The intensity of sound when integrated over any Gaussian surface ( $\int I d A$ ) that encloses the source yields
A) a conserved quantity.
B) the pressure.
C) a nonconserved quantity.
D) none of the above.

Question

Quizplus · Accepted Answer

The integral of the intensity of sound over any Gaussian surface that encloses the source yields a conserved quantity, which is the total power output by the source. This is because sound intensity is a measure of the power per unit area, and integrating this over a closed surface captures the total power emitted by the source, which is conserved as it passes through the surface.

The Intensity of Sound When Integrated Over Any Gaussian Surface ∫IdA\int I d A∫IdA

The Intensity of Sound When Integrated Over Any Gaussian Surface $\int I d A$