Question

F(x, y, z = yz i xz j (xy 4z k c is the line segment from (3, 0, ?1 to (6, 4, 1 (a find a function f such that f = ?f. f(x, y, z = (b use part (a to evaluate c ?f · dr along the given curve

c.

Answer 1

Without knowing what the obviously missing symbols could be, I'm going to guess and say that the given vector field is


`\mathbf F(x,y,z)=yz\,\mathbf i+xz\,\mathbf j+(xy+4z)\,\mathbf k`

though I'm not as confident about that last component as I am about the other two.

Now the goal is to find a function
`f(x,y,z)` such that
`\\abla f=\mathbf F`, which is to say
`f` satisfies


`\begin{cases}f_x=yz\\f_y=xz\\f_z=xy-4z\end{cases}`

Here
`f_i` denotes the partial derivative of
`f` with respect to the independent variable
`i`.

Integrating the first equation with respect to
`x` yields


`f(x,y,z)=xyz+g(y,z)`

Differentiating this with respect to
`y` gives


`f_y=xz=xz+g_y\implies g_y=0\implies g(y,z)=h(z)`

Differentiating with respect to
`z`, we get


`f_z=xy-4z=xy+h'(z)\implies h'(z)=-4z\implies h(z)=-2z^2`

This means you have


`f(x,y,z)=xyz-2z^2`

and in particular, that
`\\abla f=\mathbf F`, which means the line integral depends only on the value of
`f` at the endpoints of its path.

This means


`\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=f(6,4,1)-f(3,0,-1)=24`