Calcule \int_{C} \mathbf{F} \cdot d \mathbf{r} ao longo da curva C
\begin{aligned}& \mathbf{F}(x, y, z)=2 x y \mathbf{i}+(x-2 y) \mathbf{j} \\& C: \mathbf{r}(t)=\sin t \mathbf{i}-2 \cos t \mathbf{j} \quad 0 \leq t \leq \pi\end{aligned}

Passo 1

O enunciado pede o cálculo do integral de linha
\int_{C}\mathbf{F}\cdot d\mathbf{r}
onde
\mathbf{F}(x,y,z)=2xy\,\mathbf{i} + (x-2y)\,\mathbf{j},
e C é dada pela parametrização
\mathbf{r}(t)=\sin t\,\mathbf{i}-2\cos t\,\mathbf{j},\quad 0\le t\le\pi.
O método é usar a fórmula
\int_{C}\mathbf{F}\cdot d\mathbf{r} =\int_{0}^{\pi}\mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt.

Passo 2

Identificamos a parametrização:
x(t)=\sin t,\quad y(t)=-2\cos t.
Derivando,
\mathbf{r}'(t) =\frac{d}{dt}(\sin t\,\mathbf{i}-2\cos t\,\mathbf{j}) =\cos t\,\mathbf{i}+2\sin t\,\mathbf{j}.

Passo 3

Substituímos na campo \mathbf{F}:
\mathbf{F}(\mathbf{r}(t)) =\bigl(2x(t)y(t),\;x(t)-2y(t)\bigr) =\bigl(2\sin t\cdot(-2\cos t),\;\sin t-2(-2\cos t)\bigr) =(-4\sin t\cos t,\;\sin t+4\cos t).

Passo 4

Formamos o produto escalar
\mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t) =(-4\sin t\cos t,\,\sin t+4\cos t)\cdot(\cos t,\,2\sin t).
Isso dá
-4\sin t\cos^2 t \;+\;2\sin t(\sin t+4\cos t) =-4\sin t\cos^2 t+2\sin^2 t+8\sin t\cos t.

Passo 5

Devemos integrar de 0 a \pi:
\int_{0}^{\pi}\bigl(-4\sin t\cos^2 t+2\sin^2 t+8\sin t\cos t\bigr)\,dt.
Separamos em parcelas:
  1. \displaystyle\int_{0}^{\pi}-4\sin t\cos^2 t\,dt = -4\int_{0}^{\pi}\sin t(1-\sin^2 t)\,dt =-4\bigl[\!-\cos t+\tfrac{\cos^3 t}{3}\bigr]_{0}^{\pi} =\tfrac{16}{3}.
  2. \displaystyle\int_{0}^{\pi}8\sin t\cos t\,dt =4\bigl[\sin^2 t\bigr]_{0}^{\pi}=0.
  3. \displaystyle\int_{0}^{\pi}2\sin^2 t\,dt =2\int_{0}^{\pi}\tfrac{1-\cos2t}{2}\,dt =\int_{0}^{\pi}(1-\cos2t)\,dt =\pi.
Somamos ainda
\displaystyle\int_{0}^{\pi}-4\sin t\,dt =4[\cos t]_{0}^{\pi}=-8.

Passo 6

Somando tudo:
\frac{16}{3}+0+\pi-8 =\pi-\frac{8}{3}.

Resposta

\int_{C}\mathbf{F}\cdot d\mathbf{r} =\pi-\frac{8}{3}.

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