explanation on line integrals

Question

dbwong · Answer

could someone please explain what happened at a)?
what did the person mean by putting it all together...?

irishboy123 · Answer

Step by step:

$\vec F = \left( \begin{matrix}
  yz \
  xz \
  xy + 18z
\end{matrix} \right) = \nabla f  = \left( \begin{matrix}
  \partial_x f \
  \partial_y f \
  \partial_z f
\end{matrix} \right)
$

$\partial_x f = yz \implies f = xyz + \alpha_1(y, z)$

$\partial_y f = xz \implies f = xyz + \alpha_2(x, z)$

$\partial_z f = xy + 18z \implies f = xyz + 9 z^2 + \alpha_3(x, y)$

$\implies \alpha_1(y, z) = \alpha_2(x, z) = 9z^2 + C$ 

....and: 

$\alpha_3(x, y) = C$

irishboy123 · Answer

There's another way to do this ... which *might* be easier to follow: You're told there is a potential function so the field is path-independent/ conservative. ie you can choose any way you like to get to from $<0,0,0>$ to $$. You will always get the same result: $f(x,y,z) - f(0,0,0)$ So for the line integral $\int_C \vec F(\vec r) \cdot d\vec r$ where $\vec r = $ along this path: $\left( \begin{matrix} 0 \ 0\ 0 \end{matrix} \right) \to \left( \begin{matrix} x \ 0\ 0 \end{matrix} \right) \to \left( \begin{matrix} x \ y\ 0 \end{matrix} \right) \to \left( \begin{matrix} x \ y\ z \end{matrix} \right)$ .....you have: - Along step 1, $y,z,dy,dz = 0$ - Along 2, $z, dx, dz = 0$ - Along 3, $dx, dy = 0$ The integral becomes: $\int_0^x 0 ~ dx + \int_0^y 0 ~ dy + \int_0^z xy + 18 z ~ dz = xyz + 9 z^2 $

phi · Answer

If you have time, see https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-21-gradient-fields/ this covers 2-D problems See Lecture 30, where they extend the ideas to 3-Dimensions