# 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} \\