Question

Estou estudando calculo, preciso de uma ajuda nesta questão?

Answer 1

(a) Queremos calcular a seguinte integral de linha

$\displaystyle\oint_\Gamma\overrightarrow{\mathbf{F}}(x,\,y)\cdot d\overrightarrow{\mathbf{r}}~~~~~~\mathbf{(i)}$

sendo $\overrightarrow{\mathbf{F}}(x,\,y)$ o seguinte campo vetorial:

$\overrightarrow{\mathbf{F}}(x,\,y)=P(x,\,y)\overrightarrow{\mathbf{i}}+Q(x,\,y)\overrightarrow{\mathbf{j}}\\\\\\ \overrightarrow{\mathbf{F}}(x,\,y)=\left\langle\, P(x,\,y),\;Q(x,\,y)\,\right\rangle$


As componentes $P$ e $Q$ do campo são as seguintes funções:

$\left\{ \!\begin{array}{l} P(x,\,y)=e^x(1-\cos y)\\\\ Q(x,\,y)=-e^x(y-\mathrm{sen\,}y) \end{array} \right.$

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A curva $\Gamma$ é regular por partes, o domínio do campo $\overrightarrow{\mathbf{F}}$ é todo o $\mathbb{R}^2,$ e as componentes de $\overrightarrow{\mathbf{F}}$ são contínuas em $\mathbb{R}^2.$ Portanto vale o Teorema de Green:

$\displaystyle\oint_\Gamma\overrightarrow{\mathbf{F}}(x,\,y)\cdot d\overrightarrow{\mathbf{r}}=\iint_{\mathrm{int}(\Gamma)}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y} \right )dx\,dy~~~~~~\mathbf{(ii)}$

onde $\mathrm{int}(\Gamma)$ é a região do plano compreendida no interior da curva $\Gamma.$

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$\bullet~~\dfrac{\partial Q}{\partial x}=\dfrac{\partial}{\partial x}\!\left[-e^x\,(y-\mathrm{sen\,}y) \right ]\\\\\\ =\dfrac{\partial}{\partial x}(-e^x)\cdot (y-\mathrm{sen\,}y)-e^x\cdot \dfrac{\partial}{\partial x}(y-\mathrm{sen\,}y)\\\\\\ =-e^x\cdot (y-\mathrm{sen\,}y)-e^x\cdot 0\\\\\\ \therefore~~\dfrac{\partial Q}{\partial x}=-e^x\,(y-\mathrm{sen\,}y)$


$\bullet~~\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}\!\left[e^x\,(1-\cos y) \right ]\\\\\\ =\dfrac{\partial}{\partial y}(e^x)\cdot (1-\cos y)+e^x\cdot \dfrac{\partial}{\partial y}(1-\cos y)\\\\\\ =0\cdot (1-\cos y)+e^x\cdot (0+\mathrm{sen\,}y)\\\\\\ \therefore~~\dfrac{\partial P}{\partial y}=e^x\,\mathrm{sen\,}y$


$\bullet~~$ Dessa forma,

$\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=-e^x\,(y-\mathrm{sen\,}y)-e^x\,\mathrm{sen\,}y\\\\\\ \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=-y\,e^x+e^x\,\mathrm{sen\,}y-e^x\,\mathrm{sen\,}y\\\\\\ \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=-y\,e^x~~~~~~\mathbf{(iii)}$

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Substituindo em $\mathbf{(ii)},$ a integral pedida é

$\displaystyle\oint_\Gamma\overrightarrow{\mathbf{F}}(x,\,y)\cdot d\overrightarrow{\mathbf{r}}=\iint_{\mathrm{int}(\Gamma)}(-y\,e^x)\,dx\,dy\\\\\\ =\int_{0}^{\pi}\int_{0}^{\mathrm{sen\,}x}(-y\,e^x)\,dy\,dx\\\\\\ =-\int_{0}^{\pi}e^x\cdot \left.\left(\dfrac{y^2}{2} \right )\right|_{0}^{\mathrm{sen\,}x}\,dx\\\\\\ =-\frac{1}{2}\int_{0}^{\pi}e^x\,\mathrm{sen^2\,}x\,dx\\\\\\ =-\frac{1}{2}\int_{0}^{\pi}e^x\left(\dfrac{1}{2}-\dfrac{1}{2}\,\cos\,2x \right )dx$

$=\displaystyle-\frac{1}{4}\int_{0}^{\pi}e^x\,dx+\dfrac{1}{4}\int e^x\cos\,2x\,dx\\\\\\ =-\frac{1}{4}\,(e^x)|_0^\pi+\dfrac{1}{4}\int_0^\pi e^x\cos\,2x\,dx\\\\\\ =-\frac{1}{4}\,(e^\pi-e^0)+\dfrac{1}{4}\int_0^\pi e^x\cos\,2x\,dx\\\\\\ =-\frac{1}{4}\,(e^\pi-1)+\dfrac{1}{4}\int_0^\pi e^x\cos\,2x\,dx~~~~~~\mathbf{(iv)}$

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Avaliando a seguinte integral (via método de integração por partes)

$\displaystyle\int e^x\cos\,2x\,dx=\dfrac{2}{5}\,e^x\,\mathrm{sen\,}2x+\dfrac{1}{5}\,e^x\cos 2x+C$

(não coloquei os detalhes da resolução, pois é um passo intermediário que tomaria muito espaço aqui)


Portanto,

$\displaystyle\int_0^\pi e^x\cos\,2x\,dx=\left.\left(\dfrac{2}{5}\,e^x\,\mathrm{sen\,}2x+\dfrac{1}{5}\,e^x\cos 2x \right )\right|_0^\pi\\\\\\ =\left(\dfrac{2}{5}\,e^\pi\,\mathrm{sen\,}2\pi+\dfrac{1}{5}\,e^\pi\cos 2\pi \right )-\left(\dfrac{2}{5}\,e^0\,\mathrm{sen\,}0+\dfrac{1}{5}\,e^0 \cos 0 \right)\\\\\\ =\dfrac{1}{5}\,e^\pi-\dfrac{1}{5}$

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Substituindo em $\mathbf{(iv)},$ finalmente obtemos

$\displaystyle\oint_\Gamma \overrightarrow{\mathbf{F}}(x,\,y)\cdot d\overrightarrow{\mathbf{r}}=-\frac{1}{4}\,(e^\pi-1)+\dfrac{1}{4}\cdot \left(\dfrac{1}{5}\,e^\pi-\dfrac{1}{5} \right )\\\\\\ =-\frac{1}{4}\,e^\pi+\dfrac{1}{4}+\dfrac{1}{20}\,e^\pi-\dfrac{1}{20}\\\\\\ =\frac{-5+1}{20}\,e^\pi+\dfrac{5-1}{20}\\\\\\ =\frac{-4}{20}\,e^\pi+\dfrac{4}{20}\\\\\\ =\dfrac{1}{5}-\dfrac{1}{5}\,e^\pi$