My Math Forum requesting for explanation of surface integration

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 September 6th, 2017, 04:35 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 120 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: Southern California, USA Posts: 1,410 Thanks: 715 $n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$ 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
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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, 05:51 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 06:07 PM.

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