Question

Determine a integral ∫ C ( x d x + y d y + z d z ) com C definida pela equação paramétrica γ ( t ) = ( 2 t 2 , t 3 , t ) com 0 ≤ t ≤1. Considere a orientação do percurso no sentido de crescimento do parâmetro t.

Answer 1

Resposta:   $\displaystyle\int_\gamma x\,dx+y\,dy+z\,dz=3.$

Explicação passo a passo:

Calcular a integral de linha do campo vetorial no $\mathbb{R}^3$

    $\mathbf{F}(x,\,y,\,z)=P(x,\,y,\,z)\mathbf{i}+Q(x,\,y,\,z)\mathbf{j}+R(x,\,y,\,z)\mathbf{k}$

com

    $\begin{cases}~P(x,\,y,\,z)=x\\ ~Q(x,\,y,\,z)=y\\ ~R(x,\,y,\,z)=z \end{cases}$

sobre a curva $\gamma$ parametrizada a seguir

    $\gamma(t)=(2t^2,\,t^3,\,t)\qquad\mathrm{com~}t\in[0,\,1].$

e orientada no sentido de crescimento do parâmetro $t.$

• Calculando o vetor tangente à curva $\gamma(t):$

    $\gamma'(t)=\big\langle (2t^2)',\,(t^3)',\,(t)'\big\rangle\\\\ \Longleftrightarrow\quad \gamma'(t)=\langle 4t,\,3t^2,\,1 \rangle\qquad\checkmark$

• Reescrevendo a integral de linha em função do parâmetro $t,$ temos

    $\displaystyle\int_\gamma x\,dx+y\,dy+z\,dz\\\\\\ =\int_\gamma \mathbf{F}\cdot \mathbf{dr}\\\\\\=\int_0^1 \mathbf{F}\big(\gamma(t)\big)\cdot \gamma'(t)\,dt\\\\\\=\int_0^1 \mathbf{F}(2t^2,\,t^3,\,t)\cdot \langle 4t,\,3t^2,\,1 \rangle\,dt$

mas $\mathbf{F}(x,\,y,\,z)=\langle x,\,y,\,z \rangle.$ Portanto, $\mathbf{F}(2t^2,\,t^3,\,t)=\langle 2t^2,\,t^3,\,t\rangle.$ Substituindo acima, a integral fica

    $\displaystyle=\int_0^1 \langle 2t^2,\,t^3,\,t\rangle \cdot \langle 4t,\,3t^2,\,1 \rangle \,dt$

Desenvolvendo o produto escalar dos vetores,

    $\displaystyle=\int_0^1 (2t^2\cdot 4t+t^3\cdot 3t^2+t\cdot 1)\,dt\\\\\\ =\int_0^1 (8t^3+3t^5+t)\,dt\\\\\\ =\left.\left(\frac{8t^4}{4}+\frac{3t^6}{6}+\frac{t^2}{2}\right)\right|_0^1$

    $=\left.\left(2t^4+\dfrac{t^6}{2}+\dfrac{t^2}{2}\right)\right|_0^1\\\\\\ =\left(2\cdot 1^4+\dfrac{1^6}{2}+\dfrac{1^2}{2}\right)-\left(2\cdot 0^4+\dfrac{0^6}{2}+\dfrac{0^2}{2}\right)\\\\\\ =\left(2\cdot 1+\dfrac{1}{2}+\dfrac{1}{2}\right)-(0)$

    $=2+\dfrac{1}{2}+\dfrac{1}{2}\\\\\\\\=\dfrac{4+1+1}{2}\\\\\\=\dfrac{6}{2} \\\\\\ =3\quad\longleftarrow\quad \mathsf{resposta.}$

Dúvidas? Comente.

Bons estudos!