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In this question, I was trying to find the counter-clockwise circulation for the field $\mathbf{F}$ and curve $C$ in this image:

enter image description here

As you can see, Sorfosh kindly took the time to answer my question, but I am confused by his answer:

  1. The solution he got for the circulation was $\oint_{\partial D} (P\, dx+Q\, dy) =-4\pi$. However, according to my textbook, a counter-clockwise circulation should be a positive number; negative numbers indicate clockwise circulation. Since we're trying to find the counter-clockwise circulation, shouldn't we be getting a positive value?

  2. His parameterization is $x=\sqrt{2}\cos(t)$ and $y=\sin(t)$. But if we let $t = 0$, then we get $x = \sqrt{2}$ and $y = 0$, which is not the curve in the image? Should we not be parameterizing the curve $C$ in such a way as to get $x = 2$ and $y = 0$ when $t = 0$?

Thank you all for taking the time to help.

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  • $\begingroup$ So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow. $\endgroup$ – John Jul 23 '18 at 14:55
  • $\begingroup$ In the figure the tags on the $x$-axis should be $\sqrt{2}$ instead of $2$. $\endgroup$ – Christian Blatter Jul 23 '18 at 15:02
  • $\begingroup$ @John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $\mathbf{F}$. $\endgroup$ – Wyuw Jul 23 '18 at 15:02
  • $\begingroup$ @ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $\sqrt{2}$ instead of 2? And what about the circulation? $\endgroup$ – Wyuw Jul 23 '18 at 15:09
  • $\begingroup$ The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-\sqrt 2 \pi$ $\endgroup$ – Doug M Jul 23 '18 at 15:36
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If we attack this directly.

$x = \sqrt 2 \cos t\\y = \sin t$

$\int_0^{2\pi} (\sqrt 2\cos t + 3\sin t, 2\sqrt 2\cos t -\sin t)\cdot(-\sqrt 2 \sin t, \cos t) \ dt\\ \int_0^{2\pi} -2\cos t\sin t - 3\sqrt 2\sin^2 t + 2\sqrt 2\cos^2 t -\cos t\sin t\ dt \\ -\sqrt 2\pi$

or we can use Greens theorem.

$\oint P\ dx + Q\ dy = \iint \frac{\partial Q}{\partial x} - \frac {\partial P}{\partial y} \ dx\ dy\\ Q = 2x-y, \frac {\partial Q}{\partial x} = 2\\ P = x+3y, \frac {\partial Q}{\partial y} = 3\\ \iint \frac{\partial Q}{\partial x} - \frac {\partial P}{\partial y} \ dx\ dy = \iint -1\ dx\ dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$\oint F\cdot dr = \iint \nabla \times F \ dV\\ \nabla \times F = -1\\ \iint -1\ dA$

And the area of the ellipse is $\sqrt 2 \pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

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  • $\begingroup$ Thank you for the help Doug! $\endgroup$ – Wyuw Jul 23 '18 at 16:50

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