Calculate moments $M_x$ and $M_{yc}$ of mass of the lamina with density $p=2$ that is the shape of a quarter circle, centered at the origin, in the first quadrant. Note: $yc$ denotes $y$-centre.
I need some help with the solution of this problem. I don't believe I set it up correctly and I'm stuck.
I've got that the area of the region is $\pi r^2/4$.
$M_x = \frac{P}{2}\int_{0}^{r} (r^2-x^2)dx$
$M_y = P \int_{0}^{r}x \sqrt{(r^2-x^2)}dx$
$\bar{X} = \int_{0}^{r} x \frac{\sqrt{(r^2-x^2)}}{ ( \pi r^2/4)} \; dx$
$\bar{Y} =\frac{1}{2}\int_{0}^{r} \frac{(r^2-x^2)}{ ( \pi r^2/4)} \; dx$
When I calculated it, I got a negative number and it's not right since the area is in the first quadrant. I'm sure I went wrong somewhere.