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Math Help - Greens theorem

  1. #1
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    Greens theorem

    Hello,

    A task says:

    Evaluate I = "closed integral"C (sinx + 3y^2)dx + (2x - e^((-y)^2))dy, where C is the boundary of the half-disk x^2 + y^2 <_ a^2, y >_ 0, oriented counterclockwise.

    When you use Green's Theorem on I you get a "double integral"(2 - 6y) dyda. I am insecure about choosing the right limits for x and y. My purpose is:

    x has an upper limit "pi", lower limit a^2. y has an upper limit "pi" and it's lower limit is a^2. Is this q correct setup?

    Excuse my poor notation, I just signed up,

    Sincere regards
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  2. #2
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    Re: Greens theorem

    How did you get 2-6y for the double integral?

    - Hollywood
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  3. #3
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    Re: Greens theorem

    The "2- 6y" is the correct integrand. As for the limits of integration, " \pi" would apply if you were integrating in polar coordinates!

    To find the limits of integration in x, y coordinates, start by drawing a picture. " x^2 + y^2 <_ a^2, y\ge 0 is a semicircle with base in the x axis from x= -a to x= a. For each x y runs from 0 up to the semicircle: y= \sqrt{a^2- x^2} (from solving x^2+ y^2= a^2 for y).
    \int_{-a}^a\int_{0}^{\sqrt{a^2- x^2}} 2- 6y dydx

    You could integrate in the opposite order: Overall, y goes from 0 to a and, for each y, x goes from -\sqrt{a^2- x^2} on the left to \sqrt{a^2- y^2}
    \int_{-a}^a\int_{-\sqrt{a^2- x^2}^{\sqrt{a^2- x^2}} x- 6y dxdy

    Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle \theta goes from 0 to \pi and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. x- 6y= rcos(\theta)- 6rsin(\theta)= r(cos(\theta- 6sin(\theta). And, of course, the "differential of area" in polar coordinates is rdrd\theta
    \int_0^\pi\int_0^a r^2(cos(\theta)- 6sin(\theta))drd\theta= \int_0^\pi (cos(\theta)- 6sin(\theta))d\theta\int_0^a r^2 dr
    Thanks from hollywood
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  4. #4
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    Re: Greens theorem

    The integrand is the divergence of the F vector:

    div F = (2 - 0) - (0 + 6y) = 2 - 6y

    The first part is derived with respect to x, thus the exponential function dies. The second part is differentiated with respect to y, hence sin(x) = 0.
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  5. #5
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    Re: Greens theorem

    I thank you on this extremely adequate answer.
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  6. #6
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    Re: Greens theorem

    Quote Originally Posted by HallsofIvy View Post
    The "2- 6y" is the correct integrand
    You are correct of course - I don't know what I was thinking earlier. The integrand should be \frac{\partial{M}}{\partial{x}} - \frac{\partial{L}}{\partial{y}} where M=2x - e^{(-y)^2} and L=\sin{x} + 3y^2.

    Thanks,
    Hollywood
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  7. #7
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    Re: Greens theorem

    Quote Originally Posted by HallsofIvy View Post
    The "2- 6y" is the correct integrand. As for the limits of integration, " \pi" would apply if you were integrating in polar coordinates!

    Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle \theta goes from 0 to \pi and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. x- 6y= rcos(\theta)- 6rsin(\theta)= r(cos(\theta- 6sin(\theta). And, of course, the "differential of area" in polar coordinates is rdrd\theta
    \int_0^\pi\int_0^a r^2(cos(\theta)- 6sin(\theta))drd\theta= \int_0^\pi (cos(\theta)- 6sin(\theta))d\theta\int_0^a r^2 dr
    My thinking of the conversion from cartesian co-ordinates to polar co-ordinates is the following:

    2 - 6y = 2 - 6rsin(theta).

    By doing this I get the same result from the Calculus result list.

    Sincerely,

    Kaemper
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