# VC4.10

Let C1 denote the straight line joining (0,0,0) to (0,0,1),C2 denote the straight line joining (0,0,1) to (0,1,1) and C3 is the line joining (0,1,1)to(2,1,1).Then the given path C consists of C1,C2,C3.

$\int _{{C}}=\int _{{C}}[(2y+3)i+xzj+(yz-x)k]\cdot (dxi+dyj+dzk)\,$

=$\int _{{C}}[(2y+3)dx+xzdy+(yz-x)dz]\,$

=$\int _{{C_{1}}}[(2y+3)dx+xzdy+(yz-x)dz]+\int _{{C_{2}}}[(2y+3)dx+xzdy+(yz-x)dz]+\int _{{C_{3}}}[(2y+3)dx+xzdy+(yz-x)dz]\,$ --(1)

Along C1,x=0,y=0 so that dx=0,dy=0 and z varies from 0 t0 1.

Therefore,$\int _{{C_{1}}}[(2y+3)dx+xzdy+(yz-x)dz]=\int _{{z=0}}^{{1}}(0)dz=0\,$ --(2)

Along C2,x=0,z=1 so that dx=0,dz=0 and y varies from 0 to 1.

Therefore,$\int _{{C_{2}}}[(2y+3)dx+xzdy+(yz-x)dz]=\int _{{y=0}}^{{1}}(0)dy=0\,$ --(3)

Along C3,y=1,z=1 so that dy=0,dz=0 and x varies from 0 to 2.

Therefore,$\int _{{C_{3}}}[(2y+3)dx+xzdy+(yz-x)dz]=\int _{{x=0}}^{{2}}(5)dx=10\,$

With the help of (2),(3) and (4),(1) becomes

$\int _{{C}}F\cdot \,dr=0+0+10=10\,$