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.