Review Session Attendance quiz:

Verify the real form of Green's Theorem with 

u(x,y)=3*x+y
v(x,y)=5*x-y

and GAMMA the triangle whose vertives are [1,1],[2,1], and [2,2].


Nice Solution by Leen KHARBOUTLI

The real form of Green's Theorem is that the line integral (u*dx + v*dy) = double integral (v_x - u_y)*dx*dy

We want to show that the RHS is the same as the LHS where u(x,y) = 3x + y and v(x,y) = 5x-y on the triangle whose vertices are [1,1], [2,1], and [2,2].

We compute the right hand side: we write v_x = 5 and u_y = 1. Then v_x - u_y = 4.

The AREA integral of 4 (being a CONSTANT) over the inside of the integral , is 4 times the area of the triangle.
So RHS=4*1/2=2.

Computing the left hand side: we see that there are three line segments that must be parametrized:

From [1,1] to [2,1]: [1,1] + t([2,1] - [1,1]) = [1,1] + [t,0] = [1+t, 1]

From [2,1] to [2,2]: [2,1] + t([2,2]-[2,1]) = [2,1] + [0,t]  = [2, 1+t]

From [2,2] to [1,1]: [2,2] + t([1,1]-[2,2]) = [2,2] + [-t,-t] = [2-t,2-t]

The variable t varies from 0 to 1 in all three parametrizations.

Then we compute the line integrals along each line segment:

For the first line segment, we have u(x,y) = u(1+t, 1) = 3(1+t) + 1 = 4  + 3t and dx = 1dt and we have v(x,y) = v(1+t,1) = 5(1+t) - 1  = 4 + 5t and dy = 0dt. So the integral from t = 0 to t = 1 of u*dx + y*dy = 4t+(3t^2)/2 evaluated from 0 to 1 = 4 + 3/2 = 11/2.

For the second segment, we do similar computations and get 17/2

For the third segment, we do similar computations and get -12. So the LHS is 2.

LHS = RHS and so we are finished