Let C be the square oriented counterclockwise with corners (0,0), (1,0), (1,1), and (0,1). If we...

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Let C be the square oriented counterclockwise with corners (0,0), (1,0), (1,1), and (0,1). If we label the edges of the square as C1, C2, C3, C4 starting from the bottom edge going counterclockwise, then the edges may be linearly parameterized, with 0 t 1, by: edge C1 x1(t)= = Y1(t) = = edge C2 x2(t)= Y2(t) = edge C3 x3(t) y3(t) = = edge C4 x4(t) Y4(t) = f y [ dx dx { v dx + x dy = [ v? dx + x dy + { v dz + z dy + [ y dx + x dy + [ y dz + z dy C' =0+0+0+0 Applying Green's theorem, C C3 y dx + x dy = = dy = [I dx dy The vector field F = yi+x jis: conservative Let C be the square oriented counterclockwise with corners (0,0), (1,0), (1,1), and (0,1). If we label the edges of the square as C1, C2, C3, C4 starting from the bottom edge going counterclockwise, then the edges may be linearly parameterized, with 0 t 1, by: edge C1 x1(t)= = Y1(t) = = edge C2 x2(t)= Y2(t) = edge C3 x3(t) y3(t) = = edge C4 x4(t) Y4(t) = f y [ dx dx { v dx + x dy = [ v? dx + x dy + { v dz + z dy + [ y dx + x dy + [ y dz + z dy C' =0+0+0+0 Applying Green's theorem, C C3 y dx + x dy = = dy = [I dx dy The vector field F = yi+x jis: conservative