Thermal Probability and Trig integrals <3

[sugar_scoot]

Given the number of molecules hitting unit area of a surface per second with speeds between v and v +dv and angles between \theta and d\theta to the normal is

\frac{1}{2} v n f(v)dv sin \theta cos \theta d\theta​

show that the average value of cos \theta for these molecules is \frac{2}{3}.

I have convinced myself the answer is 4/3 instead. Can anyone show me where I am wrong?
I used P(cos \theta) = sin \theta cos \theta

Then I normalized:
1 = c \int^{\pi}_{0} sin \theta cos \theta d\theta
so that:
c = 2

<cos \theta> = 2 \int^{\pi}_{0} sin \theta cos^{2}\theta d \theta = 2 (2/3) = 4/3

 

[fzero]

Given the number of molecules hitting unit area of a surface per second with speeds between v and v +dv and angles between \theta and d\theta to the normal is

\frac{1}{2} v n f(v)dv sin \theta cos \theta d\theta​

show that the average value of cos \theta for these molecules is \frac{2}{3}.

I have convinced myself the answer is 4/3 instead. Can anyone show me where I am wrong?
I used P(cos \theta) = sin \theta cos \theta

Then I normalized:
1 = c \int^{\pi}_{0} sin \theta cos \theta d\theta
so that:
c = 2

<cos \theta> = 2 \int^{\pi}_{0} sin \theta cos^{2}\theta d \theta = 2 (2/3) = 4/3

Integrating from 0 to \pi overcounts the number of particles by a factor of 2. You only integrate from the normal to the plane, which is 0 to \pi/2. Your normalization integral actually vanishes as written.

 

[sugar_scoot]

Thank you.

Is there an intuitive reason why normalization is unnecessary in this case? Should I continue to attempt normalization as a first step in problems like these?

 

[sugar_scoot]

Actually I just did the problem over again with the new integration limits and although my <cos\theta> is now correct, I still found a normalization constant of 2.