Finding C_2 for Curve F in Triangle C

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SUMMARY

The discussion focuses on finding the parameterization of curve C_2 for the line integral of the vector field F = along the triangle defined by the vertices (1,0), (0,1), and (-1,0). The first curve, C_1, is established as x=t, y=0 for -1≤ t ≤ 1. The challenge lies in determining C_2, which connects the points (1,0) to (0,1). The slope of this line is -1, leading to the equation y = 1 - x, which can be parameterized by letting x = t and determining the corresponding y values.

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Homework Statement



F=<xy, x-y> and C is the triangle joining (1,0), (0,1) and (-1,0) in the clockwise direction.

Homework Equations



How do I have the second part of this curve?

The Attempt at a Solution


Apparently, there should only be a sum of two integrals. So I got that curve one could be
C_1: x=t, y=0, -1≤ t ≤ 1

But I have no idea how to find C_2. There's no other way, that I see, that you can take such that you keep x constant and let y vary.
 
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The reason it should only be a sum of two integrals is because the line integral is just zero on the curve you have. You're only interested in the sum of the integral over C_2 and C_3. How might you parametrize those? For example, if you wanted to find a line in x-y coordinates to go from (1,0) to (0,1), you would find its slope (-1) and its y-intercept (1). The line would be y = 1-x. So if you let x = t, then what would y be and what would t vary between?
 

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