Washer Method Formula
The washer method is used to find the volume enclosed between two functions. In this method, we slice the region of revolution perpendicular to the axis of revolution. We call it as Washer Method because the slices obtained in this way forms washers. This method is just an extension of the disk method to determine the volume of a hollow solid of revolution.
Let us learn the Washer method formula with a few solved examples.
What is Washer Method Formula?
Let \(y=f(x)\) and \(y=g(x)\) be the continuous functions and non-negative in \([a, b]\) such that \(g(x) \leq f(x)\). Let \(R_{1}\) be the region bounded by two functions \(f(x)\) and \(g(x)\) in \([a, b]\). The volume of the solid formed by revolving the region \(R\) around the \(x\)-axis is given by
\[V=\pi \int_{a}^{b} \left( [f(x)]^2 - [g(x)]^2\right)dx\]
Let \(R_{2}\) be the region bounded by two functions \(x=f(y)\) and \(x=g(y)\) in \([c, d]\) such that \(g(y) \leq f(y)\). The volume of the solid formed by revolving the region \(R\) around the \(y\)-axis is given by
\[V=\pi \int_{c}^{d} \left( [f(y)]^2 - [g(y)]^2\right)dy\]
Solved Examples on Washer Method Formula
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Example 1:
Let \(R\) be a region bounded by \(y=x\) and \(y=x^3\) in \([0, 1]\). Find the volume of the solid obtained by rotating the region \(R\) about \(x\)-axis.
Solution:
The volume is given by
\(\begin{align}V&=\pi \int_{0}^{1} { (x)^2 - (x^3)^2}dx\\&=\pi \int_{0}^{1} {x^2-x^6}dx\\&=\pi \left[\frac{x^3}{3}-\frac{x^7}{7}\right]_{0}^{1}\\&=\frac{4\pi}{21}\end{align}\)
Hence, the required volume is \(\frac{4\pi}{21}\). -
Example 2:
Let \(R\) be a region bounded by \(y=x^2\) and \(y=\sqrt{x}\). Find the volume of the solid obtained by rotating the region \(R\) about \(x\)-axis.
Solution:
The volume is given by
\(\begin{align}V&=\pi \int_{0}^{1} { (\sqrt{x})^2 - (x^2)^2}dx\\&=\pi \int_{0}^{1} {x-x^4}dx\\&=\pi \left[\frac{x^2}{2}-\frac{x^5}{5}\right]_{0}^{1}\\&=\frac{3\pi}{10}\end{align}\)
Hence, the required volume is \(\frac{3\pi}{10}\).