Setting up integrals for the moments and the center of mass of a planar region Problem
Consider the following region: a semi-circle with radius = 3 ft on top of a rectangle with height = 11. (with constant density)
a.) Set up integrals for the moments, Mx, My, and the center of mass of the region. DO NOT evaluate the integrals.
b.) Use additivity of moments to find the center of mass of the region.
Progress
For part a.) I'm assuming I have to find separate Mx and My for both the semi-circle and the rectangle and add the Mx of semi-circle and Mx of rectangle together. Is that the correct way to go at this problem?
For the semi-circle $$M_x =  \int _{-3}^3\:\frac{1}{2}\left(\sqrt{3^2-x^2}\right)^2 dx$$ $M_y= 0$ because of symmetry and constant density.
$M_i=  3\pi $  for the rectangle I think I have to find out what $f\left(x\right)$ is. How would I do this?
 A: Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=\sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.
If we take the density $\delta=1$, we have that
$\displaystyle M_x=\int_{-3}^{3}\frac{1}{2}\left((\sqrt{9-x^2})^2-(-11)^2\right)dx=-\frac{1}{2}\int_{-3}^{3}\left(112+x^2\right)dx$,
$\displaystyle M_y=\int_{-3}^{3}x\left(\sqrt{9-x^2}-(-11)\right)dx=0$,
so $\displaystyle\overline{x}=\frac{M_y}{m}=0$ $\;\;\;$ and $\;\;\;$$\displaystyle\overline{y}=\frac{M_x}{m}=\frac{-\frac{1}{2}\int_{-3}^{3}(112+x^2)dx}{\int_{-3}^{3}\left(\sqrt{9-x^2}-(-11)\right)dx}$.

For part b), we can use 
$\displaystyle M_x=\frac{1}{2}\int_{-3}^{3}(9-x^2)dx=\int_0^3(9-x^2)dx=\left[9x-\frac{x^3}{3}\right]_0^3=18$ for the semicircle
and $M_x=dm=(-\frac{11}{2})(11)(6)=-363$ for the rectangle; so 
$\;\;\;\displaystyle\overline{y}=\frac{M_x}{m}=\frac{-363+18}{\frac{1}{2}\pi(3)^2+6(11)}=-\frac{230}{3\pi+43}$
