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

Question

Quizplus · Accepted Answer

The integral of the intensity ($I$) of sound over a surface area ($dA$) that encloses the source ($\int I dA$) gives the total power ($P$) radiating from the source. This is because sound intensity is defined as power per unit area ($I = P/A$), so when you integrate intensity over an area that encloses the source, you essentially calculate the total power emitted by the source.

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 $\int I d A$