Calculating Work Against Attractive Force Along Semi-Circle Path

[dimensionless]

A semi-circle is given by the function y=R cos(\frac{x\pi}{2R}). An attractive force F at x = R and y = 0 such that F(x, y) = F(R, 0) = 0 and F(x, y) = F(-R, 0) = F_{0}. This force along the semi-circle, relative to its source, is given by F(\theta) = F_{0} sin(\theta/2). What amount of work is done against the force when a particle moves along the circle from \theta = 0 to \theta = \pi.
To find the work I have the equation
W = - \int^{\pi}_{0} F dr
For the force I have the equation
F= F_{0} sin(\theta/2)
For the distance from the particle to the orgin of the force I have
2 R sin(\theta/2)
so for dr I have
2 R sin(\theta/2) d\theta
Putting this all tegether I get
W = - \int^{\pi}_{0} F_{0} sin(\theta/2) 2 R sin(\theta/2) d\theta
This rearanges to
W = -F_{0} R \int^{\pi}_{0} \left[1-cos(\theta) \right] d\theta
Integrating I get
W = -F_{0} R \left[ \theta-sin(\theta) \right]^{\pi}_{0}
This works out to
W = -F_{0} R \pi
The answer I am suppose to get is
W = -F_{0} R
What did I do wrong?

 

[Gokul43201]

The question doesn't make very much sense. Could you please reproduce the question exactly as it appears in the text, and also provide the name of the text ?

 

[dimensionless]

I rewrote the problem a little bit. I'm afraid it's still a little unclear without the diagram though. This question is not from a textbook.

 

[Gokul43201]

I think the mistake you are making is in interpreting what dr refers to. This is an elemental distance traveled by the particle (along the semicircular path) - it is not an element of the distance of the particle from the origin. Also keep in mind that the angle between F and dr changes with the motion of the particle.