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Math Help - vector field

  1. #1
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    vector field

    The question:
    For the vector field = -y + x, find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point \( (6)/sqrt(2), (6)/sqrt(2)\). Give an exact answer.

    i think from x=0 to x=6 it is 0 but how do i get the second part? help please
    Last edited by acosta0809; November 9th 2009 at 09:02 PM.
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  2. #2
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    Quote Originally Posted by acosta0809 View Post
    The question:
    For the vector field = -y + x,
    This is not a vector field. Did you mean -y\vec{i}+ x\vec{j}?

    find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point \( (6)/sqrt(2), (6)/sqrt(2)\). Give an exact answer.

    i think from x=0 to x=6 it is 0 but how do i get the second part? help please
    Assuming the the vector field to be integrated is indeed -y\vec{i}+ x\vec{j}, on the path from (0,0) to (6, 0) we can take x= t ( 0\le t\le 6) and y= 0. Then dx= dt and y= 0dt so \math]\int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy}= \int (0dx+ x(0))= 0[/tex].

    Write the circle x^2+ y^2= 36 as x= 6 cos(t), y= 6 sin(t) (that's a standard parameterization for a circle) so that dx= -6 sin(t)dt and dy= 6 cos(t) dt. The integral becomes \int (-6 sin(t)\vec{i}+ 6 cos(t)\vec{j})\cdot(-6 sin(t)dt\vec{i}+ 6 cos(t)dt\vec{j}) = \int(36sin^2(t)dt+ 36cos^2(t))dt= 36\int dt.

    Or course, if x= 6 cos(t) then x= \frac{6}{\sqrt{2}} when cos(t)= 1/\sqrt{2} and that happens when t= \pi/4. The integral is from t= 0 to t= \pi/4.

    (Blast! I was hoping I could get back and correct my silly mistake [I mixed up x and y] before anyone had looke at it!)
    Last edited by HallsofIvy; November 10th 2009 at 04:11 AM.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    This is not a vector field. Did you mean -y\vec{i}+ x\vec{j}?


    Assuming the the vector field to be integrated is indeed -y\vec{i}+ x\vec{j}, on the path from (0,0) to (6, 0) we can take x= t ( 0\le t\le 6) and y= 0. Then dx= dt and y= 0 dt so \int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy)= \int (0dx+ x(0))= 0.

    Write the circle x^2+ y^2= 36 as x= 6 cos(t), y= 6 sin(t) (that's a standard parameterization for a circle) so that dx= -6 sin(t)dt and dy= 6 cos(t) dt. The integral becomes \int (6 cos(t)\vec{i}+ 6 sin(t)\vec{j})\cdot(-6 sin(t)dt\vec{i}+ 6 cos(t)dt\vec{j}) = \int(-36sin(t)cos(t)dt+ 36cos(t)sin(t))dt= 0!
    .
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