Find the area of the surface of the part of the sphere $x^{2}+y^{2}+z^{2}=9$ that lies within the cylinder $x^{2}+y^{2}=3 x$ and above the $x y$-plane. Round to 2 decimal places.

Step 1 ：First, we need to calculate the height of the cap. The coordinates of the top of the sphere are $(0, 0, 3)$ and the coordinates of the edge of the cap are $(\frac{3}{2}, 0, 0)$. We will use the distance formula to find the height of the cap.

Step 2 ：The height of the cap is calculated as $h = \sqrt{(x2-x1)^{2} + (y2-y1)^{2} + (z2-z1)^{2}} = 3.3541019662496847$.

Step 3 ：The radius of the sphere is given as $r = 3$.

Step 4 ：Next, we substitute the values of $r$ and $h$ into the formula for the area of a spherical cap to find the area of the part of the sphere that lies within the cylinder and above the $xy$-plane.

Step 5 ：The area of the spherical cap is calculated as $area = 2 \pi r h = 63.22$.

Step 6 ：Finally, we round the area to 2 decimal places to get the final answer.

Step 7 ：Final Answer: The area of the surface of the part of the sphere $x^{2}+y^{2}+z^{2}=9$ that lies within the cylinder $x^{2}+y^{2}=3 x$ and above the $x y$-plane is \boxed{63.22} square units.