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Evaluate line integral

  • Thread starter DryRun
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  • #1
DryRun
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Homework Statement
http://s2.ipicture.ru/uploads/20120204/iuPLuS1l.png

The attempt at a solution

[itex]x=t^2-1[/itex] and [itex]y=t^2+1[/itex]

[itex]\frac{dx}{dt}=2t[/itex] and [itex]\frac{dy}{dt}=2t[/itex]

The line integral is of the form: [itex]\int P\,.dx+Q\,.dy[/itex]

So, i use direct substitution:
[tex]\int^1_0 4t^3-2t\sin(t^2-1)+2t\cos(t^2+1)\,.dt[/tex]
Is my method correct? I can't integrate any further, as i tried integration by parts for [itex]t\sin(t^2-1)[/itex]and got stuck.
 

Answers and Replies

  • #2
DryRun
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I figured it out a while back. The integration gives:
[tex]\left[ t^4 +\cos (t^2-1)+\sin (t^2+1) \right]^{t=1}_{t=0}[/tex]
And the answer is: 1.528 (correct to 3 d.p.)

However, i am curious to know why the need for parametrization?
I know it's normally used to simply the integration process, for example, the preference of using polar coordinates when integrating over a circular region, rather than Cartesian coordinates. But in this case, i've tried to work it out differently, to see if i could get the answer by omitting ##t## in the integrand.

When t varies from 0→1, x varies from -1→0 and y varies from 1→2
Then, integrating the original integrand directly, i get:
[tex]\left[xy+\cos x\right]^{x=0}_{x=-1}+\left[ x+ \sin 2 -\sin 1 \right]^{y=2}_{y=1}[/tex]
And the answer that i get: [tex]1+y-\cos 1+x+\sin 2-\sin 1[/tex]
It's clear that this method doesn't give me a numerical answer, but i can't understand why not, as i used the values of ##t## to derive the limits for both x and y. Is there a way to get the final answer without using ##t## in the integrand? I can't quite understand the concept of parametrization, as i thought it was just a means to make reaching the solution easier/faster, but not the only way to solve/evaluate the integral.
 
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  • #3
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The first integrand is y(x) -sin (x). What is y(x) on the curve?
 
  • #4
DryRun
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The first integrand is y(x) -sin (x). What is y(x) on the curve?
I'm not sure what you mean.
[tex]y(t)=t^2+1[/tex]
y varies from 1→2
 
  • #5
LCKurtz
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I figured it out a while back. The integration gives:
[tex]\left[ t^4 +\cos (t^2-1)+\sin (t^2+1) \right]^{t=1}_{t=0}[/tex]
And the answer is: 1.528 (correct to 3 d.p.)

However, i am curious to know why the need for parametrization?
I know it's normally used to simply the integration process, for example, the preference of using polar coordinates when integrating over a circular region, rather than Cartesian coordinates. But in this case, i've tried to work it out differently, to see if i could get the answer by omitting ##t## in the integrand.

When t varies from 0→1, x varies from -1→0 and y varies from 1→2
Then, integrating the original integrand directly, i get:
[tex]\left[xy+\cos x\right]^{x=0}_{x=-1}+\left[ x+ \sin 2 -\sin 1 \right]^{y=2}_{y=1}[/tex]
And the answer that i get: [tex]1+y-\cos 1+x+\sin 2-\sin 1[/tex]
.
If you are going to evaluate the integral as written as a dx and dy, you have to have the integrand in terms of the x or y variables accordingly. This means on the dx integral you need to express y in terms of x (which depends on the path) and on the dy integral you need to express x in terms of y, which again depends on the path. Line integrals like this may not be independent of the path you use, but they are independent of the parameterization of the path. You could eliminate the parameter t and get a y-x equation for the path and work the dx and dy integrals as I indicated above. Should get the same answer.
 
  • #6
DryRun
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If you are going to evaluate the integral as written as a dx and dy, you have to have the integrand in terms of the x or y variables accordingly. This means on the dx integral you need to express y in terms of x (which depends on the path) and on the dy integral you need to express x in terms of y, which again depends on the path. Line integrals like this may not be independent of the path you use, but they are independent of the parameterization of the path. You could eliminate the parameter t and get a y-x equation for the path and work the dx and dy integrals as I indicated above. Should get the same answer.
From [itex]x=t^2-1[/itex] and [itex]y=t^2+1[/itex], [itex]y=x+2[/itex]
The original line integral becomes:
[tex]\int_C ((x+2)-\sin x)dx+((y-2)+\cos y)dy=\int^{0}_{-1} (x+2-\sin x)dx+ \int^2_1(y-2+\cos y)dy[/tex] But the answer is: -2.472 (correct to 3 d.p.), which is different!
 
  • #7
vela
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You made a mistake somewhere evaluating the integrals. Mathematica gives 1.52752.
 
  • #8
LCKurtz
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From [itex]x=t^2-1[/itex] and [itex]y=t^2+1[/itex], [itex]y=x+2[/itex]
The original line integral becomes:
[tex]\int_C ((x+2)-\sin x)dx+((y-2)+\cos y)dy=\int^{0}_{-1} (x+2-\sin x)dx+ \int^2_1(y-2+\cos y)dy[/tex] But the answer is: -2.472 (correct to 3 d.p.), which is different!
That integral is set up correctly and evaluates to 1.527524136 just like the other one.
 
  • #9
DryRun
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You are right. :redface:

Thanks for the help.
 
  • #10
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Notice also that there is nothing stopping you from getting rid of this silly parametrization by choosing t^2 = s, and doing the calculation as usual
 

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