Cfrgtkky

I'm studying for a course in electromagnetism, and I've been given an electric field for which I need to find the associated scalar potential. I was going to originally post this in the physics section but I think my problems are more calculus related. The field is the field generated by a sphere of radius $R$ with constant charge density $rho$ throughout its volume, so that the total charge $Q=dfrac{4pi r^3 rho}{3}$contained in the sphere is constant.

The electric field is given by $vec{E}_{text{in}}(vec{r})=dfrac{Q}{4pi epsilon_0 R^3}r$ and $vec{E}_{text{out}}(vec{r})=dfrac{Q}{4pi epsilon_0 r^2}$, where the former is valid for $rleq R$ and the latter for $rgeq R$. This I've calculated before and I do not have trouble with. The scalar potential $phi(vec{r})$ is defined by $vec{E}=-vec{nabla}phi$. The provided solutions to the problem are hand written but I'll type them here using the exact same notation:

$phi_{text{in}}=-int vec{E}_{text{in}}dvec{r}=-dfrac{Qr^2}{8pi epsilon_0 R^3} + C_1$

$phi_{text{out}}=-int vec{E}_{text{out}}dvec{r}=dfrac{Q}{4pi epsilon_0 r} +C_2$

This is literally all the information I've been given. I really don't know what these integrals are, nor how they follow from the above equation. I can see that the result of the first integral for example is just the indefinite integral $-int dfrac{Q}{4pi epsilon_0 R^3}r dr$ but I can't see how this stage was reached.

Any clarification would be much appreciated!

The integrals are over the surface $S$ of the sphere (at radius $R$). The confusion might be because you overlooked a $cdot$. The integrand is the dot product of $vec{E}$ with the unit normal to the surface, which your professor has written as $dvec{r}$ but wouild more commonly be written as $dS$. The integral can be read in spherical coordinates as
$$
int_S E_r(R, theta, phi) r,dtheta,dphi
$$
and the $vec{E}cdot vec{r}$ is just expressing that you need to take the $r$ component of the field.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
Phipars{vec{r}} & = -int_{vec{a}}^{vec{r}}vec{mrm{E}}pars{vec{s}}cdotddvec{s} + Phipars{vec{a}} =
overbrace{int_{vec{a}}^{vec{r}}{Q over 4piepsilon_{0}s^{2}},{vec{s} over s}cdotddvec{s}}^{ds{%
left{begin{array}{l}
mbox{Note that}
\
ds{vec{s}cdotddvec{s} = {1 over 2},ddpars{vec{s}cdotvec{s}}}
\ =
ds{{1 over 2},ddpars{s^{2}} = color{red}{s,dd s}}
end{array}right.}}
+ Phipars{vec{a}}
\[5mm] & =
-,{Q over 4piepsilon_{0}}int_{a}^{r}{dd s over s^{2}} + Phipars{vec{a}} =
{Q over 4piepsilon_{0}}pars{{1 over r} - {1 over a}} + Phipars{vec{a}}
end{align}

Set $ds{Phipars{vec{a}} = 0}$ as $ds{a to infty}$ such that
$bbx{ds{Phipars{vec{r}} = {Q over 4piepsilon_{0}r}}}$.

What the OP has given as the electric fields inside and outside the sphere are only the magnitudes of these fields, as has been emphasized in other comments and answers. I suspect that those expressions were derived using Gauss' Law, which relied on spherical symmetry to assume two things: (i) that the field at a point with position vector $vec{r}$, a distance $r = |vec{r}|$ from the center of the sphere, depended only on this distance, and (ii) that the field was directed radially outward from the center of the sphere along the unit vector $hat{r} = vec{r}/r$. So the electric field vector is a piecewise continuous function of the radial coordinate $r$ alone, with direction $hat{r}$, given by
$$begin{align}vec{E}(r) ,=, begin{cases}~~vec{E}_{in}(r) = dfrac{q}{R^3},r,hat{r}, ~~~ r leq R, \ \~~ vec{E}_{out}(r) = dfrac{q}{r^2},hat{r}, ~~~ r geq R, end{cases}end{align}$$ where I am writing $$q = dfrac{Q}{4piepsilon_0}$$ for convenience. Since $vec{E}(r)= -nabla phi(r)$ depends only on the radial coordinate $r$, it reduces to $$vec{E}(r) = -dfrac{dphi}{dr}(r),hat{r}.$$ The negative derivative of the scalar function $phi(r)$ with respect to $r$ thus gives the radial (and only) component of $vec{E}$, so from the piecewise expressions for the electric field we find the piecewise expressions for the derivative of $phi$: $$begin{align}-dfrac{dphi}{dr} = begin{cases}~~dfrac{q}{R^3},r, ~~~ r leq R, \ \~~ dfrac{q}{r^2}, ~~~ r geq R, end{cases}end{align}$$ The indefinite integral of each side in the two cases gives $$phi(r) = -intdfrac{q}{R^3},r,dr = - dfrac{q}{R^3}dfrac{r^2}{2} + C_1, ~~~~r leq R,$$ and $$phi(r) = -intdfrac{q}{r^2},dr = dfrac{q}{r} + C_2, ~~~~r geq R.$$ In the second expression with $r geq R$, in order for the potential to vanish as $r to infty$, we must set the integration constant $C_2 = 0$, hence $$phi(r) = dfrac{q}{r}, ~~~~r geq R.$$ In order for the potential to be continuous at $r = R$, the potentials must be equal at the boundary: $$-dfrac{q}{R^3}dfrac{R^2}{2} + C_1 = dfrac{q}{R},$$ from which we find $$C_1 = dfrac{3q}{2R}.$$ The scalar potential is thus given by the piecewise continuous function $$begin{align}phi(r) ,=, begin{cases}~ -dfrac{q}{R^3}dfrac{r^2}{2} + dfrac{3q}{2R}, ~~~r leq R,\ \~~ dfrac{q}{r}, ~~~r geq R.end{cases}end{align}$$ In terms of $q = Q/4piepsilon_0$, these are $$begin{align}phi(r) ,=, begin{cases}~~-dfrac{Q}{8piepsilon_0 R^3},r^2 + dfrac{3Q}{8piepsilon_0 R}, ~~~r leq R,\ \ ~~dfrac{Q}{4piepsilon_0 r}, ~~~r geq R.end{cases}end{align}$$