greens thorem in the normal form.

Question

verify green's theorem in the normal form by calculating both sides and showing they are equal if F= x²i + xyj, and C is the square with opposite vertices at (0,0) and (1,1)

Answer 1

F = Pi + Qj so P = x² and Q = xy

 

To get parametric equations for the line segments of the square we can use the formula r(t) = (1-t)r₀ + tr₁ where 0 ≤ t ≤ 1 and r₀ is starting point and r₁ is ending point.

 

Evaluate ∫c Pdx + Qdy over the four line segments of the square and add up the results.  I will do one side as an example and let you do the other sides.  We can easily get the four corner points of the square as (0,0), (1,0), (1,1) and (0,1).

 

Evaluate from (0,0) to (1,0) r(t) = (1-t)<0,0> + t<1,0> = <t,0> so x = t, dx = dt, y = 0, dy = 0.

 

∫c Pdx + Qdy = ∫_c x²dx + xydy = ∫ from 0 to 1 of t²dt = t³/3 from 0 to 1 = 1/3

 

Apply this same process for the line segments from (1,0) to (1,1), (1,1) to (0,1), and (0,1) to (0,0)

 

Evaluate ∫∫_D(∂Q/∂x -∂P/∂y)da =

 

∫ from x=0 to x= 1∫ from y=0 to y=1 of (∂(xy)/∂x -∂(x²)/∂y)dydx =

 

∫ from x=0 to x= 1∫ from y=0 to y=1 of ydydx =

 

∫ from x=0 to x= 1( y²/2 from y=0 to y=1)dx =

 

∫ from x=0 to x= 1(1/2)dx =

 

(1/2)x from 0 to 1 = 1/2