25

Comparing (3) and (4), we can conclude

that

Ceyf y )(

Hence, Cexyyxy sin),(

To find constant C, use .0)0,0(

1

01

00sin0)0,0( 0

C

C

Ce

Therefore, 1sin),( yexyyx

and Cxg )(

where C is an arbitrary constant of integration

26

Example 4

Find if

and .

),,( zyx

kji )34()23()2( 2233232 yzxzzxxyxyzy

2)0,0,0(

27

Integrating (1), (2) and (3) w.r.t. x, y and z

respectively, we obtain

Solution

We have .....(1) 2 32 xyzyx

.....(2) 23 32zxxyy

.....(4) ),()2( 32232 zyfyzxxydxxyzy

.....(3) 34 223 yzxzz

.....(5) ),(3)23( 32232 zxgyzxxyydyzxxy

.....(6) ),()34( 324223 yxhyzxzdzyzxz