Question

Calcule a integral de linha f x²dx + y²dy + z²dz ao longo do caminho C = C₁ + C₂. C₁ = (0,0,0) a (1,2,-1) e C₂ = (1,2,-1) a (3,2,0)​

Answer 1

Resposta:  $\displaystyle\int_C x^2\,dx+y^2\,dy+z^2\,dz=\frac{35}{3}.$

Explicação passo a passo:

Calcular a integral de linha do campo vetorial

    $\mathbf{F}(x,\,y,\,z)=x^2\mathbf{i}+y^2\mathbf{j}+z^2\mathbf{k}$

ao longo do caminho $C=C_1\cup C_2,$ sendo

•     $C_1$ o segmento de reta vai do ponto $(0,\,0,\,0)$ a $(1,\,2,\,-1);$

•     $C_2$ o segmento de reta vai do ponto $(1,\,2,\,-1)$ a $(3,\,2,\,0).$

Parametrizando as curvas $C_1$ e $C_2:$

    $C_1:~\begin{cases}~x=0+t\\ ~y=0+2t\\ ~z=0-t \end{cases}\\\\\\\Longleftrightarrow\quad C_1:~\begin{cases}~x=t\\ ~y=2t\\ ~z=-t \end{cases}\qquad \mathrm{com~}0\le t\le 1.$

    $C_2:~\begin{cases}~x=1+(3-1)t\\ ~y=2+(2-2)t\\ ~z=-1+(0-(-1))t \end{cases}\\\\\\ \Longleftrightarrow\quad C_2:~\begin{cases}~x=1+2t\\ ~y=2+0t\\ ~z=-1+1t \end{cases}\\\\\\ \Longleftrightarrow\quad C_2:~\begin{cases}~x=1+2t\\ ~y=2\\ ~z=-1+t \end{cases}\qquad\mathrm{com~}0\le t\le 1.$

Logo, temos $C_1(t)=(t,\,2t,\,-t)$ e $C_2(t)=(1+2t,\,2,\,-1+t),$ com $0\le t\le 1.$

Encontrando os vetores tangentes às curvas $C_1$ e $C_2:$

    $\Longrightarrow\quad \begin{cases}C_1'(t)=\langle 1,\,2,\,-1\rangle\\\\ C_2'(t)=\langle 2,\,0,\,1\rangle \end{cases}$

A integral pedida é

    $\displaystyle =\int_C\mathbf{F}\cdot d\mathbf{r}\\\\\\ =\int_C x^2\,dx+y^2\,dy+z^2\,dz\\\\\\ =\int_C \langle x^2,\,y^2,\,z^2\rangle \cdot d\mathbf{r}\\\\\\ =\int_{C_1\cup C_2} \langle x^2,\,y^2,\,z^2\rangle \cdot d\mathbf{r}$

   $\displaystyle=\int_{C_1} \langle x^2,\,y^2,\,z^2\rangle \cdot d\mathbf{r}+\int_{C_2} \langle x^2,\,y^2,\,z^2\rangle \cdot d\mathbf{r}$

   $\displaystyle=\int_0^1 \langle t^2,\,(2t)^2,\,(-t)^2\rangle \cdot C_1'(t)\,dt+\int_0^1 \langle (1+2t)^2,\,(2)^2,\,(-1+t)^2\rangle \cdot C_2'(t)\,dt$

    $\displaystyle =\int_0^1 \langle t^2,\,4t^2,\,t^2\rangle \cdot \langle 1,\,2,\,-1\rangle \,dt+\int_0^1 \langle 1+4t+4t^2,\,4,\,1-2t+t^2 \rangle \cdot \langle 2,\,0,\,1 \rangle\,dt$

Efetuando os produtos escalares dos vetores, obtemos

    $\displaystyle =\int_0^1 \big(t^2\cdot 1+4t^2\cdot 2+t^2\cdot (-1)\big)dt+\int_0^1 \big((1+4t+4t^2)\cdot 2+4\cdot 0+(1-2t+t^2)\cdot 1\big)dt$

    $\displaystyle =\int_0^1 \big(t^2+8t^2-t^2\big)\,dt+\int_0^1 \big((2+8t+8t^2)+0+(1-2t+t^2)\big)dt\\\\\\ =\int_0^1 8t^2\,dt+\int_0^1 \big(3+6t+9t^2\big)dt\\\\\\=\int_0^1 \big(8t^2+(3+6t+9t^2)\big)dt$

    $\displaystyle =\int_0^1 \big(3+6t+17t^2\big)dt\\\\\\ =\left.\left(3t+\dfrac{6t^2}{2}+\dfrac{17t^3}{3}\right)\right|_0^1\\\\\\ =\left.\left(3t+3t^2+\frac{17t^3}{3}\right)\right|_0^1$

    $=\left(3\cdot 1+3\cdot 1^2+\dfrac{17\cdot 1^3}{3}\right)-\left(3\cdot 0+3\cdot 0^2+\dfrac{17\cdot 0^3}{3}\right)\\\\\\ =\left(3+3+\dfrac{17}{3}\right)-(0)$

    $=\dfrac{9+9+17}{3}\\\\\\ =\dfrac{35}{3}\quad\longleftarrow\quad\mathsf{resposta.}$

Dúvidas? Comente.

Bons estudos! :-)