Center of Mass Physics help! PLEASE?

The Great Pyramid of Cheops at El Gizeh, Egypt, had a height H = 148.6 m before its topmost stone fell. Its base is a square with edge length L = 227.4 m. Its volume V is equal (L^2)H/3. Assuming that it has uniform density p(rho) = 1.8e3 kg/m^3.


(a) What is the original height of its center of mass above the base?





I know I have to integrate something... but thats about it. |||http://i299.photobucket.com/albums/mm286鈥?/a>


MN=h, MP=H, AB=L


Let EF=a


(H-h)/H=a/L ==%26gt; a=L(H-h)/H=L-Lh/H


Let dm be the mass of a thin layer at height h, parallel to the base


and with thickness dh ==%26gt;


dm=p*dV, where dV=a虏dh ==%26gt;


dm=p*(L-Lh/H)虏 dh, 0%26lt;h%26lt;H


Let Hc=height of the center of mass above the base


Hc={鈭玔m] h*dm}/m, where m=p*V


Hc={鈭玔0,H] hp(L-Lh/H)虏 dh}/(pL虏H/3)=


{L虏h虏/2-2L虏h鲁/(3H)+L虏(h^4)/(4H虏)}/


(L虏H/3) [h=0 to L]=


3(H虏/2-2H虏/3+H虏/4)/H=(1.5-2+0.75)H=H/4