I need to find the analytical solution when the plane $ P: z = grad\cdot y + z_{cut} $ cuts the hemi-sphere $ S: x^2 + y^2 + z^2 = r^2;\:y \leq 0 $. I constructed two 3D images in MatLab of the problem statement:

hemi-sphere sliced by a plane volume to be determined

The first image explicitly shows how the plane cuts the hemi-sphere, and the second image shows the volume that is desired to be known, i.e. the volume ($V$) included between the green shaded surfaces and the $XZ$-plane at $y = 0$.

For clarity, I will state the obvious: The plane may be thought of as a 2D line drawn on the $YZ$-axis and then symmetrically extruded in the $X$-direction. With this analogy, $z_{cut}$ is the $z$-coordinate where the plane cuts the $XZ$-plane at $y = 0$ and $grad$ is the gradient of the plane (with respect to the $YZ$-axis).

Now, $V$ can be divided into two smaller volumes:

  1. The volume $V_{horis}$, obtained when the plane $z = z_{cut}$ cuts the hemi-sphere. This volume is easy to calculate, since it is the volume of a spherical cap at $h = z_{cut} + r$. The formula is $V_{horis} = \frac{1}{2}[\frac{1}{3}\Pi h^2(3r-h)]$, and can be found on mathworld.wolfram.com/SphericalCap.html

  2. The remaining volume is $V_{slant}$. Realise the following: (a) $V_{slant}$ is not a wedge like the ones used for a door stopper, since it's "left" and "right" sides are part of spherical curvature; (b) $V_{slant}$ is not formed by a radius originating from $z_{cut}$ (I therefore did not use polar or spherical coordinates); (c) symmetry is not in our favour.

The last point (c) might not be true. However, up and till now, I could not see how to use symmetry. In fact, I attempted the construction of a symmetrical solution. Specifically, I completed the hemi-spherical drawing by extending the line into the RHS hemi-sphere. See the 2D sketch below (the $Y$-axis should actually be a $Z$-axis).

2D symmetry attempt

From the sketch, one might get the impression that $V_1 = V_2 = V_{slant}$. If this is the case, then a solutioin based on symmetry might be constructed. But this seems to not be the case, since these two figures are not even congruent in 2D. I use the word "seems", since congruent figures will have the same area and possibly volume, but figures with the same area or volume are not necessarily congruent. By the cosine formula:
\begin{eqnarray} r^2 = a^2 + z_{cut}^2 - 2az_{cut}\cos{(90^{\circ} + \theta)}\\ r^2 = b^2 + z_{cut}^2 - 2bz_{cut}\cos{(90^{\circ} - \theta)} \end{eqnarray}

The above simplifies to: \begin{equation} b - a = 2z_{cut}sin{(\theta)} \neq 0 \end{equation}

This does not necessarily mean that $V_1 \neq V_2$, but it does causes disbelief.

To determine the volume ($V$) analytically I took the following approach: Determine $V_{slant}$ by constructing discs (with thickness $\Delta z$) parallel to the $XY$-plane and then sum these discs from $z_{cut}$ to $z_{upper}$ as $\Delta z \rightarrow 0$ ($z_{upper}$ is the $z$-coordinate where the plane cuts the hemi-sphere at $x = 0$).

Consider a horisontal disc (i.e. parallel to the $XY$-plane) at $z=k$. The integration boundaries to compute the area of this disc is given as:

\begin{eqnarray} \begin{aligned} -\sqrt{r^2 - y^2 - k^2} \leq & x \leq \sqrt{r^2 - y^2 - k^2}\\ \sqrt{r^2 - k^2} \leq & y \leq \frac{k-z_{cut}}{grad} \end{aligned} \end{eqnarray}

Hence, \begin{equation} A_k = \int_{-\sqrt{r^2 - k^2}}^{\frac{k-z_{cut}}{grad}}\int_{-\sqrt{r^2 - y^2 - k^2}}^{\sqrt{r^2 - y^2 - k^2}}dxdy \end{equation}

Since z is a variable in the equation of the sphere, $A_k$ changes with different $z=k$, thus \begin{eqnarray} \begin{aligned} A(z) &= \int_{-\sqrt{r^2 - z^2}}^{\frac{z-z_{cut}}{grad}}2\sqrt{r^2 - y^2-z^2}dy\\ &= \bigg[y\sqrt{(r^2 - z^2) - y^2} + (r^2 - z^2)\arcsin{\Big(\frac{y}{\sqrt{r^2 - z^2}}\Big)}\bigg]_{-\sqrt{r^2 - z^2}}^{\frac{z-z_{cut}}{grad}}\\ &= \Big(\frac{z - z_{cut}}{grad}\Big)\sqrt{(r^2 - z^2) - \Big(\frac{z - z_{cut}}{grad}\Big)^2} + (r^2 - z^2)\arcsin{\Bigg(\frac{\frac{z - z_{cut}}{grad}}{\sqrt{r^2 - z^2}}\Bigg)} + \frac{\pi}{2}(r^2 - z^2) \end{aligned} \end{eqnarray}

I tried to integrate the above equation with respect to $z$ in both MatLab and on the Wolfram Alpha website, but both programs gave complex answers. This might be a result of the domain constraint on arcsin between -1 and 1. Whatever the case might be, volume is physically still a real value in $[units]^3$.

Does any one know whether there is either:

  1. An analytical solution to compute \begin{equation} V_{slant} = \int_{z_{cut}}^{z_{upper}}A(z)dz, \end{equation} or

  2. A more elegant solution to the find the volume $V$?


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