# Use Green's Theorem to calculate the work done by the force F on a particle that is moving counterclockwise around the closed path C. F(x,y) = (e^x − 9y)i + (e^y + 4x)jC: r = 2 cos(θ)

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**Green theorem states:**

Let `C` be positively oriented, piecewise smooth, simple closed curve and let `S` be region bounded by `C` and let `P(x,y)` and `Q(x,y)` have continuous partial derivatives over `S,` then

`oint_C P dx+Q dy=intint_S ((del Q)/(del x)-(del P)/(del y))dxdy` **(1)**

In your case `P(x,y)=e^x-9y`, `Q(x,y)=e^y+4x` and `C` ` `is circle with radius 1 and center at (1,0) (graph below)

Hence we need to calculate

`oint_C(e^x-9y)dx+(e^y+4x)dy=`

Now we use Green formula (1)

`intint_S (4-(-9))dxdy=13intint_Sdxdy=`

Since `S` is circle with radius 1 (if you don't see this you can use change of variables to polar coordinates) its area is `pi` i.e. `intint_Sdxdy=pi` hence

`oint_C(e^x-9y)dx+(e^y+4x)dy=13pi`

**So total work done is** `13pi.` **If force is in newtons (N) and length of the curve is in meters (m), then work done is in joules that is** `W=13pi"J"`