Well most likely no one will want to see this, but if anyone is interested... Today in Economics I wasn't paying attention to the teacher, but rather I was formulating and deriving the expression for moment of inertia of a hollow sphere. For those of you who don't know, moment of inertia is how hard it is to rotate something about a point or axis.
Well here we go. This probably will be really hard to understand because of the lack of symbols... sorry for the inconvenience.
int = integral; pi = pi; d = density; m = mass; r = radius; I = moment of inertia; variables = phi, theta, p;
I = mr^2
Converting to spherical coordinates and integrating we get:
I = int[int[int[mr^2,V]]] = int[int[int[(p*sin(phi))^2*p^2*sin(phi),p],phi],theta]
p = r in this case.
The tricky part was the innermost integral. I realized after thinking for a while that you can't integrate with respect to p because it is infinitesimally small and I don't need the whole radius. Therefore, I converted it into a trivial sum in order to just use the very edge. I also set up my bounds in order to account for the entire sphere.
int[int[sum[d*p(i)^4*(sin(phi))^3,i,1,1],phi,0,pi],theta,0,2*pi] where p(1) = r
Simplifying to a double integral...
now some trig subs to make this integrable.
simplifies down to:
d*r^4*int[4/3,theta,0,2*pi] = (8/3)*pi*d*r^4
We can assume a hollow sphere to only have an area: which is 4*pi*r^2
And planar density is mass over area
And now we substitute
(8/3)*pi*(m/(4*pi*r^2))*r^4 = (2/3)*m*r^2