If the radius of every shell $k$ for where k is the $k^{th}$ subinterval is,$\left(a + (k-\tfrac{1}{2}) \frac{b-a}{n}\right)$, you have a increasing radii sequence $(a+\dfrac{(b-a)}{2n}),(a+\dfrac{3(b-a)}{2n}),(a+\dfrac{5(b-a)}{2n}),....$
The sequence means the second shell includes the first shell, and is not hollow, the third shell includes the first and second shell. The third shell is not hollow.
Just imagine $a=0, b=5$, and the height of each shell is constant one.
Your radius includes $\dfrac{5}{2n},\dfrac{15}{2n},\dfrac{25}{2n}.$ Your cylinders volumes $2\pi$ times those radii. The third cylinder includes the first and second radii, and you're double counting volume.
The correct way to conceptualize the cylindrical shell is to start from a correct understanding of the Riemann sum's partition structure, then take a representative subinterval and compute its volume as it is swept. If we assume that the axis of rotation is the $y$-axis and that $0 < a < b$, then a representative subinterval would have the form $$x \in \left[a + (k-1) \frac{b-a}{n}, a + k \frac{b-a}{n}\right].$$ In fact, this is the $k^{\rm th}$ subinterval for $k \in \{1, 2, \ldots, n\}$. If we take the left endpoint for the height of the cylindrical shell, this gives us $h = f\left(a + (k-1) \frac{b-a}{n}\right)$. The outer radius of the shell is the right endpoint $r_1 = a + k \frac{b-a}{n}$, and the inner radius is the left endpoint $r_2 = a + (k-1) \frac{b-a}{n}$. So its volume is exactly $$\begin{align} \pi (r_1^2 - r_2^2) h &= \pi (r_1 + r_2)(r_1 - r_2) h \\ &= \pi \left(2a + (2k-1)\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right) h \\ &= 2\pi f\left(a + (k-1)\frac{b-a}{n}\right) \left(a + (k-\tfrac{1}{2}) \frac{b-a}{n}\right) \frac{b-a}{n}. \end{align}$$ However, in the limit as $n \to \infty$, it is unnecessary to take this Riemann sum: it is better to use the differential volume $$dV = 2 \pi x f(x) \, dx$$ of a representative shell with radius $x$. We can argue that its exact volume for a width of $\Delta x = \frac{b-a}{n}$ is $$\begin{align} \Delta V &= \pi \left((x + \Delta x)^2 - x^2\right) f(x) \\ &= \pi \left(2x \,\Delta x + (\Delta x)^2\right) f(x) \\ &= 2\pi x \, \Delta x f(x) + \pi f(x) (\Delta x)^2. \end{align}$$ Informally, as $\Delta x \to 0$, $\Delta x \to dx$ and $(\Delta x)^2 \to 0$; that is to say, the second term $\pi f(x) (\Delta x)^2$ vanishes much faster than the first term, and in doing so, we recover the formula for the differential volume as shown above.
Notably, the second term quantifies the error in approximating the exact volume of the cylindrical shell as an "unrolled" rectangular slab with height $f(x)$, width $2\pi x$ equal to the circumference of the shell, and thickness $\Delta x$.
The radius of the shell is not $a + (k-\tfrac{1}{2})\frac{b-a}{n}$.
As explained in my previous answer, there are two radii:
$$r_1 = a + k \frac{b-a}{n}, \quad r_2 = a + (k-1)\frac{b-a}{n}.$$
When we compute the volume of the cylindrical shell, we take the area of the annular base and multiply by the height. The annulus has area $$\pi (r_1^2 - r_2^2),$$ and this happens to simplify because we can factor: $$r_1^2 - r_2^2 = (r_1 + r_2)(r_1 - r_2).$$ The quantity $r_1 - r_2$ is the width of the annulus, which is just the width of the subinterval. The quantity $r_1 + r_2$ is twice the mean radius. This is not the radius of the shell and the shells do not overlap.
If I asked you to calculate the area of a circle of radius $r = 10$, you would compute $A = \pi r^2 = 100\pi$. But if I told you to do it by adding up the individual areas of a series of concentric, non-overlapping annular regions, each with width $1$, by partitioning the radius into the subintervals $$[0, 10] = [0,1) \cup [1, 2) \cup [2, 3) \cup \cdots \cup [9, 10],$$ then the area of the $k^{\rm th}$ annulus is $$\pi (k^2 - (k-1)^2) = 2\pi (k + \tfrac{1}{2}),$$ but $k + \frac{1}{2}$ is neither the inner nor outer radius, and the partition has not changed. It is still not overlapping.