Answer
The electric potential at $P$ is $V\left( P \right) = - 10096.7$ volts.
Work Step by Step
The charge distribution can be parametrized by
${\bf{r}}\left( t \right) = \left( {0,t,0} \right)$, ${\ \ \ }$ for $1 \le t \le 3$
Thus,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {0,1,0} \right)\cdot\left( {0,1,0} \right)} = 1$
$\delta \left( {{\bf{r}}\left( t \right)} \right) = - t$
The distance ${d_P}$ from $P = \left( {2,0,2} \right)$ to a point $\left( {0,y,0} \right)$ on the charge distribution is
${d_P} = \sqrt {{{\left( { - 2} \right)}^2} + {y^2} + {{\left( { - 2} \right)}^2}} = \sqrt {8 + {y^2}} $
Using Eq. (6), the electric potential at $P$ is given by
$V\left( P \right) = k\mathop \smallint \limits_C^{} \frac{{\delta \left( {x,y,z} \right){\rm{d}}s}}{{{d_P}}} = k\mathop \smallint \limits_1^3 \frac{{\delta \left( {x,y,z} \right)}}{{\sqrt {8 + {y^2}} }}{\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y,z} \right) = \delta \left( {x,y,z} \right) = - y$, we obtain
$V\left( P \right) = - k\mathop \smallint \limits_1^3 \frac{t}{{\sqrt {8 + {t^2}} }}{\rm{d}}t$
Write $u = 8 + {t^2}$. So, $du = 2tdt$.
By changing the variable from $t$ to $u$, the integral becomes
$V\left( P \right) = - \frac{k}{2}\mathop \smallint \limits_9^{17} {u^{ - 1/2}}{\rm{d}}u$
$V\left( P \right) = - \frac{k}{2}\cdot2\left( {{u^{1/2}}|_9^{17}} \right) = - k\left( {\sqrt {17} - 3} \right)$
With the charge density in units of ${10^{ - 6}}$ and $k = 8.99 \times {10^9}$, we obtain
$V\left( P \right) = - 8.99 \times {10^9}\left( {\sqrt {17} - 3} \right) \times {10^{ - 6}} = - 10096.7$
So, the electric potential at $P$ is $V\left( P \right) = - 10096.7$ volts.