Find the volume of the solid formed by rotating the region enclosed by
\[
x=0, x=1, y=0, y=9+x^{2}
\]
about the $y$-axis.

Step 1 ：First, we need to understand that the solid is formed by rotating the region enclosed by the given equations about the y-axis. This means that the solid is a series of cylindrical shells, each with a radius equal to the x-coordinate, a height equal to the difference between the upper and lower y-values, and a thickness of dx.

Step 2 ：The volume of each cylindrical shell is given by the formula \(2\pi xh dx\), where x is the radius, h is the height, and dx is the thickness.

Step 3 ：In this case, the radius x ranges from 0 to 1, the height h is given by the equation \(9+x^{2}\), and the thickness dx is a small change in x.

Step 4 ：We can find the total volume of the solid by integrating the volume of each cylindrical shell from x=0 to x=1. This gives us the integral \(\int_{0}^{1} 2\pi x(9+x^{2}) dx\).

Step 5 ：We can simplify this integral by distributing the \(2\pi x\) to get \(\int_{0}^{1} 18\pi x + 2\pi x^{3} dx\).

Step 6 ：We can then evaluate this integral by finding the antiderivative of the integrand and evaluating it at the limits of integration. The antiderivative of \(18\pi x\) is \(9\pi x^{2}\), and the antiderivative of \(2\pi x^{3}\) is \(\frac{1}{2}\pi x^{4}\).

Step 7 ：Evaluating these at the limits of integration gives \((9\pi (1)^{2} + \frac{1}{2}\pi (1)^{4}) - (9\pi (0)^{2} + \frac{1}{2}\pi (0)^{4}) = 9\pi + \frac{1}{2}\pi\).

Step 8 ：Adding these together gives the total volume of the solid as \(\boxed{\frac{19}{2}\pi}\) cubic units.