Calculus 3 - Center of Mass of a Lamina (Polar Coordinates) - I need a better method?

[SuperSaiyan]

The lamina is in the shape of r = 2 - 2cosθ and the density is p(x, y) = x² + y² or p(r, θ) = r²

For the mass, I set up the double integral:
θ from 0 to 2π, r from 0 to (2 - 2cosθ), r³ dr dθ = 1/4 r⁴ dθ = 1/4 (2 - 2cosθ)⁴ dθ

This would be a very cumbersome integral to calculate and I don't think that was the intention of this problem. Does anyone see another method to approach this problem from or did I write down something incorrectly? I suppose I could spend an hour writing everything out, substituting, integrating each part... Then when I go to do the moments with respect to x and y, there will be a higher exponent. :-(

 

[Deleted member 4993]

The lamina is in the shape of r = 2 - 2cosθ and the density is p(x, y) = x² + y² or p(r, θ) = r²

For the mass, I set up the double integral:
θ from 0 to 2π, r from 0 to (2 - 2cosθ), r³ dr dθ = 1/4 r⁴ dθ = 1/4 (2 - 2cosθ)⁴ dθ

This would be a very cumbersome integral to calculate and I don't think that was the intention of this problem. Does anyone see another method to approach this problem from or did I write down something incorrectly? I suppose I could spend an hour writing everything out, substituting, integrating each part... Then when I go to do the moments with respect to x and y, there will be a higher exponent. :-(

Factor 2 out then apply 1/2 angle formula (Θ = 2Φ).... see what you get.

 

[SuperSaiyan]

Factor 2 out then apply 1/2 angle formula (Θ = 2Φ).... see what you get.

Now I am left with integrating sin⁸θ/2 My textbook only cites using the half-angle formula multiple times for this.
Thank you for your help either way, I guess the problem is supposed to be this lengthy.

 

[tkhunny]

This is why we invented numerical methods and tables and Computer Algebra Systems.