Suppose that $f(x) \ge g(x) \ge 0$. We will discuss
rotating the region between $f$ and $g$ around some axis.
For example, if we rotate the region between the curve $y=f(x)$
and the line $y=g(x)$ around the $x$-axis, as shown below, then
each cross-section is a washer with inner radius $r=g(x)$ and
outer radius $R=f(x)$.
calculus.seas.upenn.edu
The area of the disk is the area of the bigger
circle minus the area of the smaller one: $\pi R^2-\pi
r^2$. For any $x$-value, we then have $A(x) = \pi
\left(f(x)^2 - g(x)^2\right)$ and our volume is
There is no need to memorize this integral; as usual the area
of the cross-section is your integrand.
The following video will work through some examples of the washer
method.
We can also rotate a region around the $y$-axis, as in the graph
below.
Our cross-sections are again washers, and the
volume of each slice is $A(y)\,dy$. We would then get the
volume as below, where, since we will integrate with respect to
$y$, we must invert $f$ and $g$ to get $x=f^{-1}(y)$ and
$x=g^{-1}(y)$.
Notice that, again,
there is no need for memorization - the area of a washer is
the area of the bigger circle minus the area of the smaller
one. You must, however, determine which is
the bigger and smaller circle.
The following video will work through rotating around the
$y$-axis.
