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Math Help - weird hexagons and green thm

  1. #1
    n22
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    weird hexagons and green thm

    Hi, I would appreciate if someone could help. thanks.

    Calculate the work done by the non conservattive force F=xj in moving a particle,around the regular hexagon with vertices at (√3,1),(0,2),(-√3,1),(-√3,-1),(0,-2) and (√3,-1) in that order. (use greens theorem).


    For this question I tried to use symmetry : I sketched the hexagon out then,I divided into half.
    y-y1=m(x-x1)

    and the formula m(gradient)=(y2-y1)/(x2-x1)

    Please see diagram attached :
    Attachment 28760
    For C1:
    y=(x+2x√3+6)/3For c2:y=2+(x/√3);For c3:y=1

    I am really not sure what I am doing wrong . could someone please check. obviously if its a positive gradient ,need pos sign for neg gradient slanting the other way I need a negative sign . is there anything else I have to account for ?cuz I seem to be somethimes getting wrong answers off by minus 1.

    Attached Thumbnails Attached Thumbnails weird hexagons and green thm-hexagon.png  
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  2. #2
    MHF Contributor

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    Re: weird hexagons and green thm

    I'm not sure what you are doing! In particular, I have no idea why you are looking at the individual sides of the hexagon. You were told to do this using "Green's theorem" which says that the integral of a vector function,  f(x,y)\vec{i}+ g(x,y,z)\vec{j}, around a closed path, counterclockwise, is equal to the integral of \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y} over the region bounded by the closed path.

    Here the vector function given is x^3\vec{j} so that f(x,y)= 0 and [/tex]g(x,y)= x^3[/tex] so you want to integrate 3x^2 over the hexagon. Since that is symmetric about the y- axis, it is sufficient to integrate of x= 0 to \sqrt{3} and double. The lower boundary is the line from (0, -2) to (\sqrt{3}, -1). The slope of that line is \frac{1}{\sqrt{3}} and the equation is y= \frac{1}{\sqrt{3}}x- 2. Similarly, the upper boundary is the line from (0, 2) to (\sqrt{3}, 1). The slope of that line is -\frac{1}{\sqrt{3}} and the equation is y= -\frac{1}{\sqrt{3}}x+ 2.

    So the solution is 2\int_0^{\sqrt{3}}\int_{-\frac{1}{\sqrt{3}}x+ 2}^{\frac{1}{\sqrt{3}}x- 2} 3x^2 dy dx.
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