Use greens theorem to evaluate the integral over c

F(x, y) = ‹y2cos(x), x2 + 2ysin(x)›
C is the triangle from (0, 0) to (1, 6) to (1, 0) to (0, 0).

and

F(x, y) = ‹y - ln(x2 + y2), 2arctan(y/x)›
C is the circle (x - 4)2 + (y - 5)2 = 25 oriented counterclockwise.

1) Note that C encloses the region bounded by y = 0, x = 1, and y = 6x.

∫c (y^2 cos x dx + (x^2 + 2y sin x) dy)
= ∫∫ [(∂/∂x)(x^2 + 2y sin x) - (∂/∂y)(y^2 cos x)] dA, by Green's Theorem
= ∫∫ [(2x + 2y cos x) - (2y cos x)] dA
= ∫(x = 0 to 1) ∫(y = 0 to 6x) 2x dy dx
= ∫(x = 0 to 1) 6x * 2x dx
= 4x^3 {for x = 0 to 1}
= 4.

2) ∫c ((y - ln(x^2 + y^2)) dx + 2 arctan(y/x) dy)
= ∫∫ [(∂/∂x)(2 arctan(y/x)) - (∂/∂y)(y - ln(x^2 + y^2))] dA, by Green's Theorem
= ∫∫ [(2 * (-y/x^2)/(1 + (y/x)^2) - (1 - 2y/(x^2 + y^2))] dA
= ∫∫ -1 dA
= -1 * (Area of the circle, which has radius 5)
= -25π.

I hope this helps!

For part a, the curve is oriented clockwise (hence the missing - sign from my answer).

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