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I'm studying in preparation for a Mathematical Analysis II examination and I'm solving past exam exercises. If it's any indicator of difficulty, the exercise is Exercise 4 of 4, part $c$ and graded for 10%.

Given $\vec{F}=(y,x)$, calculate the contour integral $\int_C \vec{F}\cdot\vec{dr}$, on the length of closed path $C:\vec{r}(t)=(\cos t, \sin t), 0\leq t\leq 2\pi$

I used polar coordinates $x=\cos\theta$ and $y=\sin\theta$ in an attempt to solve and end up with a single variable in the integral, but it led to a result of 0.

I know that I should use $\vec{dr}=dxi + dyj$ to try and simplify the product inside the integral, but so far nothing valid came out of it.

Maybe Green's or Stoke's theorem is supposed to be used?

Any help would be greatly appreciated.

Dimitris S.
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2 Answers2

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Solution with potential: Your field $F$ has potential $U(x,y)=xy$, i.e. $F=(U'_x,U'_y)$. Since the path is closed, your integral is zero.

Solution with definition: You want to calculate $\int_C y\,dx+x\,dy$. Using your parametrization of the curve, $x(t)=\cos t$, $y(t)=\sin t$, $0\leq t\leq 2\pi$. Thus $x'(t)=-\sin t$ and $y'(t)=\cos t$ and $$ \begin{aligned} \int_C y\,dx+x\,dy&=\int_0^{2\pi}(y(t)x'(t)+x(t)y'(t))\,dt\\ &=\int_0^{2\pi}(-\sin^2t+\cos^2t)\,dt\\ &=\int_0^{2\pi}\cos 2t\,dt\\ &=\Bigl[\frac{1}{2}\sin 2t\Bigr]_{0}^{2\pi}=0. \end{aligned} $$

Solution with Green's formula: You have $$ \int_C y\,dx+x\,dy=\iint_{|(x,y)|\leq 1}\Bigl(\frac{\partial}{\partial x}x-\frac{\partial}{\partial y}y\Bigr)\,dx\,dy=\iint_{|(x,y)|\leq 1}0\,dx\,dy=0. $$

mickep
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Note that $\nabla \times \vec F=\vec 0$. Then, using Stoke's Theorem

$$\oint_C \vec F\cdot d\vec r=\iint_S \nabla \times \vec F\cdot \hat n\,dS$$

yields

$$\oint_C \vec F\cdot d\vec r =\iint_S 0\,dS=0$$

And we are done!

If one wishes to evaluate the contour integral directly, then we can write

$$\begin{align} \oint_C \vec F\cdot d\vec r&=\int_0^{2\pi}(\hat x \sin \phi+\hat y\cos \phi)\cdot (-\hat x\sin \phi+\hat y \cos \phi)\,d\phi\\\\ &=\int_0^{2\pi} (\cos^2\phi-\sin^2\phi)\,d\phi\\\\ &=0 \end{align}$$

Mark Viola
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    Quick and elegant, this solution can explain the only-5% grade. May I ask for some clarification on $\nabla \times \vec{F} = \vec{0}$? – Dimitris S. Jan 23 '16 at 19:30
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    Thank you! Much appreciative. $$\nabla \times \vec F=\sum_{i=1}^3\sum_{j=1}^3\hat x_i\times \hat x_j \frac{\partial F_j}{\partial x_i}$$Now, for $\vec F=\hat xy+\hat yx$ you can easily show that $\nabla \times \vec F=0$ here. – Mark Viola Jan 23 '16 at 19:33