Using Green’s theorem, calculate the circulation of F around the curve:

counterclockwise around a triangle with vertices (2,0) (0,3) (-2,0)

F = (2x^2 + 3y) i + (2x + 3y^2) j

Is the answer (-12)?

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- Feb 3rd 2013, 05:40 AMKvandesterrenGreen's Theorem
Using Green’s theorem, calculate the circulation of F around the curve:

counterclockwise around a triangle with vertices (2,0) (0,3) (-2,0)

F = (2x^2 + 3y) i + (2x + 3y^2) j

Is the answer (-12)? - Feb 3rd 2013, 11:35 AMHallsofIvyRe: Green's Theorem
No, it is not. Is that all you want? Perhaps if you showed what you did we could point out errors. What did you get for the area of that triangle?

- Feb 4th 2013, 06:25 AMKvandesterrenRe: Green's Theorem
F = (2x^2 + 3y) i + (2x + 3y^2) j

triangle points: (2,0) (0,3) (-2,0)

Green's theorem: ∫c F · dr = ∫∫R [ (∂N/∂x) - (∂M/∂y) ] dA

N = 2x + 3y^2

M = 2x^2 + 3y

∂N/∂x = 2

∂M/∂y = 3

∫∫R (2-3) dy dx

limits of integration:

vertical: from 0 to 3

horizontal:

left side of triangle:

y = (3/2)x + 3

x = (2/3)y - 2

right side of triangle:

y = (-3/2)x + 3

x = (-2/3)y + 2

∫∫R (2-3) dy dx

∫ (upper limit: x = (-2/3)y + 2) (lower limit: x = (2/3)y - 2) ∫ (upper limit: 3) (lower limit: 0) (2-3) dy dx

inner integration:

∫ (upper limit: 3) (lower limit: 0) (2-3) dy

∫ (upper limit: 3) (lower limit: 0) (-1) dy

y evaluated from 0 to 3 = -3

outer integration:

∫ (upper limit: x = (-2/3)y + 2) (lower limit: x = (2/3)y - 2) -3 dx

-3x evaluated from [x = (2/3)y - 2] to [x = (-2/3)y + 2)]

-3(2+ 2)

-12