How Is the Center of Mass Calculated for a Solid Cone?

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The calculation of the center of mass for a solid cone involves integrating over the volume using the formula Z = 1/M ∫ ρ z dV. The textbook "Classical Mechanics" by John R. Taylor illustrates that for a given height z, the integration over the x and y dimensions corresponds to a circle with radius Rz/h, resulting in an area of πR²z²/h². The user sought clarification on the limits of integration for dx and dy, ultimately determining that they should be expressed in polar coordinates as dx dy = r dr dθ, with limits for r from 0 to R and θ from 0 to 2π. This approach simplifies the integration process for calculating the center of mass. Understanding these limits is crucial for accurately solving the problem.
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I am using the textbook called Classical Mechanics by John R. Taylor.

Z = 1/M ∫ ρ z dV = ρ/M ∫ z dx dy dz

On page 89, example 3.2, it says:

"For any given z, the integral over x and y runs over a circle of radius r = Rz / h, giving a factor of πr2 = πR2z2 / h2."

I wish the book would show the steps. Can someone please help me understand this? I want to know what the limits of integration for dx and dy are.
 
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I think I need to relearn the whole concept of integration again... let me give this a try.

At height h the radius of the cone is Rz / h. That is the radius of each "disk" stacked on top of another to form the cone.

The area of that disk with height dz is πR2z2 / h2 ... the z2 stays inside the integral and we get ∫ z3 dz...

But I want to know what the limits for dx and dy were.
 
Okay I figured it out

dx dy => r dr dθ

Limits for dr is 0 up to R... dθ is 0 to 2π
 
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