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A cup takes the shape of a right circular conical frustum. The bottom radius is $r_1$ and the top radius is $r_2$ and its height is $h$. It is tilted from the vertical by an angle $\theta$. Find an expression of volume of water that will fill the cup to the rim.

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My attempt:

By completing the cone to its vertex, we get the 'missing' altitude (the vertical distance between the vertex and the bottom of the cup). This distance is given by

$ Z_V = \dfrac{h r_1}{r_2 - r_1} $

Let the surface of water intersect the axis of the cone at a distance $z_0$ from its vertex, then using a simple diagram, we have

$s = (h + Z_V) \sec \theta_c$

where $\theta_c$ is the semi-vertical angle of the cone, and is given by

$ \theta_c = \tan^{-1} \dfrac{r_2 - r_1}{h} $

And from equality of vertical height above the vertex, we get

$ s \cos( \theta + \theta_c ) = z_0 \cos \theta $

From which we can calculate $z_0$

One can show that the semi-major axis $a$ and semi-minor axis $b$ of the water surface are given by

$ a = \dfrac{z0 \tan \theta_c \cos \theta }{ 1 - \sin^2 \theta \sec^2 \theta_c } $

$ b = \dfrac{z0 \tan \theta_c \cos \theta }{ \sqrt{1 - \sin^2 \theta \sec^2 \theta_c} } $

Putting all this together, we find that the volume of the cone is given by

$V = (1/3) \ \pi \ \dfrac{ z_0^3 \tan^2 \theta_c \cos^3 \theta }{(1 - \sin^2 \theta \sec^2 \theta_c)^{3/2} } $

From this volume we have to subtract the volume of the small cone whose base is the base of the cup, and this is given by

$V_0 = (1/3) \pi r_1^2 Z_V = 1/3 \pi r_1^3 \cot(\theta_c)$

Therefore the answer to the question is that volume is given by

$ V_{Water} = V - V_0 = \dfrac{\pi}{3} \ \bigg(\dfrac{ z_0^3 \tan^2 \theta_c \cos^3 \theta }{(1 - \sin^2 \theta \sec^2 \theta_c)^{3/2} } - r_1^3 \cot (\theta_c) \bigg)$

Hosam Hajeer
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  • What is the question? – DanDan0101 Jul 17 '23 at 07:37
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    How is that different from this question? – José Carlos Santos Jul 17 '23 at 07:40
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    @JoséCarlosSantos Strictly speaking not the same exercise, but indeed they are closely related. – Jean-Claude Arbaut Jul 17 '23 at 09:54
  • @JoséCarlosSantos They are different because in the previous problem $\theta$ was unknown, and the volume is given, but here $\theta$ is given, and the volume is to be found. – Hosam Hajeer Jul 17 '23 at 11:02
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    Yes, but basically you have a relation $f(\theta, V)=0$ between the max volume and the angle. If you know one, you know the other by solving an equation. – Jean-Claude Arbaut Jul 17 '23 at 11:19
  • Yes. I've expressed the volume of the cone as a function of $\theta$. In the previous problem (which is a bit harder) I had to solve for $\theta$ after calculating the volume of water. In this problem, I am assuming that $\theta$ is given, and I have to calculate the volume (using the same formula that I developed in the previous problem). I am re-iterating because I think it is an interesting problem and deserves a double take. – Hosam Hajeer Jul 17 '23 at 11:30

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