Answer:
Required solution (a) [tex]f(x,y,z)=xyz+3z^2+C[/tex] (b) 40.
Step-by-step explanation:
Given,
[tex]F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k[/tex]
(a) Let,
[tex]F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}[/tex]
Then,
[tex]f_x=yz,f_y=xz,f_z=xy+6z[/tex]
Integrating [tex]f_x[/tex] we get,
[tex]f(x,y,z)=xyz+g(y,z)[/tex]
Differentiate this with respect to y we get,
[tex]f_y=xz+g'(y,z)[/tex]
compairinfg with [tex]f_y=xz[/tex] of the given function we get,
[tex]g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)[/tex]
Then,
[tex]f(x,y,z)=xyz+h(z)[/tex]
Again differentiate with respect to z we get,
[tex]f_z=xy+h'(z)=xy+6z[/tex]
on compairing we get,
[tex]h'(z)=6z\implies h(z)=3z^2+C[/tex] (By integrating h'(z)) where C is integration constant. Hence,
[tex]f(x,y,z)=xyz+3z^2+C[/tex]
(b) Next, to find the itegration,
[tex]\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40[/tex]
