from a handpicked tutor in LIVE 1to1 classes
Shell Method Formula
Shell Method is used to find the volume by decomposing a solid of revolution into cylindrical shells. We slice the solid parallel to the axis of revolution that creates the shells. The volume of the cylindrical shell is the product of the surface area of the cylinder and the thickness of the cylindrical wall.
Let us learn the shell method formula with a few solved examples.
What is the Shell Method Formula?
Let \(R\) be the region bounded by \(x=a\) and \(x=b\). Suppose we form a solid by revolving it around a vertical axis. Let \(r(x)\) represent the distance from the axis of rotation to \(x\) and \(h(x)\) be the height of the shell.
The volume of the solid is given by
\[V=2 \pi \int_{a}^{b} {r(x) h(x)} dx\]
Solved Examples on Shell Method Formula

Example 1:
Let \(R\) be a region bounded by \(y=2x^2x^3\) and \(x\)axis. Find the volume of the solid obtained by rotating the region \(R\) about \(y\)axis.
Solution:
Here, \(h(x)=2x^2x^3\) and \(r(x)=x\).
The volume is given by
\(\begin{align}V&=\int_{0}^{2} {(2\pi x)(2x^2x^3)}dx\\&=2\pi \int_{0}^{2} {(2x^3x^4)}dx\\&=2\pi\left[\frac{x^4}{2}\frac{x^5}{5}\right]_{0}^{2}\\&=\frac{16\pi}{5}\end{align}\)
Hence, the required volume is \(\frac{16\pi}{5}\). 
Example 2:
Let \(R\) be a region bounded by \(x=y^3\), \(x=8\) and the \(x\)axis. Find the volume of the solid obtained by rotating the region \(R\) about \(x\)axis.
Solution:
Here, \(h(x)=8y^3\) and \(r(x)=y\).
The volume is given by
\(\begin{align}V&=\int_{0}^{2} {(2\pi y)(8y^3)}dy\\&=2\pi \int_{0}^{2} {(8yy^4)}dy\\&=2\pi\left[4y^2\frac{y^5}{5}\right]_{0}^{2}\\&=\frac{96\pi}{5}\end{align}\)
Hence, the required volume is \(\frac{96\pi}{5}\).
visual curriculum