Question

Considere uma curva C que é uma hélice circular parametrizada como x(t) = cos t; y (t) = sen t; z(t)=1 sobre o valor integral c (x^2+ y ^2 -z) ds é correto afirmar que do ponto A(1, 0, 0) ao ponto B(1, 0, 2π) a integral de linha possui qual valor?

Answer 1

Calcular

$\displaystyle\int_C\! (x^2+y^2-z)\,ds$

sendo $C$ a curva parametrizada dada por

$C(t)=(\cos t,\,\mathrm{sen\,}t,\,t)$

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• Para $t=0,$ o ponto inicial é

$C(0)=(\cos 0,\,\mathrm{sen\,}0,\,0)\\\\C(0) = (1,\,0,\,0);$


• Para $t=2\pi,$ o ponto final é

$C(2\pi)=(\cos 2\pi,\,\mathrm{sen\,}2\pi,\,2\pi)\\\\C(2\pi) = (1,\,0,\,2\pi).$

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• Encontrando o vetor tangente $C'(t)$ e calculando o seu módulo:

$C'(t)=\dfrac{d}{dt}\big(\cos t,\,\mathrm{sen\,}t,\,t\big)\\\\ C'(t)=\big(\!-\mathrm{sen\,}t,\,\cos t,\,1\big)$


O módulo do vetor tangente:

$\|C'(t)\|=\left\|(-\mathrm{sen\,}t,\,\cos t,\,1)\right\|\\\\ \|C'(t)\|=\sqrt{(-\mathrm{sen\,}t)^2+(\cos t)^2+1^2}\\\\ \|C'(t)\|=\sqrt{\mathrm{sen^2\,}t+\cos^2 t+1}\\\\ \|C'(t)\|=\sqrt{1+1}\\\\ \|C'(t)\|=\sqrt{2}\,,~~~~\text{para todo }t\in \left[0,\,2\pi\right].$

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Montando a nossa integral de linha:

$\displaystyle\int_C (x^2+y^2-z)\,ds\\\\\\ =\int_0^{2\pi} \big((\cos t)^2+(\mathrm{sen\,}t)^2-t\big)\cdot \left\|C'(t)\right\|dt\\\\\\ =\int_0^{2\pi} \big(\cos^2 t+\mathrm{sen^2\,}t-t\big)\cdot \sqrt{2}\,dt\\\\\\ =\int_0^{2\pi} (1-t)\cdot \sqrt{2}\,dt\\\\\\ =\sqrt{2}\cdot \left.\left(t-\frac{t^2}{2}\right)\right|_0^{2\pi}$

$=\sqrt{2}\cdot \left(2\pi-\dfrac{(2\pi)^2}{2}\right)\\\\\\ =\sqrt{2}\cdot \left(2\pi-\dfrac{4\pi^2}{2}\right)\\\\\\ =\sqrt{2}\cdot \left(2\pi-2\pi^2\right)\\\\\\ =\sqrt{2}\cdot 2\pi(1-\pi)\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int_C (x^2+y^2-z)\,ds=2\sqrt{2}\,\pi\cdot \left(1-\pi\right)\ \end{array}}$


Bons estudos! :-)