center of mass of half-disc

Let E be the upper half-disc of the disc x² + y² ≦ R in ℝ² with a constant surface-density 1. By the symmetry, its centre of mass locates on its medium radius, and therefore we only have to calculate the ordinate Y of the centre of mass. For doing that, one can use in this two-dimensional case instead a triple integral the double integral

1 ∫ ∫ Y = ----- ydx dy, ν (E ) E

where ν(E) = πR2
-2-- is the area (and the mass) of the half-disc. The region of integration is defined by

√ -------- E = { (x, y) ∈ ℝ2 ... − R ≦ x ≦ R, 0 ≦ y ≦ R2 − x2}.

Accordingly, we may write

∫ ∫ √----- ∫ ( ) 2 R R2−x2 2 R R2 − x2 2 R R2x x3 4R Y = πR2- dx y dy = πR2- ---2---dx = πR2- / -2--− 6-- = 3π-. −R 0 −R −R

Thus the centre of mass is the point (0, 4R3π- ).

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