When you draw it, the approach becomes immediately clear right? Just integrate under that section of intersection:

$\displaystyle S=2\mathop\int\int\limits_{\hspace{-15pt}D} \sqrt{(f_x)^2+(f_y)^2+1}\;dA$

where:

$\displaystyle f(x,y)=\sqrt{9-y^2-x^2}$

$\displaystyle D=\left\{(x,y): (x-3/2)^2+y^2=(3/2)^2\right\}$

Just complete the square in $\displaystyle x^2+y^2=3x$ to get D.

I haven't actually tried to do the integration though. May need to use spherical or other means to evaluate it. Also, it's easy to draw. Here's the Mathematica code:

Code:

pic1 = ContourPlot3D[{x^2 + y^2 + z^2 == 9,
x^2 + y^2 == 3*x}, {x, -3, 3},
{y, -3, 3}, {z, -3, 3}, ContourStyle ->
{Opacity[0.4], LightPurple}]