What is the center of mass of a lamina with variable density?

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To find the center of mass of a lamina with variable density in the first quadrant of the disk defined by x^2 + y^2 <= 1, the density is proportional to the square of the distance from the origin. The center of mass can be calculated using the formula M r_{cm} = ∫ r_m dm, indicating that the center of mass will lie along the line θ = π/2. It is recommended to choose a small element at (r, θ) for calculations. Additionally, for LaTeX inequalities, the correct codes are \leq and \geq. This approach provides a structured method to solve the problem effectively.
tandoorichicken
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Any hints on how to approach this problem?

A lamina occupies the part of the disk x^2 + y^2 &lt;= 1 in the first quadrant. Find the center of mass if the density at any point is proportional to the square of its distance from the origin
 
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And just for future reference, how do you do inequalities in latex?
 
\leq {and} \geq the code reference file is found by clicking on the code and then clicking the link
 
I don't know what responses you got for your earlier thread. I could not find it. Any way, use the following to find the CM.

M r_{cm} = \int r_m dm

Since the density depends only on the radius, CM should be along the \theta = \pi/2 line.

Choose a small element at (r, \theta ) and proceed.
 
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