- #176
JohnnyGui
- 796
- 51
Charles Link said:Yes, performing the integral over the complete ## dA_2 ## would yield ## P_{incident}=\sigma T_2^4 A_1 ## or equivalently ## E=\sigma T_2^4 ##.
Hmm, correct me if I'm wrong but I thought that integrating ##I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R}## over each ##dA_2## until all of the ceiling's surface has been calculated would give the total energy from the whole ceiling so that it gives ##P = \sigma T_2^4 \cdot A_2## instead of ##E = \sigma T_2^4##. Furthermore, I'd think that doing a double integration of ##I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R}## for each ##dA_1## as well as ##dA_2## would yield ##P_{incident} = \sigma T_2^4A_1##.
Charles Link said:You should now have a more complete understanding of what led us to the differential equation that we used in posts #53 and #54).
I have indeed. :)