# Line integrals

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

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

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

So is the integral to evaluate

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

regards

HallsofIvy
Homework Helper
I'm using the formula to find
$r(t) = (1-t)(x_{0},y_{0})+t(x_{1},y_{1}) =(1-t)(0,0)+t(1,1)\\ \text{ so } x = t y = t$ for 0<t<1
So x= t, y= t. Yes, that goes from (0,0) to (1,1)

So is the integral to evaluate

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

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

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

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

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?

So is the integral to evaluate

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

regards

yes your limits would be from 0 to 1

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

Sorry,

$fy(t)= t^2$

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
Homework Helper
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.

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

regards

Dick