Question

Utilizando o Teorema de Green, assinale a alternativa que contém a transformação da integral de linha

a ∫ ∫y dxdy b ∫ ∫ y^2/raiz X^2 + Y^2 dxdy c ∫ ∫ y^2 dxdy d ∫ ∫ y^2 + 2y/ raiz x^2+ y^2 dxdy e ∫ ∫ y/ raiz x^2+y^2dxdy

Answer 1

Teorema de Green:

$\boxed{\oint _{C}( Mdx+Ndy) =\int \int _{R}\left(\frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}\right) dxdy}$

Na sua questão, $M=\sqrt{x^{2} +y^{2}}$ e $N=y\left[ xy+\ln\left( x+\sqrt{x^{2} +y^{2}}\right)\right] dy$

A derivada parcial de M em relação a y é:

$\begin{array}{l}\frac{\partial M}{\partial y} =\frac{1}{2\sqrt{x^{2} +y^{2}}} \cdotp 2y\ \ \ \ \ \ [ Regra\ da\ Cadeia]\\\\=\frac{y}{\sqrt{x^{2} +y^{2}}}\end{array}$

E a derivada parcial de N em relação a x é:

$\begin{array}{l}\frac{\partial N}{\partial x} =y\left[ y+\frac{1}{x+\sqrt{x^{2} +y^{2}}} \cdotp \left( 1+\frac{1}{\cancel{2}\sqrt{x^{2} +y^{2}}} \cdotp \cancel{2} x\right)\right] \ \ \ \ \ [ Regra\ da\ Cadeia]\\\\\\=y\left[ y+\frac{1}{x+\sqrt{x^{2} +y^{2}}} \cdotp \left(\frac{\sqrt{x^{2} +y^{2}}}{\sqrt{x^{2} +y^2}} +\frac{x}{\sqrt{x^{2} +y^{2}}}\right)\right] \ \ \ \left[ \because 1=\frac{\sqrt{x^{2} +y^{2}}}{\sqrt{x^{2} +y^2}}\right]\\\end{array}$

$\\\\\begin{array}{l}=y\left[ y+\frac{1}{\cancel{x+\sqrt{x^{2} +y^{2}}}} \cdotp \frac{\cancel{x+\sqrt{x^{2} +y^{2}}}}{\sqrt{x^{2} +y^{2}}}\right]\\\\\\=y\left( y+\frac{1}{\sqrt{x^{2} +y^{2}}}\right)\\\\\\=y^{2} +\frac{y}{\sqrt{x^{2} +y^{2}}}\end{array}$

Portanto,

$\begin{array}{l}\displaystyle\oint _{C}( Mdx+Ndy) =\int \int _{R}\left( y^{2} +\frac{y}{\sqrt{x^{2} +y^{2}}} -\frac{y}{\sqrt{x^{2} +y^{2}}}\right) dxdy\\\\\\=\displaystyle\int \int _{R} y^{2} dxdy\end{array}$

Resposta ⇒ C