# Center of Mass Sector of a Circle

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?

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

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 $$\cos\alpha-\cos(-\alpha)=0$$.
• You mean to say I should have integrated from $-\alpha$ to $\alpha$ Commented Sep 23, 2020 at 20:14
• 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. Commented Sep 23, 2020 at 20:16