Given
$$F(x, y) = x^2\mathbf{i} + xy\mathbf{j}$$
$$x^2 + y^2 = 49$$
Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
I've been using $$ \int_C F(\vec{r}(t))\cdot \vec{r}'(t) dt $$ to do other similar problem but usually $\vec{r}$ is given.
Best Answer
As Sigur suggests, parametrize the circle as
$$r(t):=(7\cos t\,,\,7\sin t)\,\,,\,0\leq t\leq 2\pi\Longrightarrow$$
$$\Longrightarrow \oint_C F(r(t))\cdot r'(t)\,dt=\int_0^{2\pi}(49\cos^2t\,,\,49\cos t\sin t)\cdot(-7\sin t\,,\,7\cos t)\,dt=$$
$$=\int_0^{2\pi} 0\,dt=0$$