Finding limits of line integral

boneill3

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

Integrate along the line segment from (0,0) to [latex](\pi,-1)[/latex]
The integral

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



2. Relevant equations



3. The attempt at a solution

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

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 [latex] \pi [/latex]

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

Thanks

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?

boneill3

Thank you for your help.
I will need to go back and study more about parametrization.

boneill3

When calculating this line integral

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

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

with parameterization

I have [latex]x = \pi t [/latex]
[latex]y = 1-2t[/latex]
so
[latex]x' = \pi [/latex]
and
[latex]y' = -2 [/latex]

plugging into the integral I get

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

The question states that the integral is independant of path.

So if I integrate along the initial line segment [latex](0,1)[/latex]to [latex](0,\pi)[/latex]
I should be able to plug in the values f([latex](-1,\pi)[/latex])-f([latex](0,1)[/latex])
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 ?