Black Body Radiation: Exploring Radiated & Perpendicular Power

[dsdsuster]

Can someone help explain what's going on here?
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radpow.html#c1

Specifically the part about how radiated power is greater than perpendicular power.
I also don't understand why the power is being averaged over the various angles. Doesn't it make more sense to add the powers from light arriving at different angles to get the total energy transmitted?

 

[mikeph]

I can't follow this very well, nothing is defined and I can't work out the overall goal.

I think it's saying that the radiation going through the specified surface from some angle is not perpendicular, so you need to use cos(theta) to find the perpendicular component of the vector.