MHB Rayan's question at Yahoo Answers (Green's Theorem)

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Green's Theorem is applied to calculate the work done by the force F on a particle moving counterclockwise around the closed path C defined by r = 2 cos(θ). The force F is given as F(x,y) = (e^x − 9y)i + (e^y + 4x)j. The path C corresponds to a circle with the equation (x-1)² + y² = 1, and the area D is the disk bounded by C. The work W is computed as 13 times the area of D, resulting in W = 13π. This calculation demonstrates the effective use of Green's Theorem in evaluating work done by vector fields.
Fernando Revilla
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Here is the question:

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)j
C: r = 2 cos(θ)

Here is a link to the question:

Use Green's Theorem to calculate the work done by the force F? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Rayan,

Easily proved, $C:r=2\cos \theta$ is the circle $C:(x-1)^2+y^2=1$. If $D$ is the disk with boundary $C$, then by the Green's theorem, $$W=\int_C(e^x − 9y)dx + (e^y + 4x)dy=\iint_D(Q_x-P_y)dxdy=\iint_D(4+9)dxdy\\=13\iint_Ddxdy=13\mbox{Area }(D)=13\cdot \pi\cdot 1^2=\boxed{\;13\pi\;}$$
If you have further questions you can post them in the http://www.mathhelpboards.com/f10/ section.http://www.mathhelpboards.com/f10/
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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