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A region $R$ is bounded by the circle radius $4$ centred at $(5,0)$, the circle radius $2$ centred at $(0,0)$ and the x-axis, as shown.

What is the centroid of this region?

Finding the top point of $R:=(x_p, y_p)$:

$$\begin{align} 2^2 &=x_p^2 + y_p^2\\ 4^2 &= (5-x_p)^2+y_p^2\\ 16 &=25 -10x_p+x_p^2+y_p^2\\ &=29-10x_p\\ x_p &= {\small\frac{13}{10}} = 1.3\\ \color{gray}{y_p} &= \color{gray}{\sqrt{4-1.3^2} = \sqrt{2.31}} \end{align}$$

Now I could integrate across the left and right areas either side of $x=1.3$, find the two centroids and make an interpolation weighted by the areas either side, but is there a better way to do this?

  • $\begingroup$ Why don't you integrate along the $y$ axis? It looks easier to me than doing the calculations in two steps... But besides that, I don't see a more geometric approach (yet) $\endgroup$ – b00n heT Jul 20 '16 at 20:49

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