Editing Cross section (physics) (section)

=== Scattering from a 3D spherical mirror ===
The result from the previous example can be used to solve the analogous problem in three dimensions, i.e., scattering from a perfectly reflecting sphere of radius {{math|''a''}}. 

The plane perpendicular to the incoming light beam can be parameterized by cylindrical coordinates {{math|''r''}} and {{math|''φ''}}. In any plane of the incoming and the reflected ray we can write (from the previous example):
: <math>\begin{align}
r &= a \sin \alpha,\\
\mathrm dr &= a \cos \alpha \,\mathrm d \alpha,
\end{align}</math>
while the impact area element is
: <math> \mathrm d \sigma = \mathrm d r(r) \times r \,\mathrm d \varphi = \frac{a^2}{2} \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right) \,\mathrm d \theta \,\mathrm d \varphi.</math>
In spherical coordinates,
: <math>\mathrm d\Omega = \sin \theta \,\mathrm d \theta \,\mathrm d \varphi.</math>
Together with the trigonometric identity
: <math>\sin \theta = 2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right),</math>
we obtain
: <math>\frac{\mathrm d \sigma}{\mathrm d \Omega} = \frac{a^2}{4}.</math>
The total cross section is
: <math>\sigma = \oint_{4 \pi} \frac{\mathrm d \sigma}{\mathrm d \Omega} \,\mathrm d \Omega = \pi a^2.</math>