cdot F\sin \alpha }. \end{aligned}$$
(5.1)

Now, we consider the ring domain \(CC_{r\text { }}\), and we introduce the notations as in Fig. 6, maintaining at the same time the notations of Santaló from [17].

Fig. 6
figure 6

Notations for Theorem 5.1

Let \(R_{1}(x,y)\), \(R_{2}(x,y)\) denote the radii of curvature of C at the tangent points \(A_{1}\), \(A_{2}\), respectively.

Crofton proved the integral formula

$$\begin{aligned} \iint \limits _{{\text {ext}}C}\frac{\sin \omega }{t_{1}t_{2}}dxdy=2\pi ^{2}, \end{aligned}$$

where \({\text {ext}}C\) denotes the exterior of C, see [13]. We will prove some Crofton-type theorem, namely

Theorem 5.1

If \(C\in {\mathcal {C}}\) and \(r<ww\left( C\right) \), then the following integral formula holds:

$$\begin{aligned} \iint \limits _{CC_{r}}R_{2}F_{1}\frac{\sin ^{3}\omega }{t_{1}t_{2}}dxdy=\pi r^2. \end{aligned}$$
(5.2)

Proof

We consider a mapping \(T:\left( 0,2\pi \right) \times \left( 0,r\right) \rightarrow \) interior of \(CC_{r\text { }}\backslash \){some segment} defined by the formula

$$\begin{aligned} T\left( t,m\right) =z\left( t\right) +\lambda \left( t,\nu \left( t,m\right) -t\right) ie^{it}. \end{aligned}$$

We note that T is a bijection and

$$\begin{aligned} \frac{\partial T}{\partial m}=\frac{\partial \lambda }{\partial m}ie^{it}, \\ \frac{\partial T}{\partial t}=-\lambda e^{it}+hie^{it}\,, \end{aligned}$$

where h is some function. Thus the Jacobian JT of T at \(\left( t,m\right) \) is given by the formula

$$\begin{aligned} JT\left( t,m\right) =\left[ \frac{\partial T}{\partial t}\left( t,m\right) ,\frac{\partial T}{ \partial m}\left( t,m\right) \right] =-\lambda \left( t,m\right) \frac{\partial \lambda }{\partial m}\left( t,m\right) . \end{aligned}$$

On the other hand, since \( \alpha \left( t,m\right) =\nu \left( t,m\right) -t\), so \(\frac{\partial \nu }{\partial m}=\frac{\partial \alpha }{\partial m}\) and

$$\begin{aligned} \sin ^{2}\alpha \frac{\partial \lambda }{\partial m}= & {} \sin ^{2}\alpha \frac{\partial }{\partial m}\left( \frac{p(\nu ) -p\cos \alpha }{\sin \alpha }-{\dot{p}}\right) \\= & {} \left( {\dot{p}}(\nu ) \cdot \frac{\partial \nu }{\partial m}+p\sin \alpha \frac{\partial \alpha }{\partial m}\right) \sin \alpha -\left( p(\nu ) -p\cos \alpha \right) \cos \alpha \frac{\partial \alpha }{\partial m} \\= & {} \left( {\dot{p}}(\nu )\cdot \sin \alpha -p(\nu ) \cdot \cos \alpha +p\right) \frac{\partial \nu }{\partial m} \\= & {} -\mu \frac{\partial \nu }{\partial m}=\frac{-\mu m}{R(\nu ) \cdot F\sin \alpha }. \end{aligned}$$

Thus the Jacobian of T at \(\left( t,m\right) \) has the form

$$\begin{aligned} JT\left( t,m\right) =\frac{-\mu \lambda m}{R(\nu ) \cdot F\sin ^{3}\alpha }. \end{aligned}$$

Now we have

$$\begin{aligned} \iint \limits _{CC_{r\text { }}}R_{2}F_{1}\frac{\sin ^{3}\omega }{t_{1}t_{2}} dxdy=\int \limits _{0}^{2\pi }\int \limits _{0}^{r}R(\nu )\cdot F\frac{\sin ^{3}\alpha }{-\mu \lambda }\frac{-\mu \lambda m}{R(\nu ) \cdot F\sin ^{3}\alpha }dmdt=\pi r^2. \end{aligned}$$

\(\square \)