# Calculus

Find the center of mass for a shape consisting of three quarters of the unit circle, removing the quarter in the third quadrant

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asked by Lol
1. You know the area of the region is A = 3π/4
Also,
x̅ = ∫x dA
y̅ = ∫y dA
The region is easily expressed in polar coordinates, making these integrals
x̅ = 1/A ∫∫x r dr dθ = 1/A ∫[0..1] ∫[-π/2..π] r^2 cosθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)

y̅ = ∫∫y r dr dθ = ∫[0..1] ∫[-π/2..π] r^2 sinθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)

Note that this is the same as for the smaller area just in QI, since the parts in QII and QIV balance each other out.

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posted by oobleck

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