Evaluating Line Integral F over C: a Step-by-Step Guide

[boneill3]

Homework Statement

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

Homework Equations

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

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

 

[Cyosis]

What parametrization have you found for x(t) and y(t)?

 

[HallsofIvy]

i.e. what is a parameterization for the line from (0,0) to (1,1)?

 

[boneill3]

I'm using the formula to find
<br /> r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1})<br /> =(1-t)(0,0)+t(1,1)\\<br /> <br /> \text{ so } x = t y = t for 0<t<1

 

[boneill3]

So is the integral to evaluate

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

regards

 

[HallsofIvy]

I'm using the formula to find
<br /> r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1})<br /> =(1-t)(0,0)+t(1,1)\\<br /> <br /> \text{ so } x = t y = t for 0<t<1

So x= t, y= t. Yes, that goes from (0,0) to (1,1)

So is the integral to evaluate

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

regards

where did you get the "3" multiplying (3t+ 2t)?

 

[boneill3]

Let F=(3x+2y)i+(2x-y)j

\int_{a}^{b}[f(x(t),y(t))x&#039;(t) + g(x(t),y(t))y&#039;(t)]dt


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

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


Is that better?

 

[mthreezycrazy]

So is the integral to evaluate

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

regards

yes your limits would be from 0 to 1

 

[boneill3]

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

 

[boneill3]

Sorry,

<br /> fy(t)= t^2<br />

 

[boneill3]

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

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

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

and

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

Is this right so far?

regards
Brendan

 

[Dick]

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.

 

[boneill3]

So I need to put it as
\int_{0}^{1} [(3t^2+2t^2)2t+(2t^2-t^2)2t]dt

regards

 

[Dick]

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.

 

[boneill3]

Thanks for all your help guys