According to Green's theorem, ∫CPdx+Qdy=∬D(∂x∂Q−∂y∂P)dA.
For given vector field, ∂x∂Q−∂y∂P=6x−3y2, hence we need to evaluate surface integral∬(6x−3y2)dxdy over the circle of radius 1, centered at the origin.
It is convenient to do it in polar coordinates, where surface element is dS=rdrdφ, and 0≤r≤1, 0≤φ≤2π:
∫02πdφ∫01r(6rcosφ−3r2sin2φ)=∫02π(36cosφ−43sin2φ)dφ=
[2sinφ−43(2φ−41sin2φ)]02π=−43π.
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