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Finding limits of line integral

  • Thread starter boneill3
  • Start date
  • #1

Homework Statement

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

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

Homework Equations

The Attempt at a Solution

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

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 [itex] \pi [/itex]

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


Answers and Replies

  • #2
Science Advisor
Homework Helper
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
Thank you for your help.
I will need to go back and study more about parametrization.
  • #4
When calculating this line integral

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

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

with parameterization

I have [itex]x = \pi t [/itex]
[itex]y = 1-2t[/itex]
[itex]x' = \pi [/itex]
[itex]y' = -2 [/itex]

plugging into the integral I get

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

The question states that the integral is independant of path.

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