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Math Help - Line integral

  1. #1
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    Line integral

    This is my problem:
    Compute the following three line integrals directly around the boundary C of the part R of the interior ellipse (x^2/a^2)+(y^2/b^2)=1 where a>0 and b>0 that lies in the first quadrant:
    (a) integral(xdy-ydx)
    (b) integral((x^2)dy)
    (c) integral((y^2)dx)

    I used parametrisation (x=acost and y=bsint) for the arc of the ellipse.
    C is the curve r=(acost)i + (bsint)j (0 less than or equal to t less than or equal to pi).
    (a) integral(xdy-ydx)=integral from 0 to pi((acost)(bcost)dt)- integral from 0 to pi((bsint)(-asint)dt)=(ab)pi/2-(-ab)(pi)/2=ab(pi)

    (b) integral((x^2)dy)=integral from 0 to pi((acost)(acost)(bcost)dt)=0.

    (c) integral((y^2)dx)=integral from 0 to pi((bsint)(bsint)(-asint)dt)=-(4/3)a(b^2)

    Could anyone check these and see if they are right?
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  2. #2
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    Actually, should all the bounds be from 0 to pi/2 instead of 0 to pi (since I am looking at only the first quadrant)?
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  3. #3
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    Quote Originally Posted by Adebensjp05 View Post
    This is my problem:
    Compute the following three line integrals directly around the boundary C of the part R of the interior ellipse (x^2/a^2)+(y^2/b^2)=1 where a>0 and b>0 that lies in the first quadrant:
    (a) integral(xdy-ydx)
    (b) integral((x^2)dy)
    (c) integral((y^2)dx)

    I used parametrisation (x=acost and y=bsint) for the arc of the ellipse.
    C is the curve r=(acost)i + (bsint)j (0 less than or equal to t less than or equal to pi).
    (a) integral(xdy-ydx)=integral from 0 to pi((acost)(bcost)dt)- integral from 0 to pi((bsint)(-asint)dt)=(ab)pi/2-(-ab)(pi)/2=ab(pi)

    (b) integral((x^2)dy)=integral from 0 to pi((acost)(acost)(bcost)dt)=0.

    (c) integral((y^2)dx)=integral from 0 to pi((bsint)(bsint)(-asint)dt)=-(4/3)a(b^2)

    Could anyone check these and see if they are right?
    I am assuming it moves along the ellipse only.

    And I am also assuming it is positively oriented (abusing terminology).

    The the part can be described as x=a\cos t and y=b\sin t for 0\leq t\leq \frac{\pi}{2}. The reason why it is cosine for "x" because it starts from the right and goes to the left as I am assuming.
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