Answer
The electric potential at $P$ is $V\left( P \right) = - 44135.2$ volts.
Work Step by Step
We have the charge density $\delta \left( {x,y} \right) = - y\sqrt {{x^2} + 1} $. The curve can be parametrized by
${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$, ${\ \ \ }$ for $1 \le t \le 2$
Thus,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,2t} \right)\cdot\left( {1,2t} \right)} $
$ = \sqrt {1 + 4{t^2}} $
$\delta \left( {{\bf{r}}\left( t \right)} \right) = - {t^2}\sqrt {{t^2} + 1} $
The distance ${d_P}$ from $P = \left( {0,0} \right)$ to a point $\left( {x,y} \right)$ on the curve is
${d_P} = \sqrt {{x^2} + {y^2}} = \sqrt {{t^2} + {t^4}} = t\sqrt {1 + {t^2}} $
Using Eq. (6), the electric potential at $P$ is given by
$V\left( P \right) = k\mathop \smallint \limits_C^{} \frac{{\delta \left( {x,y} \right){\rm{d}}s}}{{{d_P}}} = k\mathop \smallint \limits_1^2 \frac{{\delta \left( {x,y} \right)}}{{\sqrt {{x^2} + {y^2}} }}{\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y} \right) = \delta \left( {x,y} \right) = - y\sqrt {{x^2} + 1} $, we obtain
$V\left( P \right) = k\mathop \smallint \limits_1^2 \frac{{\left( { - {t^2}\sqrt {{t^2} + 1} } \right)\left( {\sqrt {1 + 4{t^2}} } \right)}}{{t\sqrt {1 + {t^2}} }}{\rm{d}}t$
$V\left( P \right) = - k\mathop \smallint \limits_1^2 t\sqrt {1 + 4{t^2}} {\rm{d}}t$
Write $u = 1 + 4{t^2}$. So, $du = 8tdt$.
By changing the variable from $t$ to $u$, the integral becomes
$V\left( P \right) = - \frac{k}{8}\mathop \smallint \limits_5^{17} {u^{1/2}}{\rm{d}}u$
$V\left( P \right) = - \frac{k}{8}\cdot\frac{2}{3}\left( {{u^{3/2}}|_5^{17}} \right) = - \frac{k}{{12}}\left( {{{17}^{3/2}} - {5^{3/2}}} \right)$
With the charge density in units of ${10^{ - 6}}$ and $k = 8.99 \times {10^9}$, we obtain
$V\left( P \right) = - \frac{{8.99 \times {{10}^9}}}{{12}}\left( {{{17}^{3/2}} - {5^{3/2}}} \right) \times {10^{ - 6}} = - 44135.2$
So, the electric potential at $P$ is $V\left( P \right) = - 44135.2$ volts.
