According to calculus - an intuitive and physical approach(2nd) 466p
We have shown that if an arc of a curve is revolved around the $x$-axis and the arc extends from $x = a$ to $x = b$, the surface area generated is given by $$S = 2\pi\int_{a}^{b}y\sqrt{1+(y')^2}\;dx.\tag{48}$$ This formula is sometimes written in another form. In our work on arc length we showed that $$\frac{ds}{dx} = \sqrt{1+(y')^2}\;,$$ and this means that in differential form $$ds = \sqrt{1+(y')^2}\;dx.$$ Hence one can write (48) in the form $$S = 2\pi\int_{a}^{b}y\;ds.\tag{49}$$
Integrals in (48) and (49) have the same range. I think the process is similar to the change of variable. The relation between $x$ and $s$ is $s(x) = \int\sqrt{1+(y')^2}\;dx$ where $y'$ is a function of $x$. Isn't the right equation for (49) $S = 2\pi\int_{s(a)}^{s(b)}y\;ds$ ?
Your equation, where the bounds of integration are $s(a),s(b)$, are what I’d consider much more correct. However, the author’s original meaning was clear, even if not with the best notation, strictly speaking, since the bounds are always assumed to be with respect to the differential; here, it was $ds$, and it is always correct to make sure you’re integrating the right variable over the right bounds.
The author meant: integrate $ds$ between the coordinates $a$ and $b$, and to avoid confusion it is clearer to write $s(a),s(b)$.
This works because $s(x)$ is injective; if it weren’t, things could get more complicated and the author’s original notation may have perhaps been more appropriate.