VC4.10
From Exampleproblems
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)\,](/wiki/images/math/8/5/2/8520155cc60c5fc704c1110f6e955339.png)
=![\int_{C}[(2y+3)dx+xzdy+(yz-x)dz]\,](/wiki/images/math/d/4/2/d420bc4f987eb7a9858cfc8882051c21.png)
=![\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]\,](/wiki/images/math/c/2/b/c2b87a31f06fa9d6a7672aa1544b65b9.png)
--(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\,](/wiki/images/math/b/a/1/ba14eb6b7e8080853cefdb299ddaebfa.png)
--(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\,](/wiki/images/math/1/7/b/17bc22bdc2cfedf2997dce457a9b6eae.png)
--(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\,](/wiki/images/math/2/1/0/2105d284f376775fbe481bfa4347368d.png)
With the help of (2),(3) and (4),(1) becomes
