Wave Intensity: A^2(z,t) at (15 cm, .5s)

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Wave intensity is defined as the squared wave amplitude, represented by I(z,t) = A^2(z,t). For a wave with a wavelength of 10 cm and a period of 0.1 s, the amplitude is given as 1 mm. To find the instantaneous intensity at r = 15 cm and t = 0.5 s, the expression A^2(z,t) = (0.1 cos [2 π (15/10 - 0.5/0.1)])^2 is used. For the average intensity over a long time at that location, the integral of A^2(15,t) should be calculated from 0 to the period T and then divided by T to obtain the average. This approach ensures accurate computation of the average intensity.
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the wave intensity is defined as the squared wave amplitude.

I(z,t)=A^2 (z,t)
A(z,t)=a cos [2 \pi (\frac{r}{\lambda}-\frac{t}{T})]

the wavelength is 10cm
period is 1/10s
a=1mm


1. what is the instantaneous intensity at r=15cm and t=.5s?

is the solution:
A^2(z,t)=(.1 cos [2 \pi (\frac{15}{10}-\frac{.5}{.1})])^2

2. what is the average intensity averaged over a long time at that location?

\int A^2(15,t)=\int_{0}^{\inf} (.1 cos [2 \pi (\frac{15}{10}-\frac{t}{.1})])^2 dt

is that how I would solve #2?
 
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Integrating like that would give you a total sum. You need to divide it by the time to get the 'average'. A trick would be to integrate from 0 to any T (period of the wave) and divide it by T.
 
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