Question

Ensino superior cálculo
Calcule a integral de linha
xyds
onde C ´e o arco da par´abola y =x^2/2

de (0, 0) a (2, 2).

Answer 1

1) Lembre que $ds$ é a largura de um trozo de C, neste caso

        $C:\begin{cases} x(t)=t\\\\y(t)=\dfrac{t^2}{2} \end{cases}\\\\ \\ \texttt{Onde }t\in [0,2]$

$\texttt{Ent\~ao: }ds=\sqrt{x_t^2+y_t^2}\;dt\to ds=\sqrt{1+t^2}\,dt\\ \\ \\ \displaystyle \int_{C}xy\,ds=\int\limits_{0}^2t\left(\dfrac{t^2}{2}\right)\sqrt{1+t^2}\,dt\\ \\ \\ \int_{C}xy\,ds=\dfrac{1}{2}\int\limits_{0}^2t^3\sqrt{1+t^2}\,dt\\ \\ \\ \texttt{Seja }u^2=1+t^2\to u \, du=t\,dt\; \, \; \texttt{onde }u\in[1,\sqrt{5}]\\ \\ \\ \int_{C}xy\,ds=\dfrac{1}{2}\int\limits_{1}^{\sqrt{5}}(u^2-1)\cdot u\cdot u\,du$

$\displaystyle \int_{C}xy\,ds=\dfrac{1}{2}\int\limits_{1}^{\sqrt{5}}u^4-u^2\,du\\ \\ \\ \int_{C}xy\,ds=\dfrac{1}{2}\left.\left(\dfrac{u^5}{5}-\dfrac{u^3}{3}\right)\right|_{1}^{\sqrt{5}}\\ \\ \\ \int_{C}xy\,ds=\dfrac{1}{2}\left[\left(\dfrac{\sqrt{5}^5}{5}-\dfrac{\sqrt{5}^3}{3}\right)-\left(\dfrac{1^5}{5}-\dfrac{1^3}{3}\right)\right]\\ \\ \\ \int_{C}xy\,ds=\dfrac{1}{2}\left(\dfrac{10\sqrt{5}}{3}+\dfrac{2}{15}\right)\\ \\ \\ \boxed{\int_{C}xy\,ds=\dfrac{5\sqrt{5}}{3}+\dfrac{1}{15}}$