What is the volume of the solid of revolution obtained by rotating the region bounded by $f(x)=x+1$ and the $x$-axis over $[1,2]$ about the line $x=-2$ ? Enter your answer in terms of $\pi$.

Step 1 ：The volume of a solid of revolution can be found using the formula for the volume of a cylindrical shell, which is \(V = 2\pi \int_a^b r(x)h(x)dx\), where \(r(x)\) is the radius of the shell and \(h(x)\) is the height of the shell.

Step 2 ：In this case, the radius of the shell is the distance from the line of rotation to the function, which is \(x - (-2) = x + 2\), and the height of the shell is the value of the function, which is \(f(x) = x + 1\).

Step 3 ：Therefore, the volume of the solid of revolution is \(V = 2\pi \int_1^2 (x + 2)(x + 1)dx\).

Step 4 ：By calculating the integral, we find that the volume of the solid of revolution is \(V = \frac{53\pi}{3}\).

Step 5 ：Final Answer: The volume of the solid of revolution is \(\boxed{\frac{53\pi}{3}}\).