$V=\pi \int_{b}^{4}\left(\frac{1}{x}\right)^{2} \cdot d x$

Step 1 ：The problem is asking for the volume of a solid of revolution. The integral represents the volume of the solid obtained by rotating the curve \(y = \frac{1}{{x^2}}\) from \(x = b\) to \(x = 4\) around the x-axis. The \(\pi\) in front of the integral is because the volume of a solid of revolution is given by the formula \(V = \pi \int [f(x)]^2 dx\), where \(f(x)\) is the function being rotated.

Step 2 ：To solve this problem, we need to evaluate the definite integral. The antiderivative of \(\frac{1}{{x^2}}\) is \(-\frac{1}{x}\).

Step 3 ：We can use the Fundamental Theorem of Calculus to evaluate the definite integral, which states that the definite integral of a function from a to b is equal to the antiderivative evaluated at b minus the antiderivative evaluated at a.

Step 4 ：Substituting the values into the formula, we get \(V = \pi(-\frac{1}{4} + \frac{1}{b})\).

Step 5 ：This is the final answer. \(\boxed{V = \pi\left(-\frac{1}{4} + \frac{1}{b}\right)}\)