Question

calcule o valor da integral de linha integral c (y-3x), sendo o segmento da reta que une os pontos (1,2) (2,3)

Answer 1

Calcular a integral de linha

$\displaystyle\int_C (y-3x)\,d\mathbf{s}$

Sendo $C$ o segmento de reta que liga os pontos $(1,\,2)$ a $(2,\,3).$

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Parametrizando o segmento de reta:

$C(t)=(1,2)+t\cdot \big((2,\,3)-(1,\,2)\big)\\\\ C(t)=(1,2)+t\cdot (2-1,\,3-2)\\\\ C(t)=(1,2)+t\cdot (1,\,1)\\\\ C(t)=(1,2)+(t,t)\\\\ \boxed{\begin{array}{c}C(t)=(1+t,\,2+t)~~~~~~\text{com }0\le t\le 1 \end{array}}$


Então, temos as equações para métricas da curva $C:$

$C:~\left\{\begin{array}{l} x(t)=1+t\\\\ y(t)=2+t \end{array}\right.~~~~~~\text{com }0\le t\le 1.$

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Encontrando o vetor tangente $C'(t):$

$C'(t)=\big(x'(t),\,y'(t)\big)\\\\ C'(t)=(1,\,1)$


Para o cálculo desta integral de linha, só nos interessa o módulo do vetor tangente:

$\|C'(t)\|=\|(1,\,1)\|\\\\ \|C'(t)\|=\sqrt{1^2+1^2}\\\\ \|C'(t)\|=\sqrt{1+1}\\\\ \|C'(t)\|=\sqrt{2},~~~~~~\text{para todo }t\in [0,\,1].$

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Montando o cálculo da integral de linha, temos

$\displaystyle\int_C (y-3x)\,d\mathbf{s}=\int_0^1\big((2+t)-3(1+t)\big)\cdot \|C'(t)\|\,dt\\\\\\ =\int_0^1 (2+t-3-3t)\cdot \sqrt{2}\,dt\\\\\\ =\sqrt{2}\int_0^1 (-1-2t)\,dt\\\\\\ =\sqrt{2}\cdot \left(-t-t^2 \right )\big|_0^1\\\\ =\sqrt{2}\cdot (-1-1^2)\\\\ =\sqrt{2}\cdot (-2)\\\\ =\boxed{\begin{array}{c}-2\sqrt{2} \end{array}}$


Bons estudos! :-)