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.