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September 6th, 2017, 04:35 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 188 Thanks: 2  requesting for explanation of surface integration
Evaluate ∫∫ < x, y, 2 > * N dS, where D is given by z = 1  x^2  y^2, x^2 + y^2 <= 1, oriented up. Official Answer:  pi Cross Product: $\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( r cos \theta , r sin \theta , 2 ) * ( 2r^2 cos \theta , 2r^2 sin \theta , r ) ~ dr d \theta $ $\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( 2r^3  2r ) ~dr d \theta = \pi $ I managed to get the correct answer. but I didn't understand why the Jacobian of 'r' was missing from the integration since I used polar coordinate. 
September 6th, 2017, 05:25 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1064 
$n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$ for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral. 
September 6th, 2017, 05:49 PM  #3  
Senior Member Joined: Jan 2017 From: Toronto Posts: 188 Thanks: 2  Quote:
G(r, \theta) = < r cos \theta , r sin \theta , 1  r^2 > $ $\displaystyle dG/dr = < cos \theta , sin \theta , 2r > $ $\displaystyle dG/d \theta = <  r sin \theta , r cos \theta , 0 > $ Cross Product $\displaystyle dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r > $ Dot Product $\displaystyle \int_{0}^{2\pi} \int_{0}^{1} < r cos \theta , r sin \theta, 2 > * < 2r^2 cos \theta , 2r^2 sin \theta , r > $  
September 6th, 2017, 05:51 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1064  Quote:
You have to find the normal vector in cartesian coordinates first, dot it with the field vector, then you can convert everything over to polar coordinates for the integration. Last edited by romsek; September 6th, 2017 at 06:07 PM.  

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