Answer
The statement is true.
Work Step by Step
Recall that from the definition, if \(\sigma\) is a smooth surface \(S\), and \(f\) is identical to \(1\), then the double integral over the surface is defined as: \[ \iint_{\sigma} 1 \, dS = \lim_{n \to \infty} \sum_{i=1}^n \Delta S_i = \lim_{n \to \infty} S = S \] This means that when the function \(f(x, y, z)\) is identical to \(1\), the double integral over the surface is equal to the surface area \(S\) of the surface \(\sigma\).
Result: The statement is true.
