∫∫ (2x-3y)dy - (3x+4y)dx, where C is the unit circle
What I did-
∂/∂x(2x-3y) - ∂/∂y(3x+4y)= 2-4= -2
∫(theta= 0 to 2pi) ∫(r= 0 to 1) -2 dr dtheta
= -4pi
I don't believe this is correct. What should have I done? Thank you.
What I did-
∂/∂x(2x-3y) - ∂/∂y(3x+4y)= 2-4= -2
∫(theta= 0 to 2pi) ∫(r= 0 to 1) -2 dr dtheta
= -4pi
I don't believe this is correct. What should have I done? Thank you.
-
Careful with the negative signs!
∫c [(2x - 3y) dy - (3x + 4y) dx]
= ∫c [-(3x + 4y) dx + (2x - 3y) dy]
= ∫∫ [(∂/∂x)(2x - 3y) - (∂/∂y) -(3x+4y)] dA
= ∫∫ 5 dA
= ∫(θ = 0 to 2π) ∫(r = 0 to 1) 5 * (r dr dθ)
= 5π.
I hope this helps!
∫c [(2x - 3y) dy - (3x + 4y) dx]
= ∫c [-(3x + 4y) dx + (2x - 3y) dy]
= ∫∫ [(∂/∂x)(2x - 3y) - (∂/∂y) -(3x+4y)] dA
= ∫∫ 5 dA
= ∫(θ = 0 to 2π) ∫(r = 0 to 1) 5 * (r dr dθ)
= 5π.
I hope this helps!