September 8th, 2017, 09:29 AM  #1 
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  Stoke's theorem stuck here
Can anybody help me with this?

September 8th, 2017, 10:08 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 819  What are you having trouble with? This should be pretty similar to the previous problems. Last edited by skipjack; September 9th, 2017 at 03:27 AM. 
September 9th, 2017, 01:41 AM  #3 
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  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 03:28 AM. 
September 9th, 2017, 01:59 AM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 819  0 is what I get, yes, but you need to show the whole working.
Last edited by skipjack; September 9th, 2017 at 03:29 AM. 
September 9th, 2017, 02:16 AM  #5 
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  I did. But I was not sure about the 0.
Last edited by skipjack; September 9th, 2017 at 03:29 AM. 
September 10th, 2017, 10:25 AM  #6 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 819 
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 $\nabla \times f = \left(\sqrt{2} u1,v,0\right)$ 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, 11:03 AM  #7  
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  Quote:
 
September 10th, 2017, 11:10 AM  #8 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 819  
September 10th, 2017, 11:18 AM  #9 
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  
September 10th, 2017, 11:26 AM  #10 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 819 
oh.. they call it $p$ it's just the parametric description of the curve 

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