Let C be the line segment from $(2, 3, 0)$ to $(5, 7, 9)$ followed by the line segment from $(5, 7, 9)$ to $(1, 1, 2).$ Consider a force field $F(x, y, z) = <3x^2+yz+y,xz+x+z^2,xy+2yz+2z>$
Calculate the work done by $F$ on a particle taking the path $C$.
How would I solve this? Here's what I have so far:
Calculating the curl for the force field, I get = $(0,0,0)$ which means it's conservative. So I know then the integral must be:
$$\int _CF\cdot \:dr\:=\:f\left(1,\:1,\:2\right)-\:f\left(5,\:7,\:9\right)$$
Taking their respective partial derivatives I get:
$\frac{\partial }{\partial x}\left(3x^2+yz+y\right)$
$\frac{\partial }{\partial y}\left(xz+x+z^2\right)$
$\frac{\partial \:}{\partial \:z}\left(xy+2yz+2z\right)$
$$f(x,y,z)=6x+2y+2$$
So my final answer is:
$$\left(6\left(1\right)+2\left(1\right)+2\right)-\left(6\left(5\right)+2\left(7\right)+2\right)=-36$$
Is this the right process and answer?
Hint:
It is a conservative vector field but the potential function you have set up is not correct.
The Potential function is $f(x,y,z) = x^3 + xyz + xy + (y+1)z^2$.
$\frac{\partial f}{\partial x} = 3x^2 + yz + y$ which is part of your vector field. You can check other partial derivatives similarly.
One way to find the potential function is take the first component ($i$) of the vector field and integrate wrt $x$, similarly take the second component ($j$) wrt $y$. From there on, you should get the function.
Pls take only the end points for your calculation $(2, 3, 0)$ to $(1, 1, 2)$ because the integral will be path independent.