(3/4)*h^2/R+h IS THE SAME AS -(3h^2*(R-h/2)^2)/(4R^3-3Rh^2+h^3)
I used my program (which I know you hate) to factor the denominator... turns out, it factors to (R+h)(R-h/2)^2, which cancels part of the top term.
This all could have been ended a couple posts ago... my apologies for...
x = 3/4*h^2*(h-2*R)^2/(R^2*(4*R-3*h+R*h^3-3*R*h^2+3*h*R))
There's a bunch of extra terms in the denominator of the fraction... everything else is right. I can't simplify it anymore though.
I've switched from trying to work everything out by hand to a program that simplifies for me...
I do indeed understand how you got the area and thickness... and I hope to use a similar method when I try and find the Moment of Inertia of the missing part of the sphere.
As for the integral, I forgot a piece on the end... oops! The indefinite integral is...
I've been in classes all day... sorry for the lateness of my reply...
The final integral you left there... I did it, and got:
R(1/4(cos(phi))^4) = x_(2/3 - 1/3cos(phi)^2*((h-R)/R))
Now, I can't make out much on your diagram, but I'm pretty sure I can't find a way to get that into the...
Sadly, I'm sure of the formula... or at least, that's what's given to me. Maybe it's a typo on the page, or you missed a factor of 2...
I'm going to keep plugging at it... I left it alone for the evening hoping to have a new perspective on it, but it doesn't seem to be helping. Don't suppose...
Alright... I've been struggling with this derivation for QUITE some time, and I can't get a hold of my TA... so...
I'm trying to derive the centre of mass of a truncated sphere. The final answer is cm= -(3h^2*(R-h/2)^2)/(4R^3-3Rh^2+h^3) Where R is the radius of the full sphere, and h is the...