Work Done by Force Field

Switching to polar coordinates, we have that $\vec{F} = <-\sin(\theta), \cos(\theta) >$ . Then on the circular path $C$ , i.e., $\vec{r} = <\cos(\theta), \sin(\theta)>$ , the work done is

$W = \displaystyle\int_{C} \vec{F} \cdot \vec{dr} = \int_{0}^{2\pi} \vec{F} \cdot \dfrac{\vec{dr}}{d\theta} d\theta = \int_{0}^{2\pi} <-\sin(\theta), \cos(\theta)> \cdot <-\sin(\theta), \cos(\theta)> d\theta =$

$\displaystyle \int_{0}^{2\pi} (\sin^{2}(\theta) + \cos^{2}(\theta)) d\theta = \int_{0}^{2\pi} d\theta = 2\pi = \boxed{6.283}$ to 3 decimal places.

Relevant wiki: Uniform Circular Motion - Problem Solving

$\huge{\text{Oops the image has wrong expression of F}}$

Now see , Always Angle between $F$ and $dx$ is $0$ .

So Work done = $\large{\int^{2 \pi}_{0} F .dx}$ [ $2 \pi$ Because Circumference = $2 \pi . 1 = 2 \pi$ ]

So now, W = F $\large{\int^{2 \pi}_{0}} dx$

So $W = F * ( 2 \pi)$

Now Magnitude of $F = 1N$ , So

$W = 6.283 J$