Disk Method Formula
The disk method is used when the axis of revolution is the boundary of the plane region and the crosssectional area is perpendicular to the axis of revolution. This method is used to find the volume by revolving the curve \(y=f(x)\) about \(x\)axis and \(y\)axis. We call it as Disk Method because the crosssectional area forms circles, that is, disks. The volume of each disk is the product of its area and thickness.
Let us learn the disk method formula with a few solved examples.
What is the Disk Method Formula?
Let \(R_{1}\) be the region bounded by \(y=f(x)\), \(x=a\), \(x=b\) and \(y=0\). Suppose we form a solid by revolving it around the \(x\)axis. The volume of the solid is given by:
\[V=\pi\:\int_{a}^{b}\!\![f(x)]^2 dx\]
Let \(R_{2}\) be the region bounded by \(x=f(y)\), \(y=c\), \(y=d\) and \(x=0\). Suppose we form a solid by revolving it around the \(y\)axis.
The volume of the solid is given by:
\[V= \pi\:\int_{c}^{d}\!\! [f(y)]^2 dy\]
Solved Examples on Disk Method Formula

Example 1:
Let \(R\) be a region bounded by \(y=x^2\), \(x=2\), \(x=3\) and \(x\)axis. Find the volume of the solid obtained by rotating the region \(R\) about \(x\)axis.
Solution:
The volume is given by
\(\begin{align}V&=\int_{2}^{3} {\pi (x^2)^2}dx\\&=\pi \int_{2}^{3} {x^4}dx\\&=\pi\left[\frac{x^5}{5}\right]_{2}^{3}\\&=55 \pi\end{align}\)
Hence, the required volume is \(55 \pi\). 
Example 2:
Let \(R\) be a region bounded by \(y=x^24x+5\), \(x=1\), \(x=4\) and the \(x\)axis. Find the volume of the solid obtained by rotating the region \(R\) about \(x\)axis.
Solution:
The volume is given by
\(\begin{align}V&=\pi \int_{1}^{4} {(x^24x+5)^2}dx\\&=\pi \int_{1}^{4} {(x^48x^3+26x^240x+25)}dx\\&=\pi\left[\frac{x^5}{5}2x^4+\frac{26x^3}{3}20x^2+25x\right]_{1}^{4}\\&=\frac{78\pi}{5}\end{align}\)
Hence, the required volume is \(\frac{78\pi}{5}\).