A very long insulating cylinder has radius R and carries positive charge distributed throughout its...

Answer:
1. ${\displaystyle E\left(r\right)=\frac{\alpha }{4\pi {\xi }_0}(2-\frac{r}{R})}$
2. ${\displaystyle E\left(r\right)=\frac{1}{4\pi {\xi }_0}\frac{\alpha R}{r}}$
3.The results from part 1 and 2 agree when ${\displaystyle r=R}$.
Explanation:
The volume charge density is given as
${\displaystyle \rho \left(r\right)=\alpha (1-\frac{r}{R})}$
We will investigate this question in two parts. First ${\displaystyle r<R,thenr>R}$. We will show that at ${\displaystyle r=R}$, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
${\displaystyle \int \overrightarrow{E}d\overrightarrow{a}=\frac{Q_{enc}}{{\xi }_0}}$
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
${\displaystyle \int da=2\pi rh}$
where ‘h’ is the length of the imaginary Gaussian surface.
$Q_{enc}={\int }_0^r\rho (r)hdr=\alpha h{\int }_0^r(1-r/R)dr=\alpha h(r-\frac{r^2}{2R})\{\begin{array}{ll}r=r& \\ r=0\end{array}=\alpha h(\frac{2Rr-r^2}{2R})$
${\displaystyle E2\pi rh=\alpha h\frac{2Rr-r^2}{2R{\xi }_0}}$
${\displaystyle E\left(r\right)=\alpha \frac{2R-r}{4\pi {\xi }_{0}R}}$
${\displaystyle E\left(r\right)=\frac{\alpha }{4\pi {\xi }_0}(2-\frac{r}{R})}$
2. For ${\displaystyle r>R}$, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
$Q_{enc}={\int }_0^R\rho (r)hdr=\alpha {\int }_0^R(1-r/R)hdr=\alpha h(r-\frac{r^2}{2R})\{\begin{array}{ll}r=R& \\ r=0\end{array}=\alpha h(R-\frac{R^2}{2R})=\alpha h\frac{R}{2}$
${\displaystyle E2\pi rh=\frac{\alpha Rh}{2{\xi }_0}}$
${\displaystyle E\left(r\right)=\frac{1}{4\pi {\xi }_0}\frac{\alpha R}{r}}$
3. At the boundary where ${\displaystyle r=R}$:
${\displaystyle E(r=R)=\frac{\alpha }{4\pi {\xi }_0}(2-\frac{r}{R})=\frac{\alpha }{4\pi {\xi }_0}}$