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I'm trying to calculate the center of gravity of a sector of a circular disk with radius $a$ and vertex angle $2\alpha$ and density $\rho$ =1

I found the Mass using $$\int\int_R \rho dxdy= a^2\alpha$$

I then found the moment about the x-axis as $$2\int_{0}^{\alpha}\int_{0}^{a}r\cos\theta rdrd\theta=\frac{2a}{3\alpha}\sin\alpha$$ and then the moment about the y-axis as $$2\int_{0}^{\alpha}\int_{0}^{a}r\sin\theta rdrd\theta=\frac{2a}{3\alpha}(1-\cos\alpha)=\frac{4a}{3\alpha}(\sin\frac{\alpha}{2})$$ My x moment seems to be right but somehow the moment for the y-axis seems should be zero because the center of mass seems to be on the a-xis. Then shouldn't the calculation yield a zero too?... Where am I going wrong?

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  • $\begingroup$ $\cos(0) \neq 0$ $\endgroup$ Commented Sep 23, 2020 at 20:03

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The problem is in how you wrote your angular integral. It is not $$2\int_0^\alpha...,$$ but $$\int_{-\alpha}^{\alpha}...$$ Then you will have $\cos\alpha-\cos(-\alpha)=0$.

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  • $\begingroup$ You mean to say I should have integrated from $-\alpha$ to $\alpha$ $\endgroup$
    – Orpheus
    Commented Sep 23, 2020 at 20:14
  • $\begingroup$ Yes. Your angle is $2\alpha$ and you just moved the 2 outside the integral. That works for some calculations, such as mass, but not always for calculations involving the angle. $\endgroup$
    – Andrei
    Commented Sep 23, 2020 at 20:16
  • $\begingroup$ I got it...I got the y moment equal to zero...thanks a lot $\endgroup$
    – Orpheus
    Commented Sep 23, 2020 at 20:18

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