Differential element - Surface Area of Sphere

Can anyone explain how they get this formula in a book from this diagram:


U is the velocity of the particle, r is the radius and a is just the co-ordinate from the center of the particle, so a=r when a extends radially onto the boundary of the sphere.

I've attached a pdf with the relevant pages.Section 4.3.4, equation 4.97

I successfully got the part in brackets by using r=a.

I was right, it is just a volume integral. You know about integration in polars right? e.g. in 2D$dV=rdrd\theta$?

I remember learning something about it but that's about it. Whats the wiki article of it called?

Okay got it I think (Disclaimer: I am a Stats/Probability Guy so not an expert on this particular subject but acceptable at vector calculus)

Look at equation (4.92). This is an integral over the sphere's surface (so immediately r=a is constant). Now following the steps given on the sheet we arrive at (4.96) I believe you understand up to here.

Now in (4.97) the problem lies with the dA this is tricky since they are transforming the area into a system consisting of two angles (since r is fixed) now this can reach every point on the sphere. I will call the bit you have in brackets in 4.97 $\gamma$. So $KE=\frac{1}{2}\rho_c\int_0^\pi \gamma dA$ now it helps to think of dA as a volume dV with constant radius which allows us to apply the transformation: $dA=a^2sin\theta d\phi d\theta$ see: http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
so we get: $KE=\frac{1}{2}\rho_c\int_0^\pi \int_{\phi=0}^{\phi=2\pi} \gamma a^2sin\theta d\phi d\theta$
which gives the result.

Can anyone explain how they get this formula in a book from this diagram:



U is the velocity of the particle, r is the radius and a is just the co-ordinate from the center of the particle, so a=r when a extends radially onto the boundary of the sphere.