Finding centre of mass of a semi-circular lamina Watch

theterminator
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I know how to prove this but I've found there are multiple ways of doing this. However, one way of doing gives me half of the correct answer/equation and I don't understand why.

Consider integrating using this graph:



\displaystyle \bar{y} = \frac{ \int_a^{-a} \! \frac{p(a^2 - x^2)}{2} \, \mathrm{d} x }{\frac{\pi a^2 p}{2}} = -\frac{2a}{3\pi}
the answer should be:
\displaystyle \frac{4a}{3\pi}

If you turn this graph 90 degrees and integrate this way (see graph below), you get the correct answer but I do not understand what I did wrong in the first method. Can anyone explain this to me? Thank you

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ghostwalker
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(Original post by theterminator)
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Algebraic slip, I'd guess. I get the right answer. Post working if you wish.
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theterminator
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(Original post by ghostwalker)
Algebraic slip, I'd guess. I get the right answer. Post working if you wish.
Thanks. I spotted it. I said -(-a)^3 = -a^3 which is wrong :P Silly me
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