Respuesta :
the work done by the force F on a particle moving counterclockwise around the closed path C is 9π by integral.
Green's Theorem states that:
∮C F · dr = ∫∫D (∂Q/∂x - ∂P/∂y) dA
Where C is the boundary of the region D, P and Q are the components of the vector field F, and dr is the vector differential along the boundary C.
In this case, the region D is a circle with radius 2, so we can express the boundary C as r = 2 cos θ. We also need to calculate the components of the vector field F. These are P = e^x - 3y and Q = e^y + 6x.
Plugging these values into Green's Theorem, we get integral:
∮C F · dr = ∫∫D (∂Q/∂x - ∂P/∂y) dA
= ∫∫D (e^y + 6 - (-3)) dA
= ∫∫D (e^y + 9) dA
= ∫ 0 2π ∫ 0 2 (e^2 cosθ
+ 9) (2 dθ)
= 4π ∫ 0 2 (e^2 cosθ + 9) dθ
= 9π
Therefore, the work done by the force F on a particle moving counterclockwise around the closed path C is 9π.
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