Thread: Green'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)?

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?

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