Evaluating a Curve Integral: Is the Answer 0?

In summary: The value of a curve integral (of a differential) measures the length of the curve from a starting point to a ending point. In the example given, the curve is parametrized by the function f(t), and the integral is evaluated from the starting point (f(0)) to the ending point (f(2*pi)). Because the curve is parametrized by a function, the integral can be evaluated for any point on the curve. However, it is more common to parametrize a curve by its "eccentricity" (e.g. the "curvature" or "outline" of the curve), and then evaluate the integral at specific points along the curve that
  • #1
asif zaidi
56
0
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 [tex]\varsigma[/tex]: [0:2[tex]\pi[/tex]] -> R[tex]^{s}[/tex], defined by [tex]\varsigma(t) [/tex] = (e[tex]^{t}[/tex]cos(t), e[tex]^{t}[/tex]sin(t)).
Evaluate the curve integral:
integral of ( (x / (x[tex]^{2}[/tex] + y[tex]^{2}[/tex]) ) dx + (y / (x[tex]^{2}[/tex] + y[tex]^{2}[/tex]) ) dy )

Problem Solution:
Step1: Calculate the norm of parametrized function

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

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

(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[tex]^{t}[/tex]
g( f(t), g(t) ) = sin(t) / e[tex]^{t}[/tex]

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
 
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  • #2
Hi asif! :smile:

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 :smile:
 
  • #3
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 [itex]\int ds[/itex], a very specific differential.

In fact, I note that, because of the "symmetry"
[tex]\frac{\partial \frac{y}{x^2+ y^2}}{\partial x}= \frac{\partial \frac{x}{x^2+ y^2}}{\partial y}[/itex]
so this is an "exact" differential. It's integral along any closed path is 0.
Do you recognize that this is a closed path?
 
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1. What is a curve integral?

A curve integral is a mathematical representation of the area under a curve. It is used to find the total value of a function over a specific interval.

2. How is a curve integral evaluated?

A curve integral is evaluated by using a specific formula, known as the Fundamental Theorem of Calculus, which involves taking the anti-derivative of the function and evaluating it at the upper and lower bounds of the interval.

3. Why is the answer to a curve integral sometimes 0?

The answer to a curve integral can be 0 if the function being integrated is an odd function and the upper and lower bounds of the interval are symmetric about the origin. This is because the positive and negative areas under the curve will cancel each other out, resulting in a total area of 0.

4. Are there any other cases where the answer to a curve integral can be 0?

Yes, the answer to a curve integral can also be 0 if the function being integrated is a constant function and the upper and lower bounds of the interval are the same. This is because the area under a constant function is always 0.

5. How do I know if the answer to a curve integral is correct?

You can check the answer to a curve integral by using the formula for evaluating the integral and plugging in the upper and lower bounds to see if the result matches the given answer. You can also use a graphing calculator to graph the function and visually confirm the answer.

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