 My Math Forum requesting for explanation of surface integration

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 September 6th, 2017, 05:35 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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, 06:25 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1474 $n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$ for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral. September 6th, 2017, 06:49 PM   #3
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Quote:
 Originally Posted by romsek $n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$ for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral.
$\displaystyle G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >$
$\displaystyle dG/dr = < cos \theta , sin \theta , -2r >$
$\displaystyle dG/d \theta = < - r sin \theta , r cos \theta , 0 >$
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, 06:51 PM   #4
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Quote:
 Originally Posted by zollen $\displaystyle G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >$ $\displaystyle dG/dr = < cos \theta , sin \theta , -2r >$ $\displaystyle dG/d \theta = < - r sin \theta , r cos \theta , 0 >$ Cross Product $\displaystyle dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r >$
I really don't think you can start in polar coordinates like that. Not using the standard definition of the surface integral. There is an equivalent definition but the formula is different.

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 07:07 PM. Tags explanation, integration, requesting, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post KyVanchhay Calculus 4 July 27th, 2013 02:06 AM zell^ Calculus 13 April 13th, 2012 09:41 AM Zilee Calculus 4 November 25th, 2011 10:35 AM FataLIdea Calculus 1 March 28th, 2008 05:46 PM

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