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\appendix
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\footnotesize
%------------------
\section {Appendix}
%------------------
%-------------------------------
\subsection {Integration limits}
%-------------------------------

In this section, we show how the limits for the integrations along
the rays of a constant impact parameter $p=r~(1-\mu^2)^{1/2}$ can
be modified to account for the flared disk geometry.

Let us consider a midplane point $P$ or a boundary point $U$ lying
at the radial distance $r$ from the central star in the shaded
two-sector area shown in Fig.~2. To find the radiation moments
(19), (20), (21)
at one of the points, it is necessary to integrate the
radiative transfer equations (2) and (3)
over all rays passing through that point, for only those parts of
the rays which lie inside the cross-sectional disk area which we
describe by the angle $\alpha = \angle\:BOC.$ Each ray may have in
general from 2 to 4 intersection points with the disk boundaries.
We have to repeat the ray tracing for a set of such cross-sections
of different inclinations $\phi$ to the midplane, from $\psi/2$ to
 $\pi/2.$ An integration over all $\phi$ completes the radiation
moments at the given midplane or boundary point.

%-------------------------------
\subsubsection {Midplane points}
%-------------------------------
\label {midplane}

For a point $P$ in the disk midplane (Fig.~2),
 $\alpha$ is related to the opening angle $\psi$ and to the midplane
azimuthal angle $\phi = \angle\:EBC + \psi/2$ between the
cross-section and the midplane by
\be
  \alpha = \arcsin \Bigl[ \frac {\sin (\psi/2)}
                      {\sin \phi} \Bigr]
\label {alpha}
\,.
\ee
 The angle $\alpha$ decreases from $\pi/2$ to $\psi/2$ for
 $\phi \in [\psi/2, \pi/2].$ A symmetry of the point $P$ position in the
cross-section implies that we need to integrate over $\mu$ only from 0
to 1, as in spherical geometry. Therefore, we have to find five
intersection points with the lines $GH,\,OH,\,OCB,\,OA,$ and $AB$
(Fig.~2); one can derive the corresponding radii
\bea
   r_{\rm GH}\!&=&\! R_2, \qquad\qquad\quad\,\,\,
                        \theta \leq \arctan \,\xi
\,,
\\
   r_{\rm OH}\!&=&\! \frac {p} {\sin (\alpha-\theta)},
               \qquad \theta < \arctan \,\xi
\,,
\\
   r_{\rm OCB}\!&=&\! \min \Bigl[ \frac {p}
                  {\sin (\alpha+\theta)},\,R_2 \Bigr]
\,,
\\
   r_{\rm OA}\!&=&\! \frac {p} {\sin (\theta-\alpha)},
               \qquad \theta > \arctan \,\zeta
\,,
\\
   r_{\rm AB}\!&=&\! R_2, \qquad\qquad\quad\,\,\,
                        \theta \leq \arctan \,\zeta
\,,
\eea
 where we have defined
\bea
   \xi = \frac {R_2 \sin \alpha} {r+R_2 \cos \alpha}
\,,\qquad
 \zeta = \frac {R_2 \sin \alpha} {-r+R_2 \cos \alpha}
\,.
\eea

For a given impact parameter $p,$ one should integrate  $I_\nu^{+}$
along the ray, from $r_{\rm GH}$ to $r_{\rm OH}$ and from
 $r_{\rm OCB}$ to the radial position $r$ of the point $P.$ The
inward directed intensity $I_\nu^{-}$ should be integrated from
 $r_{\rm AB}$ or $r_{\rm OA}$ (depending on the ray direction
 $\theta)$ to $r.$

%-------------------------------
\subsubsection {Boundary points}
%-------------------------------
\label {boundary}

For a point $U$ in the disk's upper (or lower) boundary
(Fig.~2), $\alpha$ is defined by Eq.~(\ref {alpha}), while the
boundary azimuthal angle $\phi_{\rm b} = \angle\:BCT$ is related
to $\psi$ and $\phi$ by
\be
  \phi_{\rm b} = \arccos \Bigl[ \frac {\cos \phi}
                               {\cos (\psi/2)} \Bigr]
\,.
\ee
The angle $\phi_{\rm b}$ increases from 0 to $\pi/2$ for
 $\phi \in [\psi/2, \pi/2].$ For the point $U$ position in the cross-section,
there is no symmetry of the latter about the $\mu = 0$ direction.
Therefore, we have to integrate the intensities over $\mu$ from $-1$
to 0 and from 0 to 1, in each interval separately.

 (1). For $0 < \mu < 1,$ we have to find five intersection points
with the lines $FH,\,OH,\,OC,\,OA,$ and $AC$ (Fig.~2);
one can derive the corresponding radii
\bea
   &r_{\rm FH} =&\! R_2, \qquad\qquad\quad\,\,\,\,\,\,
                        \theta \leq \arctan \,\xi_1
\,,
\\
   &r_{\rm OH} =&\! \frac {p} {\sin (2\alpha-\theta)},
               \qquad \theta < \arctan \,\xi_1
\,,
\\
   &r_{\rm OC} =&\! r
\,,
\\
   &r_{\rm OA} =&\! \frac {p} {\sin (\theta-2\alpha)},
               \qquad \theta > \arctan \,\zeta_1
\,,
\\
   &r_{\rm AC} =&\! R_2, \qquad\qquad\quad\,\,\,\,\,\,
                        \theta \leq \arctan \,\zeta_1
\,,
\eea
 where we have defined
\bea
   \xi_1 = \frac {R_2 \sin (2\alpha)} {r+R_2 \cos (2\alpha)}
\,,\qquad
 \zeta_1 = \frac {R_2 \sin (2\alpha)} {-r+R_2 \cos (2\alpha)}
\,.
\eea

