Since your integral is with respect to $x$, and you’re using the shell method, you must be slicing your region into vertical strips and revolving those strips around a vertical axis. (Here vertical means ‘parallel to the $y$-axis’.) You have one of these strips, of ‘width’ $dx$, at each value of $x$ from $a$ to $b$. Now consider the strip at a particular $x$: the radius of the shell that it generates is the distance from $x$ to the axis of revolution.
Suppose, for example, that $a=2$ and $b=5$. If the axis of revolution is the line $x=1$, then for any $x \in [2,5]$ the distance from $x$ to the axis is $x-1$, and that is the radius of the corresponding shell. If, on the other hand, the axis is $x=7$, the distance from $x$ to the axis is $7-x$, and now that is the radius of the corresponding shell. And if the axis is $x=-7$, the distance from $x$ to the axis (and hence the radius of the shell) is $x-(-7) = x+7$.
Remember, the horizontal distance between any two points is simply the larger $x$-coordinate minus the smaller $x$-coordinate. If you were using the shell method with a horizontal axis of revolution, your slices would be horizontal (parallel to the $x$-axis), and your integral would be with respect to $y$. In that case the radius of the shell generated by the slice at a particular value of $y$ would be the vertical distance from $y$ to the axis of revolution.