# Integral with vector field in a circle.

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.

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As Sigur suggests, parametrize the circle as

$$r(t):=(7\cos t\,,\,7\sin t)\,\,,\,0\leq t\leq 2\pi\Longrightarrow$$

$$\Longrightarrow \oint_C F(r(t))\cdot r'(t)\,dt=\int_0^{2\pi}(49\cos^2t\,,\,49\cos t\sin t)\cdot(-7\sin t\,,\,7\cos t)\,dt=$$

$$=\int_0^{2\pi} 0\,dt=0$$

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Great! In this case, the orientation does not matter... lol – Sigur Oct 31 '12 at 22:33
Of course, had the result been different from zero you'd just have to change signs... – DonAntonio Oct 31 '12 at 22:38

Your curve is given by $C: r(t)=(7\cos (t),7\sin (t))$, $0\leq t\leq 2\pi$. But be careful with the orientation. This one is counter-clock wise oriented. But it is just a matter of signal.

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