# Finding limits of line integral

1. May 22, 2009

### boneill3

1. The problem statement, all variables and given/known data

Integrate along the line segment from (0,0) to $(\pi,-1)$
The integral

$\int_{(0,1)}^{(\pi,-1)} [y sin(x) dx - (cos(x))]dy$

2. Relevant equations

3. The attempt at a solution

I have used the parameterization of $x=\pi t$ and $y= 1-2t$
To get the integral:
$\int_{(0,1)}^{(\pi,-1)} [1-2t sin(\pi t) -(cos(\pi t))]dt$

But now because it is an integral of variable t I need to change the limits .

I'm not sure if I just have to put the limits of t just from 0 to $\pi$

I suppose I'm having trouble with getting from the limit of 2 variables (x,y) to a limit of one variable t

Thanks

2. May 22, 2009

### Dick

If x=pi*t and y=1-2t, then if you put t=0 then x=0 and y=1, right? If you put t=1 then x=pi and y=(-1), also right? As you came up with that fine parametrization what's the problem with finding limits for t?

3. May 23, 2009

### boneill3

I will need to go back and study more about parametrization.

4. May 23, 2009

### boneill3

When calculating this line integral

$\int_{(0,1)}^{(\pi,-1)} [y sin(x) dx - (cos(x))]dy$

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

with parameterization

I have $x = \pi t$
$y = 1-2t$
so
$x' = \pi$
and
$y' = -2$

plugging into the integral I get

$\int_{(0)}^{(1)} [1-2y sin(\pi t) \pi - (cos(\pi t))-2]$
$= -1$

The question states that the integral is independant of path.

So if I integrate along the initial line segment $(0,1)$to $(0,\pi)$
I should be able to plug in the values f($(-1,\pi)$)-f($(0,1)$)
And it should equal my original integral vaue of -1.

However I get 0

Could someone please check what I've done I show me where I am going wrong ?