# Brimful 2

## Problem

The $y$-axis is an asymptote for the following curves:

$$y =- \frac{1}{x} \quad\quad y= -\frac{1}{x^2} \quad\quad y^2= \frac{1}{x^3}\quad\quad y = \ln(x)-1$$

Imagine rotating the $x> 0, y< -1$ regions of these curves about the $y$-axis to form a set of hollow vessels. Which vessels are of finite volume?

Numerical extension questions

- Imagine that someone wishes accurately to engineer flasks for which the interior surfaces are given by these equations. They are to be truncated with a sealed base of radius $1$ micron, and a top opening of radius $1$cm. What would be their storage capacites and sizes? Are the resulting sizes such that you could envisage good approximations to such flasks being practically possible to make?
- Imagine that similar vessels are made with truncated and open tops (of radius $1$cm) and bottoms.They are to be designed so that they do not leak when filled with water. How long would such flasks need to be? (Assume that the diameter of a water molecule is $3$nm)
- Imagine that such flasks are made with open narrow ends of diameter $4$nm. Water is forcibly pumped into the flasks at a rate of $1$cm$^3$ s$^{-1}$, to create a jet of water consisting of a single water molecule. How fast would such jets emerge?

## Getting Started

## Student Solutions

## Part 1:

The volume generated by rotation about the $y$-axis can be found by the method of shells or the method of discs.

The method of shells is based upon filling the solid of revolution with an infinite number of thin cylindrical shells. The volume of each shell for the revolution about the $y$-axis of the curve $y = f(x)$, is equal to its circumference $2 \pi x$ multiplied by a thickness $\mathrm{d}x$ multiplied by its height $f(x)$. By allowing the thickness of each shell to approach 0 ($\mathrm{d}x \to 0$) and
summing all shells we obtain the definite integral:

$$ \textrm{[Volume]} = \int 2 \pi \ x \ f(x) \ \mathrm{d}x$$

Here we shall use the conventional method of discs; we split the solid up into a series of thins slices and by rotating a slice about the y axis we will generate a cylinder of volume $\pi f(y)^2\ \mathrm{d}y$. Letting $\mathrm{d}y \to 0$ and summing all discs we obtain the definite integral:

$$\textrm{[Volume]} = \pi \int f(y)^2 \ \mathrm{d}y$$

### Curve 1:

$y = \frac{-1}{x}$

$f(y)^2 = y^{-2}$

Limits of integration are $y=-\infty \to y=-1$

$\displaystyle \textrm{[Volume]} = \pi \int_{- \infty}^{-1} y^{-2} \ \mathrm{d}y = - \pi $

We can therefore fill this vessel.

### Curve 2:

$y = \frac{-1}{x^2}$

$f(y)^2 = \frac{-1}{y}$

Limits of integration are $y=-\infty \to y=-1$

$\displaystyle \textrm{[Volume]} = \pi \int_{- \infty}^{-1} \frac{-1}{y} \ \mathrm{d}y = \infty $

We will therefore never be able to fill this vessel.

### Curve 3:

$y = \frac{1}{x^{1.5}}$

$f(y)^2 = y^{\frac{-4}{3}}$

Limits of integration are $y=-\infty \to y= -1$

$\displaystyle \textrm{[Volume]} = \pi \int_{- \infty} ^{-1} y^{\frac{-4}{3}} \ \mathrm{d}y = 3\pi$

We can therefore fill this vessel.

### Curve 4:

$y = \ln(x)-1$

$f(y)^2 = e^{2y+2}$

Limits of integration are $y=-\infty \to y=-1$

$\displaystyle \textrm{[Volume]} = \pi \int_{- \infty}^{-1} e^{2y+2} \ \mathrm{d}y = \frac {\pi}{2} $

We can therefore fill this vessel

## Extension 1:

### Curve 1: $y = \frac{-1}{x}$

At $x = 1$, $y = -1$

At $x = 10^{-4}$, $y = -10^4$

$\displaystyle \textrm{[Volume]}= \pi \int_{-10^4}^{-1}y^{-2} \ \mathrm{d}y$

$\textrm{[Volume of flask]} = 3.14 \mathrm{\ cm^3}$

### Curve 2: $y = \frac{-1}{x^2}$

At $x = 1$, $y = -1$

At $x = 10^{-4}$, $y = -10^8$

$\displaystyle \textrm{[Volume]} = \pi \int_{-10^8}^{-1} -y^{-1} \ \mathrm{d}y$

$\textrm{[Volume of flask]} = 57.9 \mathrm{\ cm^3}$

### Curve 3: $y = \frac{1}{x^{1.5}}$

At $x = 1$, $y = -1$

At $x = 10^{-4}$, $y = 10^6$

$\displaystyle \textrm{[Volume]} = \pi \int_{-1}^{10^6} y^{\frac{-4}{3}} \ \mathrm{d}y$

$\textrm{[Volume of flask]} = 9.33 \mathrm{\ cm^3}$

### Curve 4: $y = \ln(x)-1$

At $x = 1$, $y = 0$

At $x = 10^{-4}$, $y =\ln(10^{-4})$

$\displaystyle \textrm{[Volume]} = \pi \int_{\ln(10^{-4}-1)}^{-1}e^{2y+2} \ \mathrm{d}y = \frac{\pi}{2} \Big[e^{0}-e^{\ln (10^{-4})} \Big]$

$\textrm{[Volume of flask]} = 1.57 \mathrm{\ cm^3}$

All of the above volumes appear too small to be used as real flasks, $1 \mathrm{\ litre} = 1000 \mathrm{\ cm^3}$ would be a reasonable volume for a flask.

## Extension 2:

We need to calculate the value of $y$ at which $x = \textrm{[radius of a molecule]} = 1.5 \mathrm{\ nm}$

The height is then given by the difference between this $y$ value and $y = -1$

$\textrm{[Height]} = -1 \mathrm{\ cm} - y(\textrm{evaluated at }x =-1.5 \times 10^{-7} \mathrm{\ cm}) $

$\textrm{[Height of flask 1]} = 66.667 \times 10^4 \mathrm{\ cm}= 66.67 \mathrm{\ km}$

$\textrm{[Height of flask 2]} = 4.444 \times 10^{13} \mathrm{\ cm} = 444.4 \mathrm{\ Gm}$

$\textrm{[Height of flask 3]} = 1.72 \times 10^{10} \mathrm{\ cm}= 172 \mathrm{\ Mm}$

$\textrm{[Height of flask 4]} = 15.7 \mathrm{\ cm}$