Calculating the Center of Mass of an Arc

[RoyalCat]

Homework Statement

There is an arc with an angle of 2\alpha and radius R
What is its center of mass. (It is implied that it is to be measured from the center of the circle the arc is a part of.

Homework Equations

Integration, trigonometry.

The Attempt at a Solution

Well, the formula for a center of mass of i objects is:
\vec r_{cm}=\frac{\Sigma m_i \vec r_i}{M_{tot}}

Since the arc has symmetry with respect to the y axis, all we need to find is the y coordinate of its center of mass.

We'll simply replace the discrete summation with an integral (I'm not too hot with these).

The mass of every tiny element of the arc length, dS is \sigma dS=\sigma R d\theta
Where \sigma \equiv \frac{M}{S}=\frac{M}{2R\alpha}

The distance of 'each' of these masses from the x axis, their y value, is provided by R\sin{\theta}

What I want my integral to do is 'sweep' across the arc. Would the correct boundaries for such an integral be:
\frac{\pi}{2}-\alpha \leftrightarrow \frac{\pi}{2}+\alpha

The integral would need to sum up the product of each of the mass elements corresponding to a particular angle theta with its corresponding height (The sine of that theta).

Would the correct integral be (I can't quite get the hang of boundaries in LaTeX, so refer to the boundaries I wrote above):
\int \sigma R*R\sin{\theta}\ d\theta

I'm not in the least bit skilled with integrals so I'd like to know if that's the correct expression, and if so, where the d\theta goes from the expression for the tiny mass element. Or if I got it wrong, I'd love a hint as to how to construct the proper integral. (A link with the basics of such integration would be greatly appreciated as well.

The textbook's answer is \frac{R\sin{\alpha}}{\alpha}
The way I solved the integral gave me that answer as well.

 

[alex_rodin]

There is all correct in your solution. The d\theta is not goes from the expression for the tiny mass element. It goes out after integrating:
\int\limits_{\frac \pi 2 - \alpha}^{\frac \pi 2 + \alpha} \sigma R^2 \sin \theta d\theta = \left . -\sigma R^2 \cos \theta \right |_{\frac \pi 2 - \alpha}^{\frac \pi 2 + \alpha} =.

Read this book about integrating and methods of applied mathematics: Zeldovich Ya.B., Yaglom I.M.
Higher math for beginners (mostly physicists and engineers).

 

[RoyalCat]

Alright, thank you very much! :)