surface integration

jayant

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Jun 17, 2021
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2
ds= i dydz + j dzdx + k dxdy
z=g(x,y)
F= iF1 + j F2 + k F3

F.ds= F1. dy dz + F2 . dzdx + F3 dxdy
For integrating above F3 dxdy is Ok
But to integrate about say F2dzdx
How to find dz ?
Some one told it is - dz/dy.dy partial derivative and negative also.
How it came so?

So F2dzdx becomes F2(x,y,z(x,y)) . [-dz/dx](x,y) . dxdy
 

HallsofIvy

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Jan 27, 2012
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Why is "dz" any different from "dx" or "dy"? Exactly how are you given the equation of the surface?
You say that "F= iF1+ jF2+ kF3" but what are F1, F2, and F3? A two dimensional surface must be give in terms of two parameters but what are the parameters? Are you saying that they are x and y?
 

jayant

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Joined
Jun 17, 2021
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2
I am getting some clue answer from some friend.
You see dz= delz/delx .dx partial derivative or
dz= delz/dely .dy or
dz= delz/delx .dx + delz/dely .dy

for F1. dy dz we wiill choose dz= delz/delx .dx
so that we get dxdy but for outward normal unit vector we make it negative also
as F1(x,y,z(x,y)) .dy . ( -- ) delz/delx .dx = [ F1(x,y,z(x,y)) . ( -- ) delz/delx (x,y) ] .dxdy

similarly for F2 . dzdx we take dz= -- delz/dely .dy and get dxdy in xy plane now.
 
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