This problem is from my Homework. I have the idea of the solution. But I don't know if it is right or not. I am quoting it from the question. "Find the work done in moving a particle once around 3/4 th a circle C in the xy plane, If the circle has center at the origin and radius 3 and the force field is given by, $\vec{F} = (2x-y+z) \mathbf{i} + (x+y-z^2) \mathbf{j} + (3x-4y+4z) \mathbf{k}$
My idea is, if i calculate it for the whole circle (let it be x) and then do this thing
$a=x-\frac{3}{4}x$ then the value of a which I am getting is it the correct answer? Or, I am wrong?
I think it is best to just solve this directly. Observe that the circle may be parametrized by
$$\mathbf{r}(\theta)=\cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j}, \quad \theta \in [0, \tfrac{3\pi}{2}]$$
and that the work is given by
$$\int_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{3\pi/2} \mathbf{F}(\mathbf{r}(\theta)) \cdot \mathbf{r}'(\theta) \, d\theta.$$
When you do the dot product, you will get a trigonometric integral. I trust you to take it from there.