Question

Alguem pode resolver essa integral dupla p mim pf ?

Answer 1

$I=\displaystyle\int_1^4 \int_{y^2}^y \sqrt{\dfrac{y}{x}}\,dx\,dy$


Observemos a região de integração:

$\left\{ \!\begin{array}{l} 1\le y \le 4\\\\ y\le x\le y^2 \end{array} \right.$


Como $x$ e $y$ são sempre positivos na região de integração, podemos desmembrar a raiz do quociente como o quociente das raízes:

$I=\displaystyle\int_1^4 \int_{y^2}^y \dfrac{\sqrt{y}}{\sqrt{x}}\,dx\,dy\\\\\\ =\int_1^4 \int_{y^2}^y \sqrt{y}\cdot \dfrac{1}{\sqrt{x}}\,dx\,dy\\\\\\ =\int_1^4 \int_{y^2}^y \sqrt{y}\cdot x^{-1/2}\,dx\,dy\\\\\\ =\int_1^4 \sqrt{y}\cdot \left.\left(\dfrac{x^{(-1/2)+1}}{-\,\frac{1}{2}+1} \right )\right|_{y^2}^y\,dy\\\\\\ =\int_1^4 \sqrt{y}\cdot \left.\left(\dfrac{x^{1/2}}{\frac{1}{2}} \right )\right|_{y^2}^y\,dy\\\\\\ =\int_1^4 \sqrt{y}\cdot \left.\big(2x^{1/2}\big)\right|_{y^2}^y\,dy\\\\\\ =\int_1^4 \sqrt{y}\cdot \left.\big(2\sqrt{x}\big)\right|_{y^2}^y\,dy$

$=\displaystyle\int_1^4 \sqrt{y}\cdot \big(2\sqrt{y}-2\sqrt{y^2}\big)\,dy\\\\\\ =\int_1^4 \sqrt{y}\cdot \big(2\sqrt{y}-2\left|y\right|\big)\,dy$


Como $y$ é sempre positivo no intervalo de integração, podemos dispensar o módulo:

$=\displaystyle\int_1^4 \sqrt{y}\cdot \big(2\sqrt{y}-2y\big)\,dy\\\\\\ =\int_1^4 \big(2\sqrt{y}^2-2y\sqrt{y}\big)\,dy\\\\\\ =\int_1^4 \big(2y-2y\sqrt{y}\big)\,dy\\\\\\ =\int_1^4 \big(2y-2y^{3/2}\big)\,dy\\\\\\ =\left.\left(y^2-2\cdot \dfrac{y^{(3/2)+1}}{\frac{3}{2}+1} \right )\right|_1^4\\\\\\ =\left.\left(y^2-2\cdot \dfrac{y^{5/2}}{\frac{5}{2}} \right )\right|_1^4\\\\\\ =\left.\left(y^2-2\cdot\dfrac{2}{5}\,y^{5/2} \right )\right|_1^4$

$=\left.\left(y^2-\dfrac{4}{5}\,y^{5/2} \right )\right|_1^4\\\\\\ =\left(4^2-\dfrac{4}{5}\cdot 4^{5/2} \right )-\left(1^2-\dfrac{4}{5}\cdot 1^{5/2} \right )\\\\\\ =\left(16-\dfrac{4}{5}\cdot 32 \right )-\left(1-\dfrac{4}{5}\cdot 1 \right )\\\\\\ =\left(16-\dfrac{128}{5} \right )-\left(1-\dfrac{4}{5} \right )\\\\\\ =\left(\dfrac{80}{5}-\dfrac{128}{5} \right )-\left(\dfrac{5}{5}-\dfrac{4}{5} \right )$

$=\left(\dfrac{80-128}{5} \right )-\left(\dfrac{5-4}{5} \right )\\\\\\ =-\,\dfrac{48}{5}-\dfrac{1}{5}\\\\\\ =-\,\dfrac{49}{5}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int_1^4 \int_{y^2}^y \sqrt{\dfrac{y}{x}}\,dx\,dy=-\dfrac{49}{5} \end{array}}$