Suppose that you initially have a container which has the shape of a paraboloid with equation
$ z = a x^2 + b y^2 $ where $ 0 \le z \le h $
Now you tilt this paraboloid by rotating it about any point (let's say the origin) about a horizontal axis whose unit vector is $v = (\cos \varphi, \sin \varphi, 0)$. Let the angle of rotation be $\theta$. What will be the volume of water that will fill the paraboloidal container, up to the the point on its rim with the minimum elevation (minimum $z$ coordinate) ?
My attempt:
Defining $r = [x,y,z]^T$ , the initial equation of the un-tilted paraboloid is
$ r^T Q_0 r + b_0^T r = 0 $
where $Q_0 = \begin{bmatrix} a && 0 && 0 \\ 0 && b && 0 \\ 0 && 0 && 0 \end{bmatrix} $ and $b_0 = \begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix} $
Rotating about the given axis which passes through the origin, results in the image of a point $r$ on the surface of the paraboloid becoming the point $r' = R r $, so that $ r = R^T r' $, therefore the equation of the tilted paraboloid is
$ r'^T R Q_0 R^T r' + b_0^T R^T r' = 0 $
But continuing from here seems complicated.
So instead of tilting the paraboloid. I thought I could compute the final orientation of the horizontal water surface relative to the paraboloid. The equation of the $xy$ plane is
$ z = 0 $ i.e. $n^T r = 0$ where $n = [0,0,1]$.
When rotating this plane about the given axis by an angle $-\theta$, the effect is rotating the normal vector. The rotated vector is given by the Rodrigues formula as
$ n' = \cos \theta n + (1 - \cos \theta) v (v^T n) + \sin (-\theta) (v \times n) $
Substituing $n$ and $v$, and noting that $v^T n = 0$, $ v \times n = [ \sin \varphi , -\cos \varphi , 0]^T $
$ n' = [n_1, n_2, n_3]^T = \begin{bmatrix} -\sin \theta \sin \varphi \\ \sin \theta \cos \varphi \\ \cos \theta \end{bmatrix} $
Now the rotated $xy$ plane shifted up by $ S = \begin{bmatrix} 0 \\ 0 \\ s \end{bmatrix} $ is given by
$ n_1 x + n_2 y + n_3 (z - s) = 0 $
We need to determine $s$ such that this plane intersecting the rim of the paraboloid at one point only.
The rim equation is
$ h = a x^2 + b y^2 , z = h $
And the plane intersection with $ z = h$ is
$ n_1 x + n_2 y + n_3 (h - s) = 0 $
We want this line to be tangent to $ a x^2 + b y^2 = h $
The normal vector to this ellipse is $u = [2 a x , 2 b y ] $, hence the constraint now is
$ n_1 b y - n_2 a x = 0 $
i.e. $ y = \dfrac{ n_2 a }{n_1 b} x $
Substituting in the ellipse equation,
$ a x^2 + \dfrac{ n_2^2 a^2} {n_1^2 b} x^2 = h $
So that $x = \pm \sqrt{ \dfrac{h b n_1^2} { n_1^2 a b + n_2^2 a^2 } } $
Having obtained $x$, and consequently $y$, we can now compute $s$.
Namely, from
$ n_1 x + n_2 y + n_3 (h - s) = 0 $
we have
$ s = h + \dfrac{n_1 x + n_2 y}{n_3} $
And we want to select the smaller of the two possible values.
What remains is to compute the volume integral of the region bounded by the paraboloid from below and the plane from above.
IMHO the best option needs to determine the projection of the actual intersection of the plane and the paraboloid (which is an ellipse) and to compute the integrals in the interior of its projection on $z=0$.
This is quite straightforward once you have determined $s$ because the projection of the ellipse has equation $$ ax^2+by^2=s-\frac{n_1x+n_2y}{n_3}, $$ which gives in (almost) canonical form $$ \left(\sqrt{a}x+\frac{n_1}{2n_3\sqrt{a}}\right)^2+\left(\sqrt{b}y+\frac{n_2}{2n_3\sqrt{b}}\right)^2 = s+\left(\frac{n_1}{2n_3\sqrt{a}}\right)^2+\left(\frac{n_2}{2n_3\sqrt{b}}\right)^2\equiv R^2, $$ and the interior of the domain $D$ can be parametrized with new $(r,t)$ coordinates $$ \begin{align*} \sqrt{a}x+\frac{n_1}{2n_3\sqrt{a}} &= r\cos t,\\ \sqrt{b}y+\frac{n_2}{2n_3\sqrt{b}} &= r\sin t, \end{align*} $$ for $t\in[0,2\pi]$ and $r\in[0,R]$. Then after the change of variables the two integrals $$ I_1=\iint_D (ax^2+by^2)\,dx\,dy, \quad I_2=\iint_D \left(s-\frac{n_1x+n_2y}{n_3}\right)dx\,dy $$ can be easily computed by hand (integral below the plane is the area of the ellipse times the height at the center): $$ I_1=\frac{\pi}{2\sqrt{ab}}R^2\left(3R^2-2s\right), $$ $$ I_2=\frac{\pi}{\sqrt{ab}}R^2\left(2R^2-s\right), $$ and the expected volume is $$ I_2-I_1 = \frac{\pi}{2\sqrt{ab}}R^4. $$