Volumes by Cylindrical Shells Method

Let’s consider the problem of finding the volume of the solid obtained by rotating about the \(x\)-axis or parallel to \(x\)-axis the region, where the core idea of cylindrical shells method for finding volumes. If we slice perpendicular to the \(y\)-axis, we get a cylinder. But to compute the inner radius and the outer radius of the washer, we would have to solve the cubic equation for \(x\) in terms of \(y\). The method of cylindrical shells is being used for finding the volume in this case, that is easier to use in such a case. We can see a cylindrical shell with inner radius, outer radius, and height. Its volume is calculated by subtracting the volume of the inner cylinder from the volume of the outer cylinder.

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.

\(\begin{aligned} \displaystyle \require{color}
\delta V &= \pi \Big[(1-x + \delta x)^2 – (1-x)^2 \Big] \times y \\
&= \pi \Big[(1-x)^2 + 2(1-x) \delta x +(\delta x)^2 -(1-x)^2 \Big] \times y \\
&= \pi \Big[2(1-x) \delta x \Big] \times \frac{x}{1+x^2} &\color{red} (\delta x)^2 \approx 0 \\
&= 2 \pi \Big[ \frac{x – x^2}{1 + x^2} \Big] \delta x \\
&= 2 \pi \Big[ \frac{1 – (1+x^2) + x}{1 + x^2} \Big] \delta x \\
&= 2 \pi \Big[ \frac{1}{1+x^2} – 1 + \frac{x}{1+x^2} \Big] \delta x \\
V &= 2 \pi \int_{0}^{1} \Big[ \frac{1}{1+x^2} – 1 + \frac{x}{1+x^2} \Big] dx \\
&= 2 \pi \Big[\tan^{-1} x – x + \frac{1}{2}\log_e (1 + x^2) \Big]_{0}^{1} \\
&= 2 \pi \Big(\frac{\pi}{4} – 1 + \frac{1}{2} \log_e{2} \Big) \\
\end{aligned} \\ \)

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