Originally Posted by

**Twig** Yes, true.

$\displaystyle z^{2}+(r-a)^{2}=a^{2} $ , it feels like he just changes x to r.

I am with on the moment arm: $\displaystyle \frac{1}{2}\sqrt{a^{2}-(x-a)^{2}} $

Shoudlnt it work just integrating "as usual" without polar coordinates, even if polar will make it easier?

What is wrong below?

$\displaystyle \int_{a}^{2a}\underbrace{\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}}_{z_{c}}\cdot \underbrace{\sqrt{a^{2}-(x-a)^{2}}}_{Height} \; dx $

I thought polar was $\displaystyle x=r\cdot cos(\theta) $