My Math Forum Stoke's theorem stuck here

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September 8th, 2017, 08:29 AM   #1
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Stoke's theorem stuck here

Can anybody help me with this?
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September 8th, 2017, 09:08 PM   #2
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Quote:
 Originally Posted by sarajoveska Can anybody help me with this?
What are you having trouble with?

This should be pretty similar to the previous problems.

Last edited by skipjack; September 9th, 2017 at 02:27 AM.

September 9th, 2017, 12:41 AM   #3
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Quote:
 Originally Posted by romsek What are you having trouble with? This should be pretty similar to the previous problems.
dS is the vector product of the partial derivatives of the curve given with f(r,phi). Since here we have only phi, dS will be zero. Is that correct?

Last edited by skipjack; September 9th, 2017 at 02:28 AM.

September 9th, 2017, 12:59 AM   #4
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Quote:
 Originally Posted by sarajoveska dS is the vector product of the partial derivatives of the curve given with f(r,phi). Since here we have only phi, dS will be zero. Is that correct?
0 is what I get, yes, but you need to show the whole working.

Last edited by skipjack; September 9th, 2017 at 02:29 AM.

September 9th, 2017, 01:16 AM   #5
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Quote:
 Originally Posted by romsek 0 is what I get, yes, but you need to show the whole working.
I did. But I was not sure about the 0.

Last edited by skipjack; September 9th, 2017 at 02:29 AM.

 September 10th, 2017, 09:25 AM #6 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715 Ok, I led you astray on this. The answer isn't 0. Doing the line integral directly $r = \left(\sqrt{2}\cos(t)-1,~\sin(t),~2-\sqrt{2}\cos(t)\right)$ $f = (0, ~y,~x y)$ $\dfrac{dr}{dt} = \left(-\sqrt{2} \sin (t),\cos (t),\sqrt{2} \sin (t)\right)$ $f(x(t),y(t),z(t)) = \left(0,\sin (t),\sin (t) \left(\sqrt{2} \cos (t)-1\right)\right)$ $f\cdot \dfrac{dr}{dt} = \sin (t) \left(-\sqrt{2} \sin (t)+\sin (2 t)+\cos (t)\right)$ $I=\displaystyle \int_0^{2\pi}f \cdot \dfrac{dr}{dt} ~dt = \int_0^{2\pi} \sin (t) \left(-\sqrt{2} \sin (t)+\sin (2 t)+\cos (t)\right) ~dt = -\sqrt{2} \pi$ Using Stokes Theorem $f = (0, ~y,~x y)$ $\nabla \times f = (x,~-y,~0)$ $s = \left(-1 + \sqrt{2}u,~v,~2-\sqrt{2}u\right)$ $s_u = (\sqrt{2},0,-\sqrt{2})$ $s_v = (0,1,0)$ $n = s_u \times s_v = (\sqrt{2},0,\sqrt{2})$ Now translate the curl of $f$ into $u,~v$ coordinates $\nabla \times f = \left(\sqrt{2} u-1,-v,0\right)$ and we translate this into polar coordinates $(u,v) \to (\rho,\theta)$ as $u=\rho \cos(\theta),~v=\rho \sin(\theta)$ $\nabla \times f =\left( \sqrt{2} \rho \cos (\theta)-1,-\rho \sin (\theta),0 \right)$ now dot this with the normal vector found above to form the integrand $\nabla \times f \cdot n = \sqrt{2} \left(\sqrt{2} \rho \cos (\theta)-1\right)$ and integrate over $\rho ~d\rho ~d\theta$ $I = \displaystyle \int_0^{2\pi}\int_0^1~ \nabla \times f \cdot n ~\rho ~d\rho~d\theta =\int_0^{2\pi}\int_0^1~\sqrt{2} \left(\sqrt{2} \rho \cos (\theta)-1\right)~\rho ~d\rho ~d\theta = -\sqrt{2}\pi$
September 10th, 2017, 10:03 AM   #7
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Quote:
 Originally Posted by romsek Ok, I led you astray on this. The answer isn't 0. Doing the line integral directly $r = \left(\sqrt{2}\cos(t)-1,~\sin(t),~2-\sqrt{2}\cos(t)\right)$ $f = (0, ~y,~x y)$ $\dfrac{dr}{dt} = \left(-\sqrt{2} \sin (t),\cos (t),\sqrt{2} \sin (t)\right)$ $f(x(t),y(t),z(t)) = \left(0,\sin (t),\sin (t) \left(\sqrt{2} \cos (t)-1\right)\right)$ $f\cdot \dfrac{dr}{dt} = \sin (t) \left(-\sqrt{2} \sin (t)+\sin (2 t)+\cos (t)\right)$ $I=\displaystyle \int_0^{2\pi}f \cdot \dfrac{dr}{dt} ~dt = \int_0^{2\pi} \sin (t) \left(-\sqrt{2} \sin (t)+\sin (2 t)+\cos (t)\right) ~dt = -\sqrt{2} \pi$ Using Stokes Theorem $f = (0, ~y,~x y)$ $\nabla \times f = (x,~-y,~0)$ $s = \left(-1 + \sqrt{2}u,~v,~2-\sqrt{2}u\right)$ $s_u = (\sqrt{2},0,-\sqrt{2})$ $s_v = (0,1,0)$ $n = s_u \times s_v = (\sqrt{2},0,\sqrt{2})$ Now translate the curl of $f$ into $u,~v$ coordinates $\nabla \times f = \left(\sqrt{2} u-1,-v,0\right)$ and we translate this into polar coordinates $(u,v) \to (\rho,\theta)$ as $u=\rho \cos(\theta),~v=\rho \sin(\theta)$ $\nabla \times f =\left( \sqrt{2} \rho \cos (\theta)-1,-\rho \sin (\theta),0 \right)$ now dot this with the normal vector found above to form the integrand $\nabla \times f \cdot n = \sqrt{2} \left(\sqrt{2} \rho \cos (\theta)-1\right)$ and integrate over $\rho ~d\rho ~d\theta$ $I = \displaystyle \int_0^{2\pi}\int_0^1~ \nabla \times f \cdot n ~\rho ~d\rho~d\theta =\int_0^{2\pi}\int_0^1~\sqrt{2} \left(\sqrt{2} \rho \cos (\theta)-1\right)~\rho ~d\rho ~d\theta = -\sqrt{2}\pi$
Okay, I understand everything except how did you find r?

September 10th, 2017, 10:10 AM   #8
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Quote:
 Originally Posted by sarajoveska Okay, I understand everything except how did you find r?
you're given $r$ in the problem.

September 10th, 2017, 10:18 AM   #9
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Quote:
 Originally Posted by romsek you're given $r$ in the problem.
Where?

 September 10th, 2017, 10:26 AM #10 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,410 Thanks: 715 oh.. they call it $p$ it's just the parametric description of the curve

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