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$r(t) = (2t\sqrt 3,1+2sin(t),2cos(t))$ with $t\in[0,2\pi]$

I need to solve : $\int_rxyz ds$

also I need to know how to resolve this type :

$\int_r F dr$

knowing that $F(x,y) = ({y\over1+xy},{x\over1+xy})$ and $r(t) = (cos(t),2sin(t))$ with $t\in[0,2\pi]$

1 Answers 1

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If $f$ is a scalar function then:

$$\int_C f \ ds = \int_{a}^b f(r) \|r'\| \ dt$$

If $F$ is a vector field we have:

$$\int_{C} F \cdot d\textbf{r} = \int_{a}^b F(\gamma) \cdot \gamma' \ dt$$

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    Could you also help me calculate those examples with the r(t) values that I wrote please ? i know how to calculate ||r|| but a and b is 0 and 2 pi ? – 2017-02-19
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    All you have to do is plug the values in. In the first example $f = xyz$, so what is $f(r)$=? – 2017-02-19
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    yes I know :D , but a and b is 0 and 2 pi ? – 2017-02-19
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    Yes, that is correct. – 2017-02-19
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    Thanks , and what about the second one $F(\gamma)$ ? your $\gamma$ simbol is equivalent to my $r$ ? – 2017-02-19
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    The bounds are the same again. You have $t \in [0,2\pi]$. – 2017-02-19
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    but what is $F(\gamma)$ and $\gamma'$ .your $\gamma$ simbol is equivalent to my $r$ ? – 2017-02-19
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    Yes. $\gamma$ is just a parametrization of your closed curve, as $r$ was. – 2017-02-19