Question

Olá, alguém me ajuda?

Answer 1

El enunciado del teorema de Green queda a tu cargo, solo colocaré el cómo aplicarlo

$\displaystyle \iint\limits_D \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}dxdy=\oint\limits_{ \partial D^+} Pdx+Qdy\\ \\ \text{En este caso: }\\ P=-x^2y\to \dfrac{\partial P}{\partial y} = -x^2\\ \\ Q=xy^2\to \dfrac{\partial Q}{\partial x} = y^2\\ \\ \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=y^2+x^2$

$\displaystyle \oint\limits_{\Gamma}-x^2ydx+xy^2dy=\iint\limits_{D}x^2+y^2 dxdy\\ \\ \\ \text{en coordenadas polares: }\\ \\ \displaystyle \oint\limits_{\Gamma}-x^2ydx+xy^2dy=\int_{0}^{2\pi}\int_{0}^{R}r^3drd\theta\\ \\\\ \boxed{\boxed{\displaystyle \oint\limits_{\Gamma}-x^2ydx+xy^2dy=\dfrac{\pi R^4}{2}}}$

=====================================================

$\displaystyle D=\left\{(x,y)\in\mathbb R^2: 0\leq x\leq 1 ; x^2\leq y\leq \sqrt{x}\right\}\\ \\ P=-y^2+\arctan x \to \dfrac{\partial P}{\partial y} = -2y\\ \\ Q=\ln y \to \dfrac{\partial Q}{\partial x} = 0\\ \\ \oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\iint\limits_{D}2y dxdy\\ \\ \oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\int_{0}^{1}\int_{x^2}^{\sqrt{x}}2ydydx\\ \\ \oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\int_{0}^{1}\left.(y^2)\right|_{y=x^2}^{y=\sqrt{x}}dx$


$\displaystyle \oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\int_{0}^{1}x-x^4 dx\\ \\ \oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\left.\left(\dfrac{x^2}{2}-\dfrac{x^5}{5}\right)\right|_{0}^1\\ \\ \boxed{\boxed{\oint\limits_{\Gamma}(-y^2+\arctan x)dx+\ln y dy =\dfrac{3}{10}}}$