# Line integral method calculate work done by a particle

I'm having trouble knowing how to go about solving this question:

Q: The force on a particle at a point with position vector $$r = xi + yj + zk$$ exerted by a charge at the origin is $$F(r)=\left(\frac{P(r)}{|r|^2}\right)$$ in which P is constant. Calculate the work done as the particle moves in a straight line from (1, 0, 0) to (1, 2, 3).

What I think I need to do:
$$r_1=i$$, $$\quad$$ $$r_2=i+2j+3k$$ $$\quad$$ so let $$r(t)=i+2tj+3tk$$,$$\quad$$ $$0\lt t \lt 1$$
Then $$\frac{dr}{dt}=2+3=5$$ and $$|r|^2=1+13t^2$$
Therefore, $$F(r)=\frac{P(i+2tj+3tk)}{1+13t^2}$$ As such, $$\int_C F(r) \cdot dr=\int_0^1 \frac{P(i+2tj+3tk)}{1+13t^2} \cdot5 dt=\int_0^1 \frac{5P(1+5t)}{1+13t^2}dt=5P\int_0^1 \frac{1+5t}{1+13t^2}dt$$

Is this the right way to go about answering this question, or am I doing something completely wrong? Any advice would be greatly appreciated.

• Note that $r(t)=(1,2t,3t)$ and $r'(t)$ is also a vector. – PierreCarre Feb 1 at 13:26

$$\int_C F(r)\cdot dr = \int_0^1 F(r(t))\cdot r'(t) dt = \int_0^1 \frac{P}{1+4t^2+9t^2} (1,2t,3t)\cdot(0,2,3) dt = \int_0^1 \frac{13 P \,t}{1+13t^2}dt$$