Answer
The inequality is true.
Work Step by Step
Step 1: We are given an inequality involving two surface integrals: \[ \iint_{\sigma} f(x, y, z) \, dS \geq \sigma \iint_{\sigma} g(x, y, z) \, dS \] We want to check if this inequality is true or false under the conditions: \[ f(x, y, z) \geq g(x, y, z) \] Step 2: The surface integral is defined as the limit of a sum over patches \(\Delta S_k\) of the surface \(\sigma\): \[ \iint_{\sigma} F(x, y, z) \, dS = \lim_{\Delta \to \infty} \sum_{k} F(x_k^*, y_k^*, z_k^*) \Delta S_k \] where \(F(x_k^*, y_k^*, z_k^*)\) is the value of the function at the point \((x_k^*, y_k^*, z_k^*)\) where the patch \(\Delta S_k\) is located. Step 3: Both surface integrals in the inequality are being integrated over the same surface \(\sigma\), so we can divide that surface into the same patches \(\Delta S_k\). Step 4: Comparing the sums for both integrals, since we divided \(\sigma\) into the same patches \(\Delta S_k\) and the area of \(\Delta S_k\) is always positive, we have: \[ \sum_{k} F(x_k^*, y_k^*, z_k^*) \Delta S_k \geq \sum_{k} g(x_k^*, y_k^*, z_k^*) \Delta S_k \] This inequality holds because at every point \((x_k^*, y_k^*, z_k^*)\), the condition \(f(x, y, z) \geq g(x, y, z)\) holds. Thus, the original inequality is also true. Step 5: In summary, we wanted to show that the given inequality holds, and we were able to prove it based on the definition of surface integrals and the condition \(f(x, y, z) \geq g(x, y, z)\).
Result: The inequality is true.
