Question

sider F and C below. F(x, y, z) = yz i + xz j + (xy + 4z) k C is the line segment from (1, 0, −2) to (6, 4, 1) (a) Find a function f such that F = ∇f. f(x, y, z) = xyz+2z2+c (b) Use part (a) to evaluate C ∇f · dr along the given curve C.

Answer 1

a) The function is
`f(x,y,z) = xyz+2z^2`

b) The value of the integral is 18

Explanation:

a) We are given that
`F(x,y,z) (yz,xz,xy+4z)`. We want to find a function f such that the gradient of f is F. That is
`\\ablda f = F` . Suppose that such f does exist, if that is the case, then by definition of the gradient, we have that

`F(x,y,z) = ((\partial f)/(\partial x),(\partial f)/(\partial y),(\partial f)/(\partial z))`

From here, we have that

`yz = (\partial f)/(\partial x)`

if we integrate both sides with respect to x, we get that

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

where g is a function that depens on y and z only. Now, we differentiate this equation with respect to y and make it equal to the 2nd component of F. That is

`xz + \frac{\partial g}\partial{y} = xz`

This implies that
`(\partial g)/(\partial y) =0`. This means that g actually depends only on z. Until now, f is of the form

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

If we repeat the previous step, by differentiating with respect to z and making it equall to the third component of F we get

`xy + (\partial g)/(\partial z) = xy + 4z`

This implies that
`(\partial g)/(\partial z) = 4z` . If we integrate both sides with respect to z, we get that
`g(z) = 2z^2`

So f is of the form
`f(x,y,z) = xyz+2z^2`

b) To calculate the integral over the given segment, we can use the function f. Since the path is from (1,0,-2) to (6,4,1), then the value of the integral is given by evaluatin f at the end point and the substracting the value of f at the start point, that is

`\int_C F \cdot dr = f(6,4,1) -f(1,0,-2) = 24+2(1)^2- (0+2(-2)^2)) = 18`