How to calculate work done when movement is not in the direction of the force? I was trying to solve this problem: 

but I encountred a problem while I was trying to evaluate the work done by $F$ over $AB$ which is: 
$$ W= \int_{A}^{B} \vec{F}\cdot d\vec{s} $$ 
but how to compute this kind of integral? I tried to switch the variabe $d\vec{s}$ with $ Rd\theta$:
$$W= \int_{0}^{\pi/6} FR ~d\theta $$ 
but there isn't a function of $\theta$ inside the integral, so how to evaluate this integral?
 A: You forgot the unit vector $[-\sin\theta,\cos\theta]$ along the path. In your integral $W= \int_{0}^{\pi/6} F\cdot R ~d\theta $, replace $F \cdot R ~d\theta$ by $R F\cdot [-\sin\theta,\cos\theta] ~d\theta$ and you are done ($F\cdot [-\sin\theta,\cos\theta] = -F_1\sin\theta+F_2\cos\theta$ for $F=[-F_1,F_2]$.)
A: Here $\vec{F} \cdot d\vec{s} = FR d\theta \cos(\pi-\theta)$. Because when the object is at angle $\theta$ from the vertical, the angle between $\vec{F}$ and $d\vec{s}$ is $(\pi-\theta) $. Refer to the diagram below (ds is exaggerated) : 
Now this is can be easily integrated using $\cos(\pi-\theta ) = -\cos\theta $. But the limits you have written are wrong, $\frac{\pi} {6} $ should actually be the lower limit and 0 be the upper. 
A: A particle is moving between points $\:\mathrm{A}\:$ and $\:\mathrm{B}\:$ of a  curve $\:C\:$ and a constant force $\:\mathbf{F}\:$ is applied continuously  on it. The work done by $\:\mathbf{F}\:$  between points $\:\mathrm{A}\:$ and $\:\mathrm{B}\:$ is ...
On Plane 

In Space


EDIT

OP asks : 
  this means that the work done over AB is equal to the work done over the purple path in the Figure below?
  
  Answer :
  No, the work done over AB is equal to the work done over the green path AB', the projection of the curvilinear orbit (here: circular arc) AB on the direction of the constant force $\:\mathbf{F}\:$. This is valid for constant force vector. If the force is not constant in magnitude and/or direction then you must study about line integrals.

The purple path AB (segment) is the integral :
$$
\int\limits_{\rm{arc\:AB}} \mathrm{d}\mathbf{s}
$$
