Question

(Integral de linha sobre um Campo) Calcular a integral f(x,y)=(-y^2,x^2 ) em γ(t)=(t^2,t) com t variando de 0 a 1.

Answer 1

Resposta:  − 3/10.

Explicação passo a passo:

Calcular a integral de linha do campo vetorial

    $\mathsf{\overset{\to}{\mathbf{f}}(x,\,y)=(-y^2,\,x^2)}$

sobre a curva

    $\mathsf{\gamma(t)=(t^2,\,t),\qquad com~0\le t\le 1.}$

Encontrando o vetor tangente à curva:

    $\mathsf{\gamma'(t)=\big((t^2)',\,(t)'\big)}\\\\ \mathsf{\Longrightarrow\quad \gamma'(t)=(2t,\,1)}$

A integral pedida é

    $\mathsf{\displaystyle\int_0^1 \overset{\to}{\mathbf{f}}\big(\gamma(t)\big)\cdot \gamma'(t)\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \big(\!-\!y(t)^2,\,x(t)^2\big)\cdot (2t,\,1)\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 \big(\!-\!t^2,\,(t^2)^2\big)\cdot (2t,\,1)\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 (-t^2,\,t^4)\cdot (2t,\,1)\,dt}$

Calculando o produto escalar na integral:

    $=\mathsf{\displaystyle\int_0^1 (-t^2\cdot 2t+t^4\cdot 1)\,dt}\\\\\\ =\mathsf{\displaystyle\int_0^1 (-2t^3+t^4)\,dt}\\\\\\ =\mathsf{\left.\left(\!-2\cdot \dfrac{t^4}{4}+\dfrac{t^5}{5}\right)\right|_0^1}\\\\\\ =\mathsf{\left.\left(-\,\dfrac{t^4}{2}+\dfrac{t^5}{5}\right)\right|_0^1}\\\\\\ =\mathsf{\left(-\,\dfrac{1^4}{2}+\dfrac{1^5}{5}\right)-\left(-\,\dfrac{0^4}{2}+\dfrac{0^5}{5}\right)}\\\\\\ =\mathsf{\left(-\,\dfrac{1}{2}+\dfrac{1}{5}\right)-(0)}\\\\\\ =\mathsf{-\,\dfrac{5}{10}+\dfrac{2}{10}}\\\\\\ =\mathsf{-\,\dfrac{3}{10}\quad \longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)