Find center of mass of the lamina

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SUMMARY

The discussion focuses on finding the center of mass of a lamina defined by the region where 36 <= x^2 + y^2 <= 81 and y >= 0, with density proportional to the distance from the origin. The user attempted to solve the problem using polar coordinates, identifying the mass as 45π/2 and the x-coordinate of the center of mass as 0 due to symmetry. However, they encountered issues calculating the y-coordinate, specifically with the Momentx integral, which they initially computed as 342. Another participant pointed out a mistake in the integration process, emphasizing the need to include an additional 'r' in the integrals.

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  • Understanding of polar coordinates in calculus
  • Knowledge of calculating moments and center of mass
  • Familiarity with double integrals
  • Basic principles of density functions in physics
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Students studying calculus, particularly those focusing on physics applications, as well as educators looking for examples of solving center of mass problems in laminae with variable density.

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Homework Statement



Find the center of mass of the lamina which occupies if the density at any point is proportional to the distance from the origin.

36 <= x^2+y^2 <= 81, y >= 0

Homework Equations





The Attempt at a Solution


Rewrote it in polars to get 6<r<9. The x is clearly 0 as you can see from symmetry, but I can't get y. The total mas should be 45pi/2 takeing the integral of r dr d(theta) with 0<theta<pi and 6<r<9. And to find y I need to take the Momentx/mass, but I am not getting the right answer. For momentx I took the double integral of r**2*sin(theta) with the same limits to get 342. Can you see where I'm going wrong. Also density=k*r but that just leaves an extra k on mass and Mx which cancel. Any help would be great
 
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Yes, the x coordinate is 0, by symmetry. The integration, using polar coordinates, will be with r= 6 to 9 and \theta= 0 to \pi.

I suggest you do the mass integral over again. I get much more then "45/pi/2".

I do get the same thing for Momentx as you did.
 
Last edited by a moderator:
I it turns out I just forgot an r in my integrals, just another careless mistake.
 

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