Calculate curve integral

Question

Given is curve integral $\int_{\Gamma }^{} \! udx + v dy + w dz\, $ for path $\Gamma$ which is direct distance from $O=(0,2,0)$ to $C=(5,4,3)$. The following is given: $u=-2xy^2+2z, v=-x^3+4xy, w=2x^2+4z^2$. Calculate curve integral.

I think I should just put $u,v$ and $w$ in integral that is given, but what to do with points that are given? How to write these boundaries?

Answer 1

Start

The parametric equations of the line $OC$ are

$$x=0+(5-0)t, dx=5dt$$ $$y=2+(4-2)t, dy=2dt$$ $$z=0+(3-0)t, dz=3dt$$

for $t=0$, it is $O$, and for $t=1$, we get the point $C$.

then

$$u=-2(5t)(2+2t)^2+2(3t)=-40t^3-80t^2-34t$$

do the same for $v$ and $w$ then plug them and integrate according to the variable $t$ from $0$ to $ 1.$

Answer 2

Recall that for a vector field $$F:\mathbb{R}^3\to \mathbb{R}^3, \quad F(x,y,z)=\begin{pmatrix}u(x,y,z) \\ v(x,y,z) \\ w (x,y,z) \end{pmatrix}$$ and a $C^1$ path $\gamma:[a,b]\to \mathbb{R}^3$ the integral of $F$ along $\gamma$ is defined as: $$\int_\gamma F\overset{def}{=}\int_\gamma F(\vec{x})\cdot\,d\vec{x}\overset{def}{=}\int_\gamma udx+udy+udz\overset{def}{=}\int_a^bF(\gamma(t))\cdot\gamma'(t)\,dt$$ Now you need only to calculate the integral on the very right.