Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

As Sigur suggests, parametrize the circle as

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

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

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

share|improve this answer
    
Great! In this case, the orientation does not matter... lol –  Sigur Oct 31 '12 at 22:33
1  
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.