This is a question in my test review. I've been trying to figure it out for a while now, to no avail. Please help.
There's a container comprised of a cylinder with a cone stacked on top with the same radius. The height of both the cylinder and the cone is r. There's enough water in the container to completely fill up the cylinder when upright. What is the height of the water when the container is laid on its side?
The closest I got to solving it was by trying to describe the volume of air in terms of h (the height of the air on its side, or 2r-water_height) and setting it equals to the volume of the original cone. The half in the cylinder was basically r*circle_segment_area, but I got stuck on the half in the cone. I feel like integrals should be involved but I can't figure out how to write one without putting a variable in the limits.
Best Answer
Assume that $r$= radius of cylinder = radius of cone = height of cylinder = height of cone.
Hence we can calulate the following:
We know that
hence when container is lying on its side, the level of water must be greater than the midheight, $r$, of the container (on its side).
Let the level of water in the container on its side be $r+a$ where $0<a<1$.
(a) Cylinder
Volume of air is the area of a circular segment multiplied by its height $r$, i.e. $$\begin{align} V_1&=\frac 12 r^3 \ (2\alpha - \sin 2\alpha)\cdot r\qquad \text{where $\cos\alpha=\frac ar$}\\ &=\frac 12 r^3\ (2\alpha-2\sin\alpha\cos\alpha)\\ &=r^3\left[\cos^{-1}\left(\frac ar\right) -\left(\frac ar\right)\sqrt{1-\left(\frac ar\right)^2}\ \ \right]\\ &=r^3(\cos^{-1}\lambda-\lambda\sqrt{1-\lambda^2})\qquad \text{where $\lambda=\frac ar$}\\ &=\frac{r^3}3\left[3\cos^{-1}\lambda-3\lambda\sqrt{1-\lambda^2}\right] \end{align}$$
(b) Cone
Calculate the the volume of air in the cone by integrating circle segments of infinitesimally small thickness and of different radius $u$, as $u$ varies from $r$ (the part adjacent to the cylinder) to $a$ (edge of container).
Area of circle segment with radius $u$ is given by $$\frac 12 u^2 \ (2\phi - \sin 2\phi) = u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2} \;\right]\qquad \text{where $\cos\phi=\frac au$}\\$$
Volume of air in cone is given by integrating this from $u=r$ to $u=a$, i.e. $$\begin{align} V_2&=\int_r^a u^2\ \left[ \cos^{-1} \left(\frac au\right) - \frac au \sqrt{1-\left(\frac au \right)^2}\;\right] du\\ &=\int_r^a u^2\cos^{-1} \left(\frac au\right) -a\sqrt{u^2-a^2} \; du\\ &=a^2 \int_r^a \left(\frac ua \right)^2\cos^{-1} \left(\frac au\right) -\sqrt{\left(\frac ua\right)^2-1} \; \; du\ \end{align}$$
Put $$\begin{align} \frac au = \cos\phi \Rightarrow u&=a\sec\phi\quad \Rightarrow \sec\phi=\frac ua\\ du&=a\sec\phi\tan\phi \; d\phi\\ u=r \Rightarrow \phi&=\cos^{-1}\frac ar=\cos^{-1}\lambda\\ u=a \Rightarrow \phi&=\cos^{-1}1=0 \end{align}$$
This gives $$\begin{align} V_2&=a^2\int_{\cos^{-1}\lambda}^{0}\left[(\sec^2\phi) \cdot \phi-\sqrt{\sec^2\phi-1}\right]\ a\sec\phi\tan\phi \; d\phi\\ &=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec\phi\tan^2\phi \; d\phi\\ &=a^3\int_{\cos^{-1}\lambda}^0 \phi\sec^3\phi\tan\phi-\sec^3\phi +\sec\phi \; d\phi\\ &=\cdots \text{(and after some tedious integration)}\cdots\\ &=\frac {a^3}3 \left[\phi\sec^3\phi-2\sec\phi\sqrt{\sec^2\phi-1}+\ln(\sec\phi+\ sqrt{\sec^2\phi-1})\right]_0^{\cos^{-1}\lambda}\\ &=\frac {a^3}3\left[ \left(\cos^{-1}\frac au\right)\left(\frac ua\right)^3-2\frac ua\sqrt{\left(\frac ua\right)^2-1}+\ln\left(\frac ua+\sqrt{\left(\frac ua\right)^2-1}\right)\right]_{u=a}^{u=r}\\ &=\frac {a^3}3 \left[ \left(\cos^{-1} \frac ar\right)\left(\frac ra\right)^3-2\frac ra \sqrt{\left(\frac ra\right)^2-1}+\ln\left(\frac ra+\sqrt{\left(\frac ra\right)^2-1}\right)\right]\\ &=\frac {a^3}3 \left[ \frac 1{\lambda^3} \cos^{-1}\lambda -\frac 2\lambda \sqrt{\frac 1{\lambda^2}-1}+\ln\left(\frac 1\lambda+\sqrt{\frac 1{\lambda^2}-1}\right) \right]\\ &=\frac {r^3}3 \left[ \cos^{-1} \lambda-2\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda +\sqrt{\frac 1{\lambda^2}-1}\right)\right] \end{align}$$
(c) Total
Total volume of air in container is $$\begin{align} V_1+V_2&=\frac {\pi r^3}3\\ \frac{r^3}3\left[4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)\right]&=\frac {\pi r^3}3\\ 4\cos^{-1}\lambda-5\lambda\sqrt{1-\lambda^2}+\lambda^3\ln\left(\frac 1\lambda (1 + \sqrt{1-\lambda^2})\right)-\pi&=0\\ \lambda &\approx 0.36876\\ a=\lambda r &\approx 0.36876 r \end{align}$$
Hence, height of water in container on its side is $$r+a=(1+\lambda)r\approx 1.36876r \qquad \blacksquare$$
NB:
1. See this for a useful guide to the messy integration and the volume of a vertical cut of a cone.
2. See this for a useful guide for integrating $\sec^3\theta$.