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- #1

Since it is uniform and symmetric on the x axis, x = 0.

For y, we have

$$

y_{cm} = \frac{\int y\sigma dA}{\frac{\pi R^2}{2}}

$$

In polar, $dA = rdrd\phi$.

So the integral becomes

$$

\int_0^{\pi}\int_0^R y^2\sin\phi drd\phi

$$

How did we end up with an additionally $y$ and a $\sin\phi$?