Finding the Line Integral of a Tangential Component with Parameterization

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SUMMARY

The discussion focuses on calculating the line integral of the vector field F = y/(x^2 + y^2) i - x/(x^2 + y^2) j along a specified path from (-1,0) to (0,1) to (1,1) to (1,0). The user attempted to parameterize the paths and integrate, but encountered discrepancies in the results, specifically obtaining 0 instead of the expected π. The correct approach involves using the integral formula int_0^1 F(r(t)) * dr = int_0^1 F(r(t)) * dr/dt dt, and applying proper parameterization and integration techniques to resolve the integral accurately.

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jaejoon89
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Finding line integrals -- please help!

Given
F = y/(x^2 + y^2) i - x / (x^2 + y^2) j

Find the line integral of the tangential component of F from (-1,0) to (0,1) to (1,1) to (1,0) (assuming F is NOT path independent).

---

I tried parameterizing each of the three paths using the formula
r(t) = (1-t)r_0 + tr_1

---

I've been racking my brain on this for hours. The book says the answer is pi. My teacher says it isn't.

For path 1:
x = -1 + t, y = t
=> dx = dt = dy

For path 2:
x = t, y = 1
=> dy = 0

For path 3:
x = 1, y = 1-t
=> dx = 0

I then substituted these values in the original equation for F and integrated from t = 0 to t = 1 for each step. I get

[tan^-1 (1) - tan^-1 (1)] + [tan^-1 (1) - tan^-1 (0)] + [tan^-1 (0) - tan^-1 (1)] = 0 + pi/4 - pi/4 = 0

Perhaps I take one of the tan^-1 (0) to be pi instead? What am I doing wrong? Please help!
 
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I did the first path and got pi/2, not 0. What integral did you get?

Remember, the integral is int_0^1 F(r(t)) * dr = int_0^1 F(r(t)) * dr/dt dt, where * denotes the dot product. For this path, since r = (-1+t, t), dr/dt is (1,1). After working it out, it came down to int_0^1 1/((-1+t)^2 + t^2) dt which you solve by completing the square on the bottom so you'll end up with an tan^-1. Hopefully you can read this-- unfortunately TeX isn't working.
 

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