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.

  • $\begingroup$ I've used your tip, is it correct? $\endgroup$ – SABOY Mar 23 at 11:34
  • $\begingroup$ Looks good to me. $\endgroup$ – jmerry Mar 23 at 11:36

You can use $t\mapsto\begin{pmatrix}0\\1\\t\end{pmatrix}$ for $t\in[0,2\pi]$.


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