Do you mean
F(x,y) = e^(2x) + x^(2)yi + e^(2y) − xy^2 j ?
If so, then
F•dr = (e^(2x) + x^2y) dx + (e^(2y) − xy^2) dy
That makes
My = x^2
Nx = -y^2
∫F.dr = -∫∫x^2+y^2 dy dx
Using polar coordinates, that is just
-∫∫r^2 * r dr dθ
inside the circle:
r = [0,2]
θ = [0,2pi]
Use Green's Theorem to evaluate
C (F · dr)
(Check the orientation of the curve before applying the theorem.)
F(x, y) = e^(2x) + x^(2)y, e^(2y) − xy^2
C is the circle x^2 + y^2 = 4 oriented clockwise
1 answer