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Line integrals

  • Thread starter boneill3
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  • #1
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Homework Statement



Let F=(3x+2y)i+(2x-y)j Evaluate [itex] \int_{C}F . dr [/itex] where C is the line segment from (0,0) to (1,1)

Homework Equations



[itex]\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt [/itex]

The Attempt at a Solution



How do I choose the correct values of x and y as a function of t?
And do I just integrate over a = 0 to b = 1 ?

regards
 

Answers and Replies

  • #2
Cyosis
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What parametrization have you found for x(t) and y(t)?
 
  • #3
HallsofIvy
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i.e. what is a parameterization for the line from (0,0) to (1,1)?
 
  • #4
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I'm using the formula to find
[itex]
r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1})
=(1-t)(0,0)+t(1,1)\\

\text{ so } x = t y = t [/itex] for 0<t<1
 
  • #5
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So is the integral to evaluate

[itex]

\int_{0}^{1} [(3t+2t)3+(2t-t)]dt

[/itex]

regards
 
  • #6
HallsofIvy
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I'm using the formula to find
[itex]
r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1})
=(1-t)(0,0)+t(1,1)\\

\text{ so } x = t y = t [/itex] for 0<t<1
So x= t, y= t. Yes, that goes from (0,0) to (1,1)

So is the integral to evaluate

[itex]

\int_{0}^{1} [(3t+2t)3+(2t-t)]dt

[/itex]

regards
where did you get the "3" multiplying (3t+ 2t)?
 
  • #7
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Let F=(3x+2y)i+(2x-y)j

[itex]\int_{a}^{b}[f(x(t),y(t))x'(t) + g(x(t),y(t))y'(t)]dt [/itex]


I got the three from the partial derivative dx of 3x+2y = 3

But I think I may have stuffed up because the formula
says x'(t) which would be 1.

So the integral would be

[itex]\int_{0}^{1} [(3t+2t)+(2t-t)]dt[/itex]


Is that better?
 
  • #8
So is the integral to evaluate

[itex]

\int_{0}^{1} [(3t+2t)3+(2t-t)]dt

[/itex]

regards

yes your limits would be from 0 to 1
 
  • #9
127
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Hi Guys

If C instead of a line segment is y= x^2 can I substitute x = t so that C is now just
[latex}
fy(t)= t^2
[/itex]

regards
 
  • #10
127
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Sorry,

[itex]
fy(t)= t^2
[/itex]
 
  • #11
127
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I've used the new parameterization of x = t^2 y = t^2 ( for C y=x^2)

On the original function

So is the integral to evaluate now

[itex]\int_{0}^{1} [(3t+2t)2t+(2t-t)2t]dt[/itex]

Where x'(t) = x'(t^2) = 2t

and

Where y'(t) = y'(t^2) = 2t

Is this right so far?

regards
Brendan
 
  • #12
Dick
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You need to put x=t^2 and y=t^2 into the definition of the vector F as well, right? Or if you are doing y=x^2, x=t, y=t^2, dx=dt, dy=2tdt. You seem to be mixing up your parametrizations.
 
  • #13
127
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So I need to put it as
[itex]\int_{0}^{1} [(3t^2+2t^2)2t+(2t^2-t^2)2t]dt[/itex]

regards
 
  • #14
Dick
Science Advisor
Homework Helper
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Yes, that's right for parametrizing along the line between (0,0) and (1,1) using x=t^2 and y=t^2. And you should get the same answer as using x=t and y=t.
 
  • #15
127
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Thanks for all your help guys
 

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