Question

Mude a integral cartesiana para uma integral polar e escolha a alternativa que contenha o resultado do cálculo da integral dupla:

Answer 1

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\!\!\int_{R(x,y)}\!\!\!\!\!\!\!\!\! f(x,y)\,dx\,dy=\int\!\!\!\int_{R(r,\theta)}\!\!\!\!\!\!\!\!\!f(rcos(\theta),rsen(\theta)) r\,dr\,d\theta\end{array}}$

$\large\boxed{\begin{array}{l}\sf -1\leqslant x\leqslant 1\\\sf-\sqrt{1-x^2}\leqslant y\leqslant\sqrt{1-x^2}\\\sf y=\sqrt{1-x^2}\\\sf y^2=1-x^2\\\rm x^2+y^2=1\\\sf r^2=1\\\sf r=1\\\sf trata-se\,de\,um\,c\acute irculo\,de\,centro\,na\,origem\\\rm e\,raio\,1\\\sf portanto\,0\leqslant r\leqslant1\\\sf sabe-se\,que\, x=rcos(\theta)\\\sf se~r=0\implies x=0\\\sf se~r=1\implies x=cos(\theta)\\\sf mas~-1\leqslant x\leqslant 1\\\sf -1\leqslant cos(\theta)\leqslant1\end{array}}$

$\large\boxed{\begin{array}{l}\sf como\,o\,c\acute irculo\,inteiro\,mede\,360^\circ\\\sf podemos\,dizer\,que\\\sf0\leqslant \theta\leqslant 2\pi\\\sf note\,tamb\acute em\,que\,o\,integrando\,2\dfrac{1}{(1+x^2+y^2)^2}\,dy\,dx\\\\\sf torna-se\,2~\dfrac{1}{(1+r^2)^2} r\,dr\,d\theta.\\\sf portanto\\\\\displaystyle\sf\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1+x^2}}2\dfrac{1}{(1+x^2+y^2)^2}dydx=\int_{0}^{2\pi}\!\!\!\int_{0}^1 2~\dfrac{1}{(1+r^2)^2}r\,dr\,d\theta\end{array}}$