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Thread: help with greens theorem

  1. #1
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    help with greens theorem

    just trying figure this example out using greens theorem:


    $\displaystyle
    I = \oint_R {9y^3 dx - 9x^3 dy}
    $


    mapped counter-clockwise, is the circle

    $\displaystyle
    x^2 + y^2 = 81
    $

    enter as a multiple of pi

    so using substitution,

    $\displaystyle
    \int {udx + vdy = \int_{} {\int_R {(\frac{{\delta y}}
    {{\delta x}} - \frac{{\delta u}}
    {{\delta x}}} } } )dxdy
    $

    so

    $\displaystyle
    u = 9y^3
    $
    $\displaystyle
    v = - 9x^3
    $

    and

    $\displaystyle
    - 27\int {\int_R {x^2 + y^2 dxdy} }
    $

    $\displaystyle
    - 27\int\limits_0^{2\pi } {\int\limits_0^1 {r^2 rdrd\theta } }
    $

    so i got this far how does this end bit work?? do i integrate? where do i go from here...
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  2. #2
    Eater of Worlds
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    Doesn't r vary from 0 to 9?.

    $\displaystyle -27\int_{0}^{2\pi}\int_{0}^{9}r^{3}drd{\theta}$
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  3. #3
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    oops yeah my mistake, so how would i proceed, from here, i know im supposed to change polars, but i dont know exactly how to solve this
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  4. #4
    Eater of Worlds
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    Just integrate. You should know how to do this simple integration since you're in Calc III.

    $\displaystyle \int_{0}^{9}r^{3}dr=\frac{1}{4}r^{4}$

    $\displaystyle \frac{1}{4}(9)^{4}-\frac{1}{4}(0)^{4}=\frac{6561}{4}$

    Now, integrate wrt to theta:

    $\displaystyle \int_{0}^{2\pi}\frac{6561}{4}d{\theta}=\frac{6561} {4}(2{\pi})-\frac{6561}{4}(0)=\frac{6561\pi}{2}$

    Now, multiply by -27: $\displaystyle \frac{-177147\pi}{2}$
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  5. #5
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    any help with this?

    help...

    int {[xy-4x^4y-ysqrt(x^2+y^2+1/2y^2sinx]dx

    + [4xy^4+8/3x^3y^2+sqrtx^2+y^2-ycosx]dy}

    where r is the path mapped counter-clockwise: along y=o,
    0<x<1; then along x^2+y^2=1, from (1,0)to (1/sqrt2,1/sqrt2);
    finally y=x from (1/sqrt2,1sqrt2) to (0,0)
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