Mass of a Cardioid: Find Mass & Center of Mass

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Homework Help Overview

The discussion revolves around finding the mass and center of mass of a region defined by the cardioid r = 1 + cos(θ), with a density function of ρ = r. Participants are exploring the setup of integrals necessary for calculating these properties in polar coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of double integrals for mass and center of mass, questioning the limits of integration and the appropriate forms of the integrands. There is a focus on the integration in polar coordinates and the implications of the density function.

Discussion Status

The conversation is ongoing, with some participants providing guidance on maintaining limits of integration and the structure of the integrands. There is recognition of potential mistakes in calculations, but no consensus has been reached on the final values or methods.

Contextual Notes

Participants are navigating the complexities of integrating in polar coordinates, particularly regarding the volume element and the implications of the density function on the mass calculations. There is a mention of confusion regarding the functions to integrate, indicating a need for clarification on the setup.

Locoism
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Homework Statement



Find the mass and center of mass of the region bounded by the cardioid r =
1 + cos(θ), assuming the density function is given by ρ = r.

The Attempt at a Solution



The first part is simple enough, I set up the integral

\int_{0}^{2\pi}\int_{0}^{1+cos(\theta)}r dr d(\theta)

I find the mass is 3pi/2

Now for the centre of, how do I set up the integrals? I know it would be the double integral of xρ drdθ, and then yρ drdθ, but do I keep the same limits? x would be rcos(θ) and y is rsin(θ), but this gives me some really weird functions to integrate.
 
Last edited:
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I'm not sure what's so weird about the functions to integrate, but you are making a mistake already. In polar coordinates to find area you integrate the volume element dxdy=r*dr*d(theta). So I think you just computed the area. If the density is also r, that should give you an extra factor of r.
 
Oh so in that case I get 5/6 pi. But still, what about the centre of mass? I need My and Mx, do I keep the same limits of integration?
 
Locoism said:
Oh so in that case I get 5/6 pi. But still, what about the centre of mass? I need My and Mx, do I keep the same limits of integration?

Well, I don't get 5*pi/6. I'd check that. But, yes, keep the same limits of integration. Why not? Just put a factor of x or y into the integrand and work them out. Then you divide by total mass, right?
 
sorry 5pi/3. Thank you!
 

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