Results 1 to 5 of 5

Math Help - Evaluate Line Integral

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Evaluate Line Integral

    Q: Evaluate \int_{\gamma} y^2 e^{2\sin{x}} \cos{x} dx + ye^{2\sin{x}} dy where \gamma is the arc of the curve y = \cos{x} from the point (0,1) to the point (\frac{\pi}{2},0).

    A: We know that the integral is exact, meaning that we can choose any path from (0,1) to (\frac{\pi}{2},0). In the solutions they have gone through the origin, along the coordinate axes. So the path taken is from (0,1) to (0,0), and then to (\frac{\pi}{2},0)

    The answers then state that:

    The integral is then

    \int^0_1 ye^{0} dy + \int^{\frac{\pi}{2}}_0 0 dx

    Not sure how they get this? It looks like they've just subbed x=0 and y=\frac{\pi}{2} into the first and second bit of the original integral?

    I know it's going to be something obvious but I'd appreciate it if someone could help me out with this.

    Thanks in advance

    Craig
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by craig View Post
    Q: Evaluate \int_{\gamma} y^2 e^{2\sin{x}} \cos{x} dx + ye^{2\sin{x}} dy where \gamma is the arc of the curve y = \cos{x} from the point (0,1) to the point (\frac{\pi}{2},0).

    A: We know that the integral is exact, meaning that we can choose any path from (0,1) to (\frac{\pi}{2},0). In the solutions they have gone through the origin, along the coordinate axes. So the path taken is from (0,1) to (0,0), and then to (\frac{\pi}{2},0)

    The answers then state that:

    The integral is then

    \int^0_1 ye^{0} dy + \int^{\frac{\pi}{2}}_0 0 dx

    Not sure how they get this? It looks like they've just subbed x=0 and y=\frac{\pi}{2} into the first and second bit of the original integral?
    On the line from (0, 1) to (0, 0), x is identically 0. And on the line from (0, 0) to (\pi/2, 0), y is identically 0. You could, for example, take as parametric equations for (0, 1) to (0, 0), x= 0, y= t with t from 1 to 0 and for (0, 0) to (\pi/2, 0), x= t, y= 0 for t from 0 to \pi/2.

    I know it's going to be something obvious but I'd appreciate it if someone could help me out with this.

    Thanks in advance

    Craig
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Thanks for the reply.

    I understand most of what you're saying, when you sub in y = 0 then that's where you get the 0 from etc.

    I'm just confused how you get:

    \int^0_1 ye^{0} dy ?

    What happens to the y^2 from the original integral?

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by craig View Post
    Thanks for the reply.

    I understand most of what you're saying, when you sub in y = 0 then that's where you get the 0 from etc.

    I'm just confused how you get:

    \int^0_1 ye^{0} dy ?

    What happens to the y^2 from the original integral?

    Thanks again
    On the first leg, from (0, 1) to (0, 0) x is identically 0 so x does not change. That means that dx is also 0.

    More specifically, if we let x= 0, y= t, then dx= 0dt and dy= dt.
    \int y^2e^{2sin(x)}dx+ ye^{2sin(x)}dy becomes \int t^2e^0(0dt)+ te^0(dt)= \int t dt
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Of course! Thankyou for that, always amazes me how simple most things are when you know the answer...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Evaluate the line integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 18th 2011, 08:38 AM
  2. [SOLVED] Evaluate line integral (parabola)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 4th 2010, 08:19 PM
  3. Evaluate line integral with 3 dimensions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 15th 2010, 02:31 AM
  4. Evaluate the line integral...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2008, 01:04 PM
  5. Evaluate the line integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 8th 2007, 08:15 AM

Search Tags


/mathhelpforum @mathhelpforum