Results 1 to 5 of 5

Thread: calculate line integral and draw curve C

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    26

    calculate line integral and draw curve C

    I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

    This one is also a tuffy, perhaps someone may want to show me the light?


    Question:

    Let C be the closed curve C1 + C2 , where C1 is given by

    r1 (t) = t i + t^2 j, o <= t <= 1

    r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2



    a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

    b) Calculate the line integral $\displaystyle \int_C
    \mathbf{F}\cdot\,dr$ , where F is the vector field: F ( x, y ) = x^2 i + x y j
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by mohammadmurtaza View Post
    I am truly astonished to the amount of help I received, I'd like to thank Scott, Mr. Fantastic, Chris, and TheEmptySet for all this help you guys have given me. I now feel like I'm no longer alone...

    This one is also a tuffy, perhaps someone may want to show me the light?


    Question:

    Let C be the closed curve C1 + C2 , where C1 is given by

    r1 (t) = t i + t^2 j, o <= t <= 1

    r2 (t) = (2 - t) i + (2 - t) j , 1 <= t <= 2



    a) Draw a sketch of the curve C, with arrows to indicate orientation (im not sure if it's possible to do this online, but a description will be very very much appreciated)

    b) Calculate the line integral $\displaystyle \int_C
    \mathbf{F}\cdot\,dr$ , where F is the vector field: F ( x, y ) = x^2 i + x y j
    $\displaystyle C_1$ is the parabola $\displaystyle y = x^2$ starting at $\displaystyle (0,0)$ and going to $\displaystyle (1,1). $ $\displaystyle C_2 $ is a straight line starting at $\displaystyle (1,1)$ and going back to $\displaystyle (0,0)$. I'll do the first and you the second.

    If $\displaystyle x =t$ and $\displaystyle y = t^2$ then $\displaystyle dx = dt $ and $\displaystyle dy = 2t\,dt.$ The line integral becomes

    $\displaystyle
    \int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}
    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    26

    Thanks Danny!!!!!

    Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by mohammadmurtaza View Post
    Thanks Danny, ur the MANN!!! I really appreciate it However I am stuck on the second one. Since I don't know how to write in the correct notation, can you show me the second one as well?
    Quote Originally Posted by Danny View Post
    $\displaystyle C_1$ is the parabola $\displaystyle y = x^2$ starting at $\displaystyle (0,0)$ and going to $\displaystyle (1,1). $ $\displaystyle C_2 $ is a straight line starting at $\displaystyle (1,1)$ and going back to $\displaystyle (0,0)$. I'll do the first and you the second.

    If $\displaystyle x =t$ and $\displaystyle y = t^2$ then $\displaystyle dx = dt $ and $\displaystyle dy = 2t\,dt.$ The line integral becomes

    $\displaystyle
    \int \limits_{c_1} x^2 dx + xy\,dy = \int_0^1 t^2\, dt + t \cdot t^2 \cdot2t\,dt = \int_0^1 r^2 + 2t^4 \, dt = \frac{11}{15}
    $
    Here $\displaystyle x = 2 - t, y = 2 - t, t = 1 \to 2$ so $\displaystyle dx = dt, dy=dt$ so the line integral becomes

    $\displaystyle
    \int_1^2 (2-t)^2 dt + (2-t)(2-t)dt
    $

    although what I might suggest is to follow the line $\displaystyle y = x$ ($\displaystyle dy=dx$) from $\displaystyle x = 1 \to 0$ so

    $\displaystyle
    \int_1^0 2x^2dx = - \frac{1}{3}
    $
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2009
    Posts
    26

    I love you!!!!

    Thanks soo much danny!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Line integral along the curve
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Mar 1st 2011, 03:12 AM
  2. draw curve
    Posted in the Geometry Forum
    Replies: 8
    Last Post: Jan 26th 2011, 09:22 AM
  3. Line Integral over Curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 28th 2010, 02:49 AM
  4. Replies: 2
    Last Post: Jan 27th 2010, 10:06 AM
  5. Replies: 7
    Last Post: Aug 29th 2009, 09:09 AM

Search Tags


/mathhelpforum @mathhelpforum