Calculate the intersection area of a spherical shell segment and a plane I'd like to calculate the intersection of a conical segment of a spherical shell with a plane. Here is a 2D cut of the situation:
Sketch of a 2D cut of the intersection of interest
That is, the volume object should be defined by an intersection of a cone with a spherical shell. The cone has a (2D) opening angle of $\phi$. The spherical shell is the region of a ball between two concentric spheres of radii $r - \Delta r/2$ and $r + \Delta r/2$. The sketch shows a 2d cut through this object as a blue area.
This object should now be cut with a plane that is perpendicular to the radius, with a distance of $d$ to the origin of the spheres. This area is indicated by the red dashed line in the sketch. For the shown situation, we would try to find the area of the solid red line (rotated around the axis aligned with the green line).
I'd like to find an equation for the intersection area as a function of the shown parameters, i.e. $A_{\mathrm{intersection}}(r, \Delta r, \phi, d)$.
I am unsure how to approach this problem, as there are special cases like the one shown, where the intersection area is a ring, while for smaller values of $d$ it would first be a circle.
I can define the volume of the object as an integral expression like this:
$$
\int_0^{2\pi}\int_0^{\phi}\int_{r-\Delta r/2}^{r+\Delta r/2} R^2\sin(\varphi)\,\mathrm{d}R\,\mathrm{d}\varphi\,\mathrm{d}\theta
$$
But I am unable to approach the intersection with an integral expression. Any help appreciated!
 A: 
This object should now be cut with a plane that is perpendicular to the radius, with a distance of $d$ to the origin of the spheres.

In that case, the intersection is an annular ring.  (Note, your diagram has $r$ and $d$ swapped.)
When a plane intersects a ball of radius $R$ at (minimum) distance $d$, $d \le R$, the intersection is a circular disc with radius $\sqrt{R^2 - d^2}$ (based on Pythagorean theorem; $R$ being the hypotenuse and $d$ a side of the right triangle).  Because the area of a circular disc with radius $r$ is $\pi r^2$, the area of the intersection is $\pi\left(\sqrt{R^2 - d^2}\right)^2 = \pi\left(R^2 - d^2\right)$.
Let's use $r_1$ for the outer radius, and $r_2$ for the inner radius, i.e. $r_1 = r + \Delta r / 2$, $r_2 = r - \Delta r / 2$, $r_1 \ge r_2$.
This means that the area of intersection $A$ between a plane and a spherical shell of outer radius $r_1$, inner radius $r_2$, at (minimum) distance $d$, is
$$A = \begin{cases}
0, & d \ge r_1 \\
\pi \left(\sqrt{r_1^2 - d^2}\right)^2, & r_2 \le d \le r_1 \\
\pi \left(\sqrt{r_1^2 - d^2}\right)^2 - \pi \left(\sqrt{r_2^2 - d^2}\right)^2, & 0 \le d \le r_2 \\
\end{cases}$$
i.e.
$$A = \begin{cases}
0, & r_1 \le d \\
\pi \left( r_1^2 - d^2 \right), & r_2 \le d \lt r_1 \\
\pi \left( r_1^2 - r_2^2 \right), & 0 \le d \lt r_2 \\
\end{cases}$$
Note that the relative orientation between the plane and the sphere (spherical shell) does not matter.  The distance between a point and the plane is always measured perpendicular to the plane, so no matter the orientation, $d$ is measured perpendicular to the plane, and the above applies.

If the axis of the cone is perpendicular to the plane, we can easily adjust the above to work for the conical section of a spherical shell as well.
[Note: The following has been edited to correct the expressions. In particular, the radius of the intersection between a right circular cone with angle $\phi$ and a plane perpendicular to the cone axis at distance $d$ from the apex is $d \tan \phi$, and not $d \sin \phi$ as in my earlier version.  Apologies for the error.]
Consider the following diagram, with $r_1$ and $r_2$ the outer and inner radius of the spherical shell, and $d_1$ and $d_2$ the perpendicular distances where the cone surface intersects the outer and inner surfaces of the shell,
$$\begin{aligned}
d_1 &= r_1 \cos \phi \\
d_2 &= r_2 \cos \phi \\
\end{aligned}$$

We have quite a few regions to handle here, depending on the plane distance $d$ from the sphere center/cone apex:

*

*$r_1 \lt d$: The plane is too far from the sphere to intersect.


*$r_1 = d$: The plane intersects the sphere at a single point.


*$d_1 \le r_2$: Left side of the diagram.

*

*$r_2 \le d \lt r_1$: Top interval on the left side.
The intersection is a circular disc with radius $\sqrt{r_1^2 - d^2}$.


*$d_1 \le d \lt r_2$: Second interval from top on the left side.
The intersection is an annular ring with outer radius $\sqrt{r_1^2 - d^2}$, and inner radius $\sqrt{r_2^2 - d^2}$, so the area of the intersection is therefore constant: $\pi \bigl( (r_1^2 - d^2) - (r_2^2 - d^2) \bigr) = \pi r_1^2 - \pi r_2^2$.


*$d_2 \lt d \lt d_1$: Third interval from top on the left side.
The intersection is an annular ring with outer radius $d \tan \phi$, and inner radius $\sqrt{r_2^2 - d^2}$.


*$d_2 = d$: The intersection is circle with zero area.


*$d \lt d_2$: The plane is too close to the sphere center or cone apex to intersect.




*$r_2 \lt d_1$: Right side of the diagram.

*

*$d_1 \le d \lt r_1$: Top interval on the right side.
The intersection is a circular disc with radius $\sqrt{r_1^2 - d^2}$.


*$r_2 \le d \lt d_1$: Second interval from top on the right side.
The intersection is a circular disc with radius $d \tan \phi$.


*$d_2 \lt d \lt r_2$: Third interval from top on the right side.
The intersection is an annular ring with outer radius $d \tan \phi$, and inner radius $\sqrt{r_2^2 - d^2}$.


*$d = d_2$: The intersection is a circle with zero area.


*$d \lt d_2$: The plane is too close to the sphere center or cone apex to intersect.
Therefore, the area of intersection $A$ can be defined as a function of plane distance $d$ as
$$A = \begin{cases}
0, & r_1 \le d \\
\pi \left(r_1^2 - d^2\right), & \max\left(r_2, d_1\right) \le d \lt r_1 \\
\pi \left(d^2 \tan^2 \phi\right), & r_2 \le d \le d_1 \\
\pi \left( r_1^2 - r_2^2 \right), & d_1 \le d \lt r_2 \\
\pi \left( d^2 \tan^2 \phi - r_2^2 + d^2 \right), & d_2 \lt d \lt \min\left(r_2, d_1\right) \\
0, & d \le d_2 \\
\end{cases}$$
