Tilted paraboloidal cup - new water depth

Align the tumbler axis along the $x$ -axis. The parabola is of the form $z^2 = ka$ . Let $x_0$ be the $x$ -coordinate of the bottom of the tumbler. Then

$\begin{cases} 2^2 = kx_0 \\ 4^2 = k(x_0 + 10) \end{cases} \implies k = 1.2$ and the equation of the parabola is $z^2 = 1.2x$ and $x_0 = \frac {10}3$ .

We note that the radius $r(x)$ of the cross-sectional circle is given by the parabola $r(x) = \sqrt{1.2 x}$ and the volume of half-glass of water is given by:

$\begin{aligned} V_\frac 12 & = \int_{x_0}^{x_0+5} \pi r^2\ dx = \int_{\frac {10}3}^{\frac {25}3} 1.2 \pi x \ dx = 0.6\pi x^2 \ \bigg|_{\frac {10}3}^{\frac {25}3} = 35\pi \approx 109.9557429 \end{aligned}$

The water surface in the tumbler is an ellipse. Align the major semiaxis in the $xz$ -plane. Let the $x$ -coordinates of the ends near and farther away from the $z$ -axis be $a$ and $b$ respectively. For any $a$ , the volume of water in the tumbler has two parts, for $x < a$ , $V_1(a) = 0.6\pi \left(a^2 - \dfrac {100}9 \right)$ volume of the paraboloidal frustrum, and for $x > a$ , $V_2$ volume of a "wedge" whose element $\delta x$ has a cross-sectional of a major circle segment of radius $r(x)$ , center on the $x$ -axis and its top edge at a height $h(x)$ from the center. The cross-sectional area is given by $A= \left(\pi - \cos^{-1} \left(\dfrac hr \right) \right) r^2 + h \sqrt{r^2-h^2}$ and the volume of the wedge is given by:

$V_2(a) = \int_a^b \left(\left(\pi - \cos^{-1} \left(\dfrac hr \right) \right) r^2 + h \sqrt{r^2-h^2}\right) dx$

For a given $a$ , $h(x)$ is given by the blue line or $h(x) = \sqrt 3 (a-x) + \sqrt{1.2a}$ and $b = \dfrac {5a+\sqrt{40a}+2}5$ . With $r(x)$ , $h(x)$ , $a$ , and $b$ known, we can find $V_2(a)$ by integration and $V(a) = V_1(a) + V_2(a)$ . We need to find the value of $a$ such that $V(a) = V_\frac 12$ . Since it is quite formidable to solve the integral manually, I used a Python code. At first I thought to write one that can do the iterations to find the required $a$ but failed. So I did the iteration manually and get $a \approx 6.60759$ . From the $z$ -intercepts of the red dash line and black line, we get $d = (z_1 - z_ 2) \sin 30^\circ \approx 5.243523044$ , $\implies \lfloor 10^4 d \rfloor = \boxed{52435}$ .

If we take the origin at the apex of the paraboloid, then its equation is

$z = a (x^2 + y^2) = a r^2$

since $r_1 = 2$ at the bottom and $r_2 = 4$ at the top and

the difference in $z$ is $10$ , then

$10 = a (16 - 4)$

so that $a = 5/6$

The volume of water in the cup plus the volume of the missing tip is

The z-value of the water surface in the cup before tilting is

$z_L = 1/2 a (r_1 ^2 + r_2 ^2 ) = 5/12 (4 + 16) = 100/12 = 25/3$

The corresponding value of r is $r_L = \sqrt{ z_L / a }$

$V_1 = 2 \pi \displaystyle \int_{r= 0}^{r= r_L} a r^2 \cdot r d r$

$= 1/2 a \pi r_L^4 = 1/(2 a) \pi z_L^2$

Next, the water surface in the tilted cup will cross the axis of the cup at

an elevation of $z = z_0$ . This is shown in the attached image which shows

the cup after untilting.

We need the find the volume under the blue line (plane).

If we assume the x-axis is pointing towards the reader, then the y-axis is

pointing to the right and the z-axis is pointing vertically up.

The equation of the plane of the surface of the water is $z = \tan \dfrac{\pi}{6} y + z_0$

where $z_0$ is yet to be determined.

To find the volume of the paraboloid under this plane, we'll find its intersection

with the plane. Let $t = \tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$ , then

$t y + z_0 = a (x^2 + y^2)$

$t y / a + z_0 / a = x^2 + y^2$

$x^2 + ( y - t/(2 a) )^2 = t^2/(2a)^2 + z0/a = R^2$

This is the equation of the projection of the elliptical surface of water onto

the xy plane. Now we can find the volume by the following volume integral

$V_2 = \displaystyle \iint_D (t y + z_0) - a (x^2 + y^2) dx dy$

At this point, we'll make a change of variables $x = r \cos \theta , y = t/(2a) + r \sin \theta$

The determinant of the jacobian of this change of variables is simply $r$ . Hence, the integral becomes

$V_2 = \displaystyle \int_{\theta = 0}^{2 \pi} \int_{r = 0}^{R} (t (t/(2a) + r \sin \theta) + z_0) - a (r^2 + t^2/(4a^2) + t r /(a) \sin \theta ) ) r dr d\theta$

Integrating over $\theta$ first, eliminates the terms containing $\sin \theta$ , and multiplies the rest by $2 \pi$ . Hence,

$V_2 = \displaystyle 2 \pi \int_{r = 0}^{R} ( t^2/(4 a) + z_0 - a r^2 ) r dr$

Integrating yields,

$V_2 = \pi ( (t^2/(4 a ) + z_0 )(t^2/(4 a^2) + z_0/a ) - \frac{1}{2} a (t^2/(4a^2) + z_0/a )^2 )$

$= \frac{1}{2a} \pi ( t^2 / (4 a) + z_0 )^2 )$

Setting $V_2$ equal to $V_1$ , gives us $z_0$

$1/(2 a) \pi z_L^2 = \frac{1}{2a} \pi ( t^2 / (4 a) + z_0 )^2 )$

implies that $z_0 = z_L - t^2/(4 a)$

The numerical value of $z_0 = 25/3 - (1/3) (6)/(20) = 25/3 - 1/10 = (250- 3)/30 = 247/30$

The depth of water is given the length of the green line in the figure which is related to the length of the red line by

${L_{\text{Green}}} = \cos 30^{\circ} {L_{\text{Red}}}$

And we have

$L_{\text{Red}} = z_0 + t r_1 - a r_1^2$

Hence, finally, the new depth of water is

$d = \cos 30^{\circ} ( z_0 + t r_1 - a r_1^2 ) = \dfrac{\sqrt{3}}{2} ( 247/30 + 2/\sqrt{3}- 20/6 ) ) ) \approx 5.24352$

therefore, the answer is $\lfloor 52435.2 \rfloor = \boxed{52435}$ .