Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ I was finding the moment of inertia of a hollow sphere, and got stuck on the integration of:
$${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$$
Any hint to as to solve this integral? Please help.
Substituting $t = R^2 - y^2$
$${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy = {M\over2R^2}\int(t)^{3\over2}dy$$
$$={M\over2R^2}{2\over5}t^{5\over2}$$
$$={M\over(R^2)}{1\over5}t^{5\over2}$$
Will I get the integral upon substituting value of $t$? 
Moment of Inertia:
Considering the hollow sphere to be made of small rings, of thickness dy, radius x and of mass dm:
$$dI = dm(x)^2$$
$$dm = {Mxdy\over2R^2}$$
Upon drawing a figure:
$$x^2 = R^2 - y^2$$
$$dI = {Mxdy\over2R^2}(x)^2$$
$$I = {M\over2R^2}\int(x^3)dy$$
$$I = {M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$$

Let p be the areal mass density
$$p = \frac{M}{4\pi R^2}$$
$$dm = p * 2\pi R dy$$
$$dm = {Mxdy\over2R^2}$$
 A: To make my hint explicit, with $y = R\sin t$, $dy = R\cos t \ dt$ and the integral equals
$$\frac{M}{R^2}\int (R^2(1-\sin^2 t))^{3/2} . R\cos t \ dt = MR^2 \int \cos^4 t \ dt$$
Now $\cos^2 t = \frac{1}{2}(1 + \cos(2t))$, one of the 'half-angle' relations of trig. Hence $$\cos^4 t = \left( \frac{1}{2}(1 + \cos(2t)) \right)^2 = \frac{1}{4} \left( 1 + 2\cos(2t) + \cos^2(2t)\right)$$ Integrating the first two of those terms is straightforward. Use the 'half-angle' formula for $\cos^2$ again to integrate the last term.
A: If you slice your hollow sphere with horizontal slices of ${\rm d}y$ thickness the mass of the sphere is
$$ M = \int \limits_0^R w {\rm d}y $$ where $w$ is some kind of linear mass density, found to be $w=\frac{M}{R}$
The mass moment of inertia is similarly equal to
$$ I = \int \limits_0^R w x^2 {\rm d}y = \int \limits_0^R \frac{M}{R} \sqrt{R^2-y^2} {\rm d}y$$
The above evaluates to $I=\frac{2}{3} M R^2$ which is the correct answer according to http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html
I used $(x,y)=(R \cos \theta,R \sin \theta)$ substitution in the integral to make it super easy.
