# Shortcuts for $\int_{\kappa}F dx$

Let $$F(x,y,z):=\begin{pmatrix} x^{2}+5y+3yz \\ 5x +3xz -2 \\ 3xy -4z \end{pmatrix}$$

and $$\kappa: [0, 2\pi] \to \mathbb R^{3}, t\mapsto\begin{pmatrix}\sin t\\ \cos t \\ t \end{pmatrix}$$

I was asked to find $$\int_{\kappa}Fdx$$.

I have tried to calculate it directly:

$$\int_{\kappa}Fdx=\int_{0}^{2\pi}\begin{pmatrix} \sin (t)^{2}+5\cos t+3\cos {(t)}t \\ 5\sin{(t)} +3\sin{(t)}t -2 \\ 3\sin{(t)}\cos{(t)} -4t \end{pmatrix}\cdot \begin{pmatrix}\cos t\\ -\sin t \\ 1 \end{pmatrix}dt$$

and basically I get something that I cannot calculate.

I have been given the tip of using path independence.

First I have seen that $$DF(x,y,z)$$ is symmetrical, so I can use path independence.

I am new to curve integrals, so I am unsure what curve $$\overline{\kappa}:[a,b]\to \mathbb R^{3}$$ (where $$\overline{\kappa}(a)=\kappa(0)$$ and $$\overline{\kappa}(b)=\kappa(2\pi)$$) I am supposed to use, in order to have a better integral to calculate.

Using $$\overline{\kappa}(t)=\begin{pmatrix} 0 \\ 1 \\ t \end{pmatrix}$$

we get $$\int_{0}^{2\pi}\begin{pmatrix} 5+3t \\ -2 \\ -4t \end{pmatrix}\cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}dt=-4\int_{0}^{2\pi}t dt=-8\pi^{2}$$

From $$(0,1,0)$$ to $$(0,1,2\pi)$$? It's hard to go wrong with a straight line.
And with $$x$$ and $$y$$ the same between the endpoints and $$x$$ zero throughout, lots of things just zero out cleanly.
You can use $$t\mapsto\begin{pmatrix}0\\1\\t\end{pmatrix}$$ for $$t\in[0,2\pi]$$.