Great Pyramid of Cheops (Center of Mass Problem)

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The discussion focuses on calculating the center of mass and work required to lift the blocks of the Great Pyramid of Cheops. The original height of the center of mass above the base needs to be determined using integration techniques, considering the pyramid's dimensions and uniform density. For calculating the work, the center of mass height is used as the distance for lifting all blocks, but the integration must account for the varying heights of each mass element. Clarification is provided that the integrals for mass and height are not the same, emphasizing the need for careful calculation. The conversation highlights the complexities of multi-dimensional integration in physics problems.
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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?
(b) What is the work required to lift all the blocks into place from the base level?

Kay, so I'm good with finding center of mass in one dimension, but in two and three dimensions, it is confusing the hell out of me. Am I supposed to integrate? If so, what? I just need a push in the right direction (I think).

Also, for finding the work, I'm guessing I'm supposed to use the CM height as the distance for all of the blocks traveled, but I'm not sure about that either. Any help would be appreciated!
 
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Yes, you need to integrate.

If z is the height above the base of an element of mass of thickness dz then the element of mass at height z is proportional to (h-z)^2 dz. You need to find the average of z which means summing over all elements of mass - and dividing by the total mass.
 
But wouldn't that just be dividing the total mass by the total mass? I'm really confused. I need to have the answer tonight, so if anyone can give me a bit more information, it would really help.
 
No. The integrand of the integral in the numerator will have an additional factor of z in it so the integrals are not the same.
 

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