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July 28th, 2017, 03:50 AM  #1 
Newbie Joined: Jun 2017 From: Earth Posts: 17 Thanks: 0  Does this equation hold for line integral?
$\displaystyle \int_C fdx_1=\int_{C'} f' dx_1$ in which C is a curve in 3D space and C' is its projection onto $\displaystyle x_1 x_2$ plane. f' is the same projection of f. Other condition are assumed to be true when needed such as C is sufficiently smooth or f has definition and smooth enough. If this equation is true, can you give me an intuitive explanation? Thank you. 
July 28th, 2017, 04:48 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,944 Thanks: 797 
The notation seems odd. You have $\displaystyle f(x_1, x_2, x_3)$, a function in three variables, and a path, C, in three dimensions. Them $\displaystyle f'(x_1, y_1)$ is that function "projected" into the $\displaystyle (x_1, x_2)$ plane and C' is C projected to the $\displaystyle (x_1, y_1)$ plane, right? There seems to be a simple counter example to this claim. Take $\displaystyle f(x_1, x_2,x_3)= x_3$ and the path C to be the line x_1= 1, x_2= t, x_3= t with t from 0 to 1. Then $\displaystyle \int_C f dx_1=\int_0^1 1 dt= 1$ while $\displaystyle \int_{C'} f' dx_1= \int_0^1 0 dx_1= 0$ 

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