Solid Angle Rutherford Scattering

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Purple Baron
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Homework Statement


derive an equation for the solid angle for a Rutherford scattering detector given a detcor window area of [itex]A[/itex] and a distance to the detector of [itex]D[/itex] for some scattering angle [itex]\phi[/itex] given that [itex]d\Omega =2\pi sin\phi d\phi[/itex]

Homework Equations


[itex]d\Omega =2\pi sin\phi d\phi[/itex]
[itex]A=Dd\phi[/itex]

The Attempt at a Solution


integrating [itex]d\Omega =2\pi sin\phi d\phi[/itex] to get solid angle gives [itex]\Omega =\frac{2\pi A}{D}\int sin\phi d\phi[/itex] howver this gives a negative value due to the integral of sine and shouldn't soild angle be positive? Is this correct or am i missing a step? Thank You
 
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Purple Baron said:
A=Ddϕ
That's dimensionally incorrect. You have an area on the left and a distance on the right. And the dϕ looks wrong.
Shouldn't it be ##A = D^2d\Omega##?
As BvU posted, your problem with the negative sign will resolve itself when you put in the bounds on ϕ.