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By Green's theorem
∫cPdx+Qdy=∫∫R(∂Q∂x−∂P∂y)dxdy......(A)
Here p=3x2−8y2Q=4y−6xy
∂p∂y=−1by∂Q∂x=−6y
Consider L.H.S. of Green's theorem
(a) Along C1: y = 0 dy = 0 x varies to 0 to 1
∫c1Pdx+Qdy=1∫x=03x2dx=3[x33]10=1
(b) Along C2: y = 1-x, dy = -dx x varies 1 to 0
∫c2Pdx+Qdy=0∫x=13x2−8(1−x)2dx−4[(1−x)−6x(1−x)]dx∴∫c2Pdx+Qdy=[3x33+8(1−x)33−4(1−x)22(−1)+6(x22−x33)]01∴∫c2Pdx+Qdy=[0+83+2−1−(6∗16)]∴∫c2Pdx+Qdy=[83+2−2]=83
(c) Along C3: x = 0, dx = 0, y varies 1 to 0
∫C3Pdx+Qdy=0∫y=10+4ydy∴∫C3Pdx+Qdy=2[y2]01=−2∗1∴∫C3Pdx+Qdy=−2
∴∫CPdx+Qdy=∫C1Pdx+Qdy+∫C2Pdx+Qdy+∫C3Pdx+Qdy=1+8/3−2∫CPdx+Qdy=5/3..................................................(1)
Now R.H.S. of equation (A): y: 0 to 1 - x & x: 0 to 1
∫∫R(∂Q∂x−∂P∂y)dxdy=1∫x=01−x∫y=0(−6y)y+16dxdy∫∫R(∂Q∂x−∂P∂y)dxdy=1∫x=01−x∫y=010ydxdy∫∫R(∂Q∂x−∂P∂y)dxdy=1∫x=01−x∫y=010(y22)1−x0dx∫∫R(∂Q∂x−∂P∂y)dxdy=1∫x=0102(1−x2)dx∫∫R(∂Q∂x−∂P∂y)dxdy=1∫x=05[(1−x)33(−1)]∫∫R(∂Q∂x−∂P∂y)dxdy=−5/3.............................................(2)
∴ From (1) and (2) Green's theorem is verified i.e.
∫cPdx+Qdy=∫∫R(∂Q∂x−∂P∂y)dxdy