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Verify Green's theorem for (3x28y2)dx+(4y6xy)dy where c is boundary of the region defined by x = 0, y = 0 & x+ y = 1
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By Green's theorem

cPdx+Qdy=R(QxPy)dxdy......(A)

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Here p=3x28y2Q=4y6xy

py=1byQx=6y

Consider L.H.S. of Green's theorem

(a) Along C1: y = 0 dy = 0 x varies to 0 to 1

c1Pdx+Qdy=1x=03x2dx=3[x33]10=1

(b) Along C2: y = 1-x, dy = -dx x varies 1 to 0

c2Pdx+Qdy=0x=13x28(1x)2dx4[(1x)6x(1x)]dxc2Pdx+Qdy=[3x33+8(1x)334(1x)22(1)+6(x22x33)]01c2Pdx+Qdy=[0+83+21(616)]c2Pdx+Qdy=[83+22]=83

(c) Along C3: x = 0, dx = 0, y varies 1 to 0

C3Pdx+Qdy=0y=10+4ydyC3Pdx+Qdy=2[y2]01=21C3Pdx+Qdy=2

CPdx+Qdy=C1Pdx+Qdy+C2Pdx+Qdy+C3Pdx+Qdy=1+8/32CPdx+Qdy=5/3..................................................(1)

Now R.H.S. of equation (A): y: 0 to 1 - x & x: 0 to 1

R(QxPy)dxdy=1x=01xy=0(6y)y+16dxdyR(QxPy)dxdy=1x=01xy=010ydxdyR(QxPy)dxdy=1x=01xy=010(y22)1x0dxR(QxPy)dxdy=1x=0102(1x2)dxR(QxPy)dxdy=1x=05[(1x)33(1)]R(QxPy)dxdy=5/3.............................................(2)

From (1) and (2) Green's theorem is verified i.e.

cPdx+Qdy=R(QxPy)dxdy

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