F(x,y,z) = i+(x+yz)j+(xy-squareroot(z))k, C is the boundary of the part of the plane 3x+2y+z=1.
I know integral F dot dr = double integral curl F dot ds. I found the curl of F to be -j+k. I don't know how to set that up in the integral or what ds is. Can somebody help please??
Update:C is in the first octant
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Verified answer
If you want the region S to be between the given plane, and x = 0, y = 0, z = 0:
Projecting S onto the xy-plane gives a region between 3x + 2y = 1, x = 0, and y = 0.
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So, ∫c F · dr
= ∫∫s curl F · dS, by Stokes' Theorem
= ∫∫ <x-y, -y, 1> · <-z_x, -z_y, 1> dA
= ∫∫ <x-y, -y, 1> · <3, 2, 1> dA, since z = 1 - 3x - 2y
= ∫(x = 0 to 1/3) ∫(y = 0 to (1 - 3x)/2) (3x - 5y + 1) dy dx
= ∫(x = 0 to 1/3) (3xy - (5/2)y^2 + y) {for y = 0 to (1 - 3x)/2} dx
= ∫(x = 0 to 1/3) (3x(1 - 3x)/2 - (5/2)((1 - 3x)/2)^2 + (1 - 3x)/2) dx
= ∫(u = 0 to 1) (u(1 - u)/2 - (5/2)((1 - u)/2)^2 + (1 - u)/2) du/3, letting u = 3x
= (1/24) ∫(u = 0 to 1) (4u(1 - u) - 5(1 - u)^2 + 4(1 - u)) du
= (1/24) ∫(u = 0 to 1) (-9u^2 + 10u - 1) du
= (1/24) (-3u^3 + 5u^2 - u) {for u = 0 to 1}
= 1/24.
I hope this helps!