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My doubt is in the part where he sets $V=V_s - \int{E dr}$. My idea is that he is summing two potentials, from the surface, and the one he's trying to calculate with the integral. Is that right? If so:

1) Why the potential at the surface? 2) The first $E$ calculated is valid for the whole sphere (which includes the interior of the shell)?

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1) Why the potential at the surface?

This approach is probably used because part (a) of the problem gives you an explicit expression for $V_S$ and so the expression for $V(r)$ is self-contained without having to consider what happens inside the shell. Basically the solution makes use of the fact that $$ V_S = \int_0^{r_2}{E dr} = \int_0^{r}{E dr} + \int_{r}^{r_2}{E dr} = V(r) + \int_{r}^{r_2}{E dr} $$ Compare this to the alternative, which would be to calculate $$ V(r) = \int_0^{r}{E dr} = \int_0^{r_1}{E dr} + \int_{r_1}^{r}{E dr} $$ You'd have to do the 2nd integral, which is similar to $\int_{r}^{r_2}{E dr}$, but you'd also have to solve part (c) to tell what is the 1st integral. Trivial as that may be, the solution you have shows that you don't have to look at part (c) yet.

2) The first E calculated is valid for the whole sphere (which includes the interior of the shell)?

The expression for E is calculated strictly for $r_1 \le r \le r_2$, meaning it does not apply to $r < r_1$ or $r > r_2$. If you ask whether that expression can be extended to the interior of the shell, the answer is no, it cannot. If you ask whether it takes into account the interior of the shell, yes, it does - through the nature of the charge distribution.

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