The right volume
Can you rotate a curve to make a volume of 1?
Problem
A segment of the curve $y=f(x)$ starts at $(0, 0)$ and ends at $(1, 1)$. It is rotated about the $x$ axis to form a volume. Can you find a curve which will result in a volume of $1$?
That volumes of revolution are often used in mechanics calculations. They were used by Archimedes around 250 BC to calculate volumes of solids created by rotating conic sections about an axis.
Getting Started
Curves such as $y=x, y=x^{3/2}, y=x^{5.4}$ pass through the required points. Can you think of a more general curve to try to revolve?
Student Solutions
If we suppose that the curve $y=f(x)$ is integrable then the volume created will be
$$
V = \int^1_0 \pi y^2 dx\;.
$$
To get a feel for the sort of curve we might need, first consider the special case $y=x$, which clearly passes through the two points. Then,
$$
V = \int^1_0 \pi x^2 = \pi\left[\frac{x^3}{3}\right]^1_0 = \frac{\pi}{3}\;.
$$
This is slightly larger than $1$, so we could consider a family of curves which are beneath $y=x$.
We could look for a curve like the blue one in the diagram below:
This looks like a section of a polynomial which has a root at $0$.
How about $y=x^4$?
Then,
$$
V = \int^1_0 \pi x^8 = \pi\left[\frac{x^9}{9}\right]^1_0 = \frac{\pi}{9}\;.
$$
This is still not right, but I think a polynomial could work.
Let's try $y=x^a$ where $a$ is a real number.
Then,
$$
V = \int^1_0 \pi x^{2a} = \pi\left[\frac{x^{2a+1}}{2a+1}\right]^1_0 = \frac{\pi}{2a+1}\;.
$$
If we let $a=(\pi-1)/2$ then the volume is $1$.
Therefore a solution is
$$
y=x^{(\pi-1)/2}\;.
$$
There are, of course, others!
Teachers' Resources
Why do this problem?
This is an interesting inverse style problem where students must think very carefully to create the correct starting conditions to achieve a given end result. This target motivates experimentation leading to deeper understnding of the principles of volumes of revolution.
Possible approach
This problem would make suitable challenge at the end of a lesson on volumes of revolution to inspire practice on a variety of functions. As such students could work independently or have a mini competition to achieve a volume of exactly one the most quickly or with the simplest function or alternatively to get the volume numerically closest to one in a fixed, short period of time.
In setting the problem, before any time pressure is introduced, students may like to consider what sorts of graphs pass through the two points specified and what volumes are determined by some of these graphs. A cone created by rotating the graph of $y=x$ would be very simple to visualise, sketch and calculate the volume of, even without calculus. This gives a starting point from which to suggest functions whose graphs still pass through $(0,0)$ and $(1,1)$ and which determine a volume closer to $1$.
Some time for experimentation should follow, which could be competitive or collaborative, with everyone trying to find the "best" solution by the group's own criteria. The challenge could be extended as a homework assignment and possible solutions compared and discussed.
Possible extension
Brimful and Brimful 2 are extended problems in the context of volumes of revolution for specified functions and about the $y$-axis.
Possible support
The problem is greatly simplified if you do not require the function to go through the point $(1,1)$. This could be a starting point for all or a simplification for those who struggle.