Results 1 to 9 of 9

Thread: Line integral

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    32

    Line integral

    To calculate the integral of $\displaystyle ydx+(x^2+y^2)dy$ over the curve $\displaystyle y=\sqrt{4-x^2}$ between (-2,0) and (2,0), can I parameterize the curve to $\displaystyle x=2\cos t$, $\displaystyle y=2\sin t$, $\displaystyle \pi \leq t \leq 2\pi$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,136
    Thanks
    2610

    Re: Line integral

    yep
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: Line integral

    Quote Originally Posted by polarbear73 View Post
    To calculate the integral of $\displaystyle ydx+(x^2+y^2)dy$ over the curve $\displaystyle y=\sqrt{4-x^2}$ between (-2,0) and (2,0), can I parameterize the curve to $\displaystyle x=2\cos t$, $\displaystyle y=2\sin t$, $\displaystyle \pi \leq t \leq 2\pi$?
    Note that the contour is an semicircle in the first and second quadrants in a clockwise direction.
    Therefore $y= 0\to 2\to 0~\&~x= -2\to 0\to 2$

    So your parametric path will not work. But $x=2\cos(t),~y=-2\sin(t),~\pi\le t\le 2\pi$ will work.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,136
    Thanks
    2610

    Re: Line integral

    You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: Line integral

    Quote Originally Posted by romsek View Post
    You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.
    I taught complex integrals so long that I recognize a standard trick question. The $y=\sqrt{4-x^2}$ is the upper half-circle in I & II.
    So $0\le y\le 2~\&~-2\le x\le 2$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2010
    Posts
    32

    Re: Line integral

    Hey thanks. I did think that the curve went through the third and fourth quadrants because I thought the standard convention of angular displacement was counterclockwise. I failed to notice that the curve, as written, had to be in the first and second quadrant.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,136
    Thanks
    2610

    Re: Line integral

    Plato keeps us all honest... or tries anyway.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2010
    Posts
    32

    Re: Line integral

    So if I integrate $\displaystyle ydx+(x^2+y^2)dy$ over the curve $\displaystyle y=\sqrt{4-x^2}$ and parameterize the curve as $\displaystyle x=2\cos t, y=2\sin t, 0\leq t \leq\pi$ I get $\displaystyle \int(-4\sin^2 t +8\cos t) dt$ unless I'm doing something wrong. And if I do that integration I get $\displaystyle -2t+8\sin t+\sin 2t+ C$. If I evaluate that from $\displaystyle 0$ to$\displaystyle \pi$ I get $\displaystyle -2\pi$. This is not the answer in the text book. Could someone point me in the right direction?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: Line integral

    Quote Originally Posted by polarbear73 View Post
    So if I integrate $\displaystyle ydx+(x^2+y^2)dy$ over the curve $\displaystyle y=\sqrt{4-x^2}$ and parameterize the curve as $\displaystyle x=2\cos t, y=2\sin t, 0\leq t \leq\pi$ I get $\displaystyle \int(-4\sin^2 t +8\cos t) dt$ unless I'm doing something wrong. And if I do that integration I get $\displaystyle -2t+8\sin t+\sin 2t+ C$. If I evaluate that from $\displaystyle 0$ to$\displaystyle \pi$ I get $\displaystyle -2\pi$. This is not the answer in the text book. Could someone point me in the right direction?
    That not at all correct. Read reply #3.
    In the above you have gone from (2,0) to (-2,0) clockwise.
    That is not at all what the OP is asking for.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Dec 11th 2011, 11:30 PM
  2. [SOLVED] Line Integral along a straight line.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 11th 2011, 07:18 PM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Sep 16th 2009, 11:50 AM
  5. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 15th 2009, 01:28 PM

Search Tags


/mathhelpforum @mathhelpforum