Diameter $ACE$ is divided at $C$ in the ratio $2:3$.  The two semicircles, $ABC$ and $CDE$, divide the circular region into an upper (shaded) region and a lower region.  The ratio of the area of the upper region to that of the lower region is

$\text{(A)}\ 2:3 \qquad \text{(B)}\ 1:1 \qquad \text{(C)}\ 3:2 \qquad \text{(D)}\ 9:4 \qquad \text{(E)}\ 5:2$
Draw $\overline{AE}$ to divide the big circle in half.  Assign $AC = 4$ and $CE = 6$ so that the radii work out to be integers.  ($4x$ and $6x$ can be used instead, but the $x$ will cancel in the ratio.)  
The shaded region is equal to the area of semicircle $\overarc{AE}$ on top, plus the area of the semicircle $CDE$ on the bottom, minus the area of semicircle $\overarc{ABC}$ on top.
The radii of those three semicircles are $5, 3,$ and $2$, respectively.Thus, the area of the shaded region is $\frac{1}{2}\pi\cdot5^2 + \frac{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi$
The total area of the circle is $\pi \cdot 5^2 = 25\pi$.Thus, the unshaded area is $25\pi - 15\pi = 10\pi$.Therefore the ratio of shaded:unshaded is $15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}$.