Let’s consider the problem of finding the volume of the solid obtained by rotating about the \(x\)-axis or parallel to \(x\)-axis of the region, where the core idea of cylindrical shells method for finding volumes. We get a cylinder if we slice perpendicular to the \(y\)-axis. But to compute the washer’s inner and outer radius, we would have to solve the cubic equation for \(x\) in terms of \(y\). The method of cylindrical shells is used to find the volume in this case, which is easier to use in such a case. We can see a cylindrical shell with an inner radius, outer radius, and height. Its volume is calculated by subtracting the inner cylinder’s volume from the outer cylinder’s volume.
Worked Example of Finding a Volume by Cylindrical Shells Method
The diagram shows the graph of \(\displaystyle f(x) = \frac{x}{1 + x^2} \).
The area bounded by \(y = f(x) \), the line \(x = 1\) and the \(x\)-axis is rotated about the line \(x=1\) to form a solid volume. Use the cylindrical shells method to find the volume of the solid.
\(\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\delta V &= \pi \Big[(1-x + \delta x)^2-(1-x)^2 \Big] \times y \\
&= \pi \Big[(1-x)^2 + 2(1-x) \delta x +(\delta x)^2-(1-x)^2 \Big] \times y \\
&= \pi \Big[2(1-x) \delta x \Big] \times \frac{x}{1+x^2} &\color{red} (\delta x)^2 \approx 0 \\
&= 2 \pi \Big[ \frac{x-x^2}{1 + x^2} \Big] \delta x \\
&= 2 \pi \Big[ \frac{1-(1+x^2) + x}{1 + x^2} \Big] \delta x \\
&= 2 \pi \Big[ \frac{1}{1+x^2}-1 + \frac{x}{1+x^2} \Big] \delta x \\
V &= 2 \pi \int_{0}^{1} \Big[ \frac{1}{1+x^2}-1 + \frac{x}{1+x^2} \Big] dx \\
&= 2 \pi \Big[\tan^{-1} x-x + \frac{1}{2}\log_e (1 + x^2) \Big]_{0}^{1} \\
&= 2 \pi \Big(\frac{\pi}{4}-1 + \frac{1}{2} \log_e{2} \Big)
\end{aligned} \)
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