# weird hexagons and green thm

• Jul 8th 2013, 04:47 PM
n22
weird hexagons and green thm
Hi, I would appreciate if someone could help. thanks.

Calculate the work done by the non conservattive force F=x³j 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=(x²+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.

• Jul 8th 2013, 05:52 PM
HallsofIvy
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, $\displaystyle f(x,y)\vec{i}+ g(x,y,z)\vec{j}$, around a closed path, counterclockwise, is equal to the integral of $\displaystyle \frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}$ over the region bounded by the closed path.

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

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