Finding moments $M_x$ and $M_y$ and center mass of a region 
Set up integrals for the moments, $M_x$ and $M_y$, and the center of mass of the region (constant density, do not evaluate integrals)

The region is a tall rectangle with a semi-circle on top. $r=3,\;h=11$
From what I've learned, moments $M_x$ and $M_y$ can be found by:
$$\int_a^b \frac12[f(x)]^2 dx \;\;\;\text{ and } \;\;\int_a^b xf(x) dx \;\;\;\;\;\text{respectively}$$
I realize that I can use $x^2+y^2=1 \rightarrow \sin^2t + \cos^2t = 1$ for the semi-circle, but what do I do for the rectangle? (if that's even the correct method...)
The center can be found be dividing each moment by the area, which in this case the area of the semi-circle is $\frac{9\pi}{2}$(center at $(0,\frac\pi4))$ and rectangle is $66$
 A: Moments typically take the form
\begin{equation}
I_x=\sum\limits_{i=1}^{n}\left(I_x^{\left(i\right)}\right), \:\text{and}\:I_y=\sum\limits_{i=1}^{n}\left(I_y^{\left(i\right)}\right),
\end{equation}
keeping in mind the equation $I=Mr^2$ for a point mass. So what the sums above are doing is basically adding up all of the bits of mass along the radius squared (in this case discretely, so if we had a bunch of points containing the mass the above would sorta be fine). For a continuous distribution of mass we must use integrals and rewrite the above as
\begin{equation}
I_x=\int\limits_{0}^{M}r^2\:dM,\:\text{and}\:I_y=\int\limits_0^{M}r^2\:dM.
\end{equation}
Notice that these integrals are with respect to $dM$ instead of the length, however. The above equation could be rewritten as
$$ dI=r^2\:dM$$
for a point mass then, and what we want to do is add up all of the $dI$'s along $r$, so we must rewrite the $dM$, or the mass of each individual point, in terms of the geometry of the object. So for a "rod" (line) it will simply be the mass distribution along the length, and if the density is uniform and does not depend on the point within the rod then our $dM$ may be rewritten as
$$dM=\frac{M}{L}\:dr,$$
because the mass-distribution for a uniform "line" is simply the mass of the rod divided by the length of the rod. If density were to be taken into consideration, then we would rewrite our $dM$ as $$dM=\rho\left(r\right)dV=\rho\left(r\right)\:dr$$
for a rod. This allows us to find that, for a rod, the sum of all of the $dI$'s is thus
$$I=\int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{M}{L}r^2\:dr=\frac{L^2M}{12}.$$ Notice also I've changed the limits of integration so that we have the rotational axis positioned at the center.
Now to answer your question. Since we will be finding the moment of a rectangular plate we could apply what we know about the rod, or we could add up all of the $dI$'s in both directions using a double-integration to sum them up in one direction first, and then sum them in the orthogonal direction.
The first method might be slightly easier (though I would prefer the second). Using the parallel-axis theorem we could just sum up all of the rods we've already calculated at different distances from the axis of rotation:
$$dI=\frac{dM\:L^2}{12}+dM\:x^2.$$
As stated above, we must again rewrite our $dM$ in terms of the geometry of the object. Now we're dealing with an area instead of just a length, so we have
$$dM=\rho\left(r\right)\:dV$$
and our "volume" could be thought of as the area of a tiny rectangular region, $dV=dX\cdot dY$. Since the density distribution is assumed to be uniformly $1,$ we don't have to necessarily worry about $dX$ and $dY$ (I'll come back to this in a bit), so we have $dM=\frac{M}{L^2}\cdot L\cdot dx$, or the mass of the entire plate divided by the area of the plate (our mass-distribution) times the area of that piece of mass (which for the rod was just $dr$), and we can replace this to get
$$dI=\frac{ML\:dx}{12}+\frac{Mx^2}{L}\implies I=\int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\left(\frac{ML\:dx}{12}+\frac{Mx^2}{L}\right)=\frac{ML^2}{6}.$$
Notice this is twice the result of just one rod. If we were to use a double integral, considering our $dV=dX\cdot dY$ we would arrive at the same conclusion if the density is uniform:
\begin{equation}
\int_{\frac{-L}{2}}^{\frac{L}{2}}\int_{\frac{-L}{2}}^{\frac{L}{2}}\left(X^2+Y^2\right)\rho\left(x,\:y\right)\:dX\:dY\bigg|_{\rho\left(x,\:y\right)=1}=2\cdot \int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{M}{L}r^2\:dr=\frac{ML^2}{6},
\end{equation}
and if we were to try and find the $I$ of a cube we would write
\begin{equation}
\int_{\frac{-L}{2}}^{\frac{L}{2}}\int_{\frac{-L}{2}}^{\frac{L}{2}}\int_{\frac{-L}{2}}^{\frac{L}{2}}\left(X^2+Y^2+Z^2\right)\rho\left(x,\:y,\:z\right)\:dX\:dY\:dZ\bigg|_{\rho\left(x,\:y,\:z\right)=1}=3\cdot \int\limits_{\frac{-L}{2}}^{\frac{L}{2}}\frac{M}{L}r^2\:dr=\frac{ML^2}{4}.
\end{equation}
So you may change the limits and manipulate the equations above to fit your needs for a rectangular plate or a rectangular prism of any dimensions. Just note that things get a lot more complicated as more dimensions and different situations are considered (such as rotating about an axis at various angles instead of perpendicular to the sides). It's neat to think about, however, such as what the moment might look like for a tumbling asteroid or something of the sort, which would thus change periodically as the object rotated and we must consider that it's not a perfect shape,such as a box but has many, many sides.