For a given impact parameter $p,$ one should integrate  $I_\nu^{+}$
along the ray, from $r_{\rm FH}$ to $r_{\rm OH}.$ The inward
directed intensity $I_\nu^{-}$ should be integrated from the radial
position $r_{\rm OC}$ of the point $U$ to $r_{\rm OA}$ or
 $r_{\rm AC}$ (depending on $\theta).$

 (2). For $-1 < \mu < 0,$ we have to find five intersection points
with the lines $AC,\,OA,\,OC,\,OH,$ and $FH$ (Fig.~2);
the latter two intersections exist only for the cross-sections with
 $2\alpha > \pi/2.$ The corresponding radii are
\bea
   &r_{\rm AC} =&\! R_2, \qquad\qquad\quad\,\,\,\,\,\,
                        \theta \geq \arctan \,\xi_2
\,,
\\
   &r_{\rm OA} =&\! \frac {p} {\sin (\theta-2\alpha)},
               \qquad \theta < \arctan \,\xi_2
\,,
\\
   &r_{\rm OC} =&\! r
\,,
\\
   &r_{\rm OH} =&\! \frac {p} {\sin (2\alpha-\theta)},
               \qquad \theta > \arctan \,\zeta_2
\,,
\\
   &r_{\rm FH} =&\! R_2, \qquad\qquad\quad\,\,\,\,\,\,
                        \theta \leq \arctan \,\zeta_2
\,,
\eea
 where we have defined
\bea
   \xi_2 = \frac {R_2 \sin (2\alpha)} {r-R_2 \cos (2\alpha)}
\,,\qquad
 \zeta_2 = \frac {R_2 \sin (2\alpha)} {-r-R_2 \cos (2\alpha)}
\,.
\eea

For a given impact parameter $p,$ one should integrate $I_\nu^{+}$
along the ray, from $r_{\rm AC}$ or $r_{\rm OA}$ (depending on
 $\theta)$ to the radial position $r_{\rm OC}$ of the point $U.$
The inward directed intensity $I_\nu^{-}$ should be integrated from
$r_{\rm FH}$ to $r_{\rm OH}.$

%-------------------------------------------
\subsubsection {Emergent intensity and flux}
%-------------------------------------------

To calculate the model emergent spectrum, we have to specify the
direction of the line of sight which we describe by the viewing angle
 $\theta_{\rm v}$ relative to the midplane of the disk; it may have
any value between 0 and $\pi/2.$ We define a set of the disk
cross-sections by different planes containing the line of sight and
passing through the stellar center (for an example of the disk
cross-section, see Fig.~2). A separate ray tracing
in the cross-section areas along this direction gives us the
emergent intensity distribution; integrating the latter over the
disk projection (over the azimuthal angle $\phi_{\rm v})$ we get
the flux.

The angle $\phi_{\rm v}$ is defined using the angle $\phi$ between
the cross-section and the disk midplane as
\bea
    \phi_{\rm v} = \arccos \Bigl[ \frac {\cos \phi}
                                 {\cos \theta_{\rm v}} \Bigr]
\,.
\eea
The angle $\alpha$ describing the cross-section geometry is given
by Eq.~(\ref {alpha}). An expression for the angle $\omega$ between
the viewing direction and the line $GB$ in the disk midplane
(Fig.~2) depends on whether $\theta_{\rm v}$ is smaller or
larger than $\psi/2.$

 (1). For $\theta_{\rm v} \leq \psi/2, \,\phi \in [\psi/2, \pi/2]$
and one can see that $\alpha \in [\pi/2, \psi/2];$ in this case,
 $\,\omega$ is defined by the expression
\be
    \omega = \arcsin \Bigl[ \frac {\sin \theta_{\rm v}}
                                 {\sin \phi} \Bigr]
\,.
\ee

 (2). For $\theta_{\rm v} > \psi/2, \,\phi \in [\theta_{\rm v},
\pi/2]$ and one can easily verify that $\alpha \in [\alpha_{\rm max},
\psi/2];$ the maximum angle $\alpha_{\rm max}$ of the
cross-sectional area is defined by Eq.~(\ref {alpha}) with
 $\phi = \theta_{\rm v}.$ In this case, $\omega$ can be computed from
the formula
\be
    \omega = \arccos \Bigl[ \cos \theta_{\rm v} \,\frac
                {\sin \phi_{\rm v}} {\sin \phi} \Bigr]
\,.
\ee
Defining appropriate values of the ray direction $\theta$ for
the midplane or boundary points (as a function of $\alpha$ and
 $\omega),$ it is possible to use the same expressions for the
integration limits (Sects.~\ref {midplane},\,\ref {boundary}) to
calculate emergent intensity and flux. Of course, it is necessary
to integrate $I_\nu^{+}$ over the whole body of the disk, from the
``background'' boundaries to those seen by the observer.

\end {document}
