# Bowl Over Solution

The following is a solution to Puzzle #12 – Bowl Over. Take a look if you haven’t already.

The angle $\theta$ is defined to be that which the bowl’s axis of symmetry subtends with the vertical direction. This angle measures the extent to which the bowl is tipped. When $\theta$ is equal to 0, the bowl is untipped; when $\theta$ is $\frac{\pi}{2}$, the bowl has been tipped completely. If $\theta$ were allowed to go to $\pi$, the bowl would be upside-down, but we won’t need to worry about this case.

By the law of ‘those angles look the same therefore they are’, the angle $\theta$ is also that which the plane of the liquid’s surface makes with the rim of the bowl.

Trigonometry allows us to work out $h$:

$h=r\sin\theta$

How does the distance $h$ help? We can work out the volume of the liquid remaining in the bowl using integration. The shape filled by the liquid is a spherical cap.

A differential element of volume occupied by a thin slice of the liquid is

$dV=\pi x^2 dy$

where the $y$ axis is assumed to run vertically and the $x$ axis in the plane of the liquid’s surface. The first factor is the cross-sectional area of the slice, the second is its thickness. To find the total volume of the liquid, we integrate:

$\displaystyle V=\pi\int_h^r x^2 dy$

So $h$ is one of the limits of integration. The surface of the spherical cap is described  by the Cartesian equation

$x^2+y^2=r^2$

Hence

$\displaystyle V=\pi\int_h^r (r^2-y^2)dy$

This integral can be solved using the substitution

$y=r\sin\varphi$

$dy=r\cos\varphi\ d\varphi$

$\displaystyle V=\pi r^3\int_{\varphi(h)}^{\varphi(r)} (\cos\varphi-\cos\varphi\sin^2\varphi)d\varphi$

$\displaystyle V=\pi r^3 \Bigg[\sin\varphi-\frac{1}{3}\sin^3\varphi\Bigg]_{\varphi(h)}^{\varphi(r)}$

$\displaystyle V=\pi r^3 \Bigg[\frac{y}{r}-\frac{1}{3}\Big(\frac{y}{r}\Big)^3\Bigg]_h^r$

$\displaystyle V=\frac{1}{3}\pi r^3\Bigg(2+\frac{h}{r}\Bigg[\Big(\frac{h}{r}\Big)^2-3\Bigg]\Bigg)$

and given that $h=r\sin\theta$,

$V(\theta)=\frac{1}{3}\pi r^3(2+\sin\theta[\sin^2\theta-3])$

Given this expression for the volume of liquid, we can work out how fast liquid is poured from the bowl:

$\displaystyle \frac{dV}{dt}=\frac{dV}{d\theta}\frac{d\theta}{dt}$

$\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi r^3 \frac{d}{d\theta}(2+\sin\theta [\sin^2\theta -3]) \frac{d\theta}{dt}$

$\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi r^3 \frac{d}{d\theta}(\cos\theta [\sin^2\theta -3]+\sin\theta [2\sin\theta\cos\theta]) \frac{d\theta}{dt}$

$\displaystyle \frac{dV}{dt}=\pi r^3(\sin^2\theta - 1)\cos\theta \frac{d\theta}{dt}$

$\displaystyle \frac{dV}{dt}=-\pi(r\cos\theta)^3 \frac{d\theta}{dt}$

We assumed for simplicity that the angle $\theta$ changes at a constant rate $\omega$ as the bowl is tipped. Hence

$\displaystyle \frac{dV(t)}{dt}=-\pi\omega(r\cos\theta)^3$

Neat!

The flow rate has a maximum when $t=0$; tipping the bowl only a small distance from its equilibrium position spills a large amount of liquid.

The flow rate goes to 0 as the last of the liquid is poured from the bowl.

It can be shown that the flow rate is monotonically decreasing – it is at all times decreasing.

The animation to the right gives a visualisation of the flow rate. All hail GeoGebra!