Question

Calcular as Integrais Duplas Iteradas:

ec0673979b5f6d4a19bca41fc80b3a82.docx

Answer 1

c) $\displaystyle\int\limits_{1}^{e}\int\limits_{0}^{y}{\dfrac{1}{x^{2}+y^{2}}\,dx\,dy}$

$=\displaystyle\int\limits_{1}^{e}\int\limits_{0}^{y}{\left.\dfrac{1}{y}\,\mathrm{arctg\,}\!\!\left(\dfrac{x}{y} \right )\right|_{0}^{y}dy}\\ \\ \\ =\int\limits_{1}^{e}\int\limits_{0}^{y}{\dfrac{1}{y}\cdot \left(\mathrm{arctg\,}\dfrac{y}{y}-\mathrm{arctg\,}\dfrac{0}{y} \right )dy}\\ \\ \\ \int\limits_{1}^{e}{\dfrac{1}{y}\cdot \left(\mathrm{arctg\,}1-\mathrm{arctg\,}0 \right )dy}\\ \\ \\ \int\limits_{1}^{e}{\dfrac{1}{y}\cdot \dfrac{\pi}{4}\,dy}\\ \\ \\ =\dfrac{\pi}{4}\int\limits_{1}^{e}{\dfrac{1}{y}\,dy}$

$=\left.\dfrac{\pi}{4}\cdot (\mathrm{ln\,}y)\right|_{1}^{e}\\ \\ \\ =\dfrac{\pi}{4}\cdot (\mathrm{ln\,}e-\mathrm{\ell n\,1})\\ \\ \\ =\dfrac{\pi}{4}\cdot (1-0)\\ \\ \\ =\dfrac{\pi}{4}$


e) $\displaystyle\int\limits_{1}^{4}\int\limits_{y}^{y^{2}}{\sqrt{\dfrac{y}{x}}\,dx\,dy}$

Como $y$ varia de $1$ a $4,$ temos que $y\geq 0.$ Portanto, podemos separar as raízes quadradas assim:

$=\displaystyle\int\limits_{1}^{4}\int\limits_{y}^{y^{2}}{\dfrac{\sqrt{y}}{\sqrt{x}}\,dx\,dy}\\ \\ \\ =\int\limits_{1}^{4}\int\limits_{y}^{y^{2}}{\sqrt{y}\cdot x^{-1/2}\,dx\,dy}\\ \\ \\ =\int\limits_{1}^{4}{\sqrt{y}\cdot \left.2x^{1/2}\right|_{y}^{y^{2}}\,dy}\\ \\ \\ =2\int\limits_{1}^{4}\int\limits_{y}^{y^{2}}{\sqrt{y}\cdot \left((y^{2})^{1/2}-y^{1/2}\right)\,dx\,dy}\\ \\ \\ =2\int\limits_{1}^{4}{y^{1/2}\cdot \left(y-y^{1/2}\right)\,dy}\\ \\ \\ =2\int\limits_{1}^{4}{\left(y^{3/2}-y\right)\,dy}$

$=2\displaystyle\int\limits_{1}^{4}{\left(y^{3/2}-y\right)\,dy}\\ \\ \\ =2\cdot \left.\left(\dfrac{2}{5}\,y^{5/2}-\dfrac{y^{2}}{2} \right )\right|_{1}^{4}\\ \\ \\ =2\cdot \left[\left(\dfrac{2}{5}\cdot 4^{5/2}-\dfrac{4^{2}}{2} \right )-\left(\dfrac{2}{5}\cdot 1^{5/2}-\dfrac{1^{2}}{2} \right ) \right ]\\ \\ \\ =2\cdot \left[\left(\dfrac{64}{5}-\dfrac{16}{2} \right )-\left(\dfrac{2}{5}-\dfrac{1}{2} \right ) \right ]\\ \\ \\ =2\cdot \left[\left(\dfrac{128}{10}-\dfrac{80}{10} \right )-\left(\dfrac{4}{10}-\dfrac{5}{10} \right ) \right ]$

$=2\cdot \left[\dfrac{48}{10}-\left(-\dfrac{1}{10} \right ) \right ]\\ \\ \\ =2\cdot \left[\dfrac{48}{10}+\dfrac{1}{10} \right ]\\ \\ \\ =2\cdot \dfrac{49}{10}\\ \\ \\ =\dfrac{49}{5}$