Let's find work of the field directly:
W=C∫x2dx−xydy
Parametric equation of a circle:
x=2cost⇒dx=−2sintdt,y=2sint+2⇒dy=2costdt,−2π<t<2π
Then
W=C∫x2dx−xydy=−2π∫2π4cos2t(−2sint)dt−4cost(sint+1)⋅2costdt=8−2π∫2πcos2tdcost+8−2π∫2πcos2tdcost−8−2π∫2πcos2tdt=316cos3t∣−2π2π−2−2π∫2π(1+cos2t)d2t=0−2(2t∣−2π2π+sin2t∣−2π2π)=−4π
Let's find
∂y∂P=∂y∂x2=0;∂x∂Q=∂x∂(−xy)=−y
Then, by Green's theorem, work is
W=∫D∫(∂x∂Q−∂y∂P)dxdy=−∫D∫ydxdy
Let's move on to polar coordinates:
x=rcosφ,y=rsinφ,0<φ<2π
x2+(y−2)2=4⇒r2cos2φ+r2sin2φ−4rsinφ+4=4⇒r2=4rsinφ⇒r=4sinφ
Then
0<r<4sinφ
So
W=−∫D∫ydxdy=−0∫2πdφ0∫4sinφrsinφ⋅rdr=−310∫2πsinφ⋅r3∣∣04sinφdφ=−310∫2π64sin4φdφ=−3640∫2π(21−cos2φ)2dφ=−3160∫2π(1−2cos2φ+cos22φ)dφ=−380∫2π(1−2cos2φ)d2φ−380∫2π(1+cos4φ)dφ=−38(2φ∣02π−2sin2φ∣02π)−32(4φ∣02π+sin4φ∣02π)=−38π−32⋅2π=−312π=−4π
Since in both cases we obtained the same value of the work, Green's theorem holds.
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