Evaluating a Curve Integral: Is the Answer 0?

[asif zaidi]

Hi:

I am evaluating the curve integral below and I am getting an answer of 0. I have looked at my solution many times and cannot see that I have done anything wrong.

My concern is that a value of 0 for a curve integral does not make sense - a curve integral measures the distance from point A to point B on a curve so how can it be 0.

Problem Statement:
Consider the parametrized curve \varsigma: [0:2\pi] -> R^{s}, defined by \varsigma(t) = (e^{t}cos(t), e^{t}sin(t)).
Evaluate the curve integral:
integral of ( (x / (x^{2} + y^{2}) ) dx + (y / (x^{2} + y^{2}) ) dy )

Problem Solution:
Step1: Calculate the norm of parametrized function

f(t) = e^{t}cos(t) ; f'(t) = e^{t}cos(t) - e^{t}sin(t)
g(t) = e^{t}sin(t) ; g'(t) = e^{t}sin(t) + e^{t}cos(t)

Therefore norm of || f'(t) + g'(t) || = sqrt(2) * e^{t}

(I am not showing the intermediate steps)

Step2:
Evaluate f(x) at f(t), g(t) and g(x) at f(t), g(t)

f( f(t), g(t) ) = cos(t) / e^{t}
g( f(t), g(t) ) = sin(t) / e^{t}

Step3:
Multiply step2 and step 3 = cos(t) + sin(t)


Step4:
Evaluate integral of Step3

integral (0-2*pi) of (cos(t) + sin(t) dt ) = 0

Plz advise what I am doing incorrectly


Thanks

Asif

 

[tiny-tim]

Hi asif!

I'm not really following what you've done, but …

i] shouldn't it be √(x² + y²)?

ii] Hint: xdx + ydy = d(x² + y²)/2

 

[HallsofIvy]

No, the curve integral of a given differential does NOT " measure the distance from point A to point B". That is true only for the path integral \int ds, a very specific differential.

In fact, I note that, because of the "symmetry"
\frac{\partial \frac{y}{x^2+ y^2}}{\partial x}= \frac{\partial \frac{x}{x^2+ y^2}}{\partial y}[/itex]<br /> so this is an &quot;exact&quot; differential. It&#039;s integral along <b>any</b> closed path is 0.<br /> Do you recognize that this is a closed path?