Finding the Line Integral of a Tangential Component with Parameterization

In summary, the conversation is about finding the line integral of the tangential component of a vector field F from (-1,0) to (0,1) to (1,1) to (1,0) using parameterization and integration. The speaker has tried solving the problem and has gotten a different answer from the book and their teacher. They are looking for clarification on their approach and potential mistakes they may have made.
  • #1
jaejoon89
195
0
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!
 
Last edited:
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  • #2


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.
 

Related to Finding the Line Integral of a Tangential Component with Parameterization

1. What is a line integral?

A line integral is a type of integral used in multivariable calculus to calculate the total value of a function along a given path or curve.

2. How is a line integral different from a regular integral?

A line integral involves integrating a function along a specific curve, whereas a regular integral involves integrating a function over a given interval.

3. What is the purpose of finding a line integral?

The purpose of finding a line integral is to calculate the total value of a function along a specific path or curve, which can be useful in various applications such as physics and engineering.

4. How is a line integral calculated?

A line integral is calculated by using a specific formula, depending on whether it is a line integral of the first or second kind. The line integral of the first kind involves integrating a scalar function, while the line integral of the second kind involves integrating a vector function.

5. What are some real-world applications of line integrals?

Line integrals have various real-world applications, such as calculating work done by a force along a specific path, calculating the flux of a vector field through a curve, and calculating the circulation of a vector field around a closed curve.

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