Review 5: Solids of Revolution - Volume of Revolution

Suppose you have the function $f(x)=x-x^2$ with $0 \leq x \leq 1$.

If we were to rotate the graph of the function around the $x$-axis, the resulting volume would take on the shape of sort of a smushed egg. 

Our question is:  What is the volume of this solid?

The game plan we will follow is the same as in many prior cases:  We will choose a direction to discretize, produce a Riemann sum approximation, then take the limit to form a definite integral. For this example, let us choose to discretize the $x$-axis (which is the same as the axis of rotation).  Notice, from the picture below, that this discretization will result in simple to compute volumes.  For simplicity, we will call them "discs."

Now the volume of one of these discs will be given by the expression $\pi r^2 h$ where $r$ is the radius of the disc and $h$ is the height.  Now, for the $i$-th disc created by the discretization, we have $r=f(x_i^*)$ and $h = \Delta x$.  Therefore we can create the Riemann sum:

$\sum_{i=1}^n \pi \left( x_i^* - (x_i^*)^2  \right)^2 \Delta x.$