Complicated moment of inertia question

[fredrick08]

Homework Statement

a sphere of radius R, with centre at r=0, has a density of rho(r)=rho(0)(1-r/R), using sphericals.

a. what is the mass of the sphere?
b. write down the intergral for the moment of inertia, about the z axis.
c. Solve it.

Homework Equations

I=$$\int$$$$\int$$$$\int$$rho(r)*r^2 dV

The Attempt at a Solution

a. M=rho*V=rho(0)(1-r/R)*4*pi*r^3/3

i need help on, i am getting somehting like..
triple intergral form 0-2pi,0-pi,0-R of (rho(0)(1-r/R)*r^3*sin(theta) dr dtheta dphi

ca anyone help me? by makign sure my intergal is right, or fixing it, I've only ever done uniform density before, so I am a bit confused about this.

 

[fredrick08]

srry. made a mistake.

 

[fredrick08]

for that integral i get (pi/5)rho(0)(R^4)... this doesn't make sense, since to get the answer in terms of M and R, i need to sub in the value of mass...

wel maybe if I just take out all the things i know, i get (3/20)MR^2... rofl I DOUBT THAT IS RIGHT..

 

[HallsofIvy]

No, M is NOT "M=rho*V=rho(0)(1-r/R)*4*pi*r^3/3" because "r" is a variable. M is the integral of the density function over the volume of the sphere:
$$M= \int_{\phi= 0}^{\pi}\int_{\theta= 0}^{2\pi}\int_r^{R} rho(0)(1-\frac{r}{R})dV$$.

 

[fredrick08]

ok yes thankyou, that makes much more sense.

 

[fredrick08]

ok i tried that integral, and i got (1/3)(pi*rho/R)(R^4-4R*r^3+3r^4)... i have no idea, what this means... I am confused, because isn't mass constant... over the entire volume? r u sure i am meant to intergrate from r to R dr.. and not 0 to R?.. and if this this is right what shoudl my integral be for the moment of inertia?

 

[fredrick08]

i think the moment of inertia should be what it is for a solid sphere.. 2/5MR^2.. but doign it the way you said, i get I=(1/5)(pi*rho/R)(R^5-5*R*r^4+4r^4) ... if i divide that for what I got form my M value... i just get a heap of mess...

 

[fredrick08]

ok i got the right answer, being the right answer is 2/5MR^2... forgot then r=0 lol