Question

Помогите решить двойной интеграл!!!!!!!(очень срочно)

Первое задание:∫∫(x^2y^2+y^2)dxdy ,область D:{1/x≤y≤2/x;x≤y≤3x}

Answer 1

Точки пересечения:

$D:\; \; \left \{ {{\frac{1}{x}\leq y\leq 2} \atop {x\leq y\leq 3x}} \right.\\\\\left \{ {{y=3x} \atop {y=1/x}} \right.\; \; \to \; \; 3x=\frac{1}{x}\; ,\; \; \frac{3x^2-1}{x}=0\; ,\; \; 3x^2=1\; ,\; x=\pm \frac{1}{\sqrt3}\\\\\left \{ {{y=3x} \atop {y=2/x}} \right.\; \; \to \; \; 3x=\frac{2}{x}\; ,\; \; \frac{3x^2-2}{x}=0\; ,\; \; 3x^2=2\; ,\; \; x=\pm \sqrt{\frac{2}{3}}\\\\\left \{ {{y=x} \atop {y=1/x}} \right.\; \; \to \; \; x=\frac{1}{x}\; ,\; \; \frac{x^2-1}{x}=0\; ,\; \; x^2=1\; ,\; \; x=\pm 1$

$\left \{ {{y=x} \atop {y=2/x}} \right. \; ,\; \; x=\frac{2}{x}\; ,\; \; \frac{x^2-2}{x}=0\; ,\; \; x^2=2\; ,\; \; x=\pm \sqrt2\\\\\\\iint \limits _{D}\, (x^2y^2+y^2)dx\, dy=\int\limits_{1/\sqrt3}^{\sqrt{2/3}}\, dx\int \limits _{1/x}^{3x}(x^2y^2+y^2)\, dy+\\\\+\int\limits_{\sqrt{2/3}}^1\, dx\int \limits _{1/x}^{2/x}\, (x^2y^2+y^2)\, dy+\int\limits^{\sqrt2}_1\, dx\int \limits _{x}^{2/x}\, (x^2y^2+y^2)dx\, dy=\\\\=\int\limits^{\sqrt{2/3}}_{1|\sqrt3}\, ((x^2+1)\cdot \frac{y^3}{3})\Big |_{1|x}^{3x}\, dx+\int\limits^1_{\sqrt{2/3}}\, ((x^2+1)\cdot \frac{y^3}{3})\Big |_{1|x}^{2|x}\, dx+\\\\+\int\limits_1^{\sqrt2}\, ((x^2+1)\cdot \frac{y^3}{3})\Big |_{x}^{2|x}\, dx=$

$=\frac{1}{3}\int\limits^{\sqrt{2/3}}_{1/\sqrt3}\, (x^2+1)(27x^3-1)\, dx+\frac{1}{3}\int\limits^1_{\sqrt{2/3}}\, (x^2+1)(\frac{8}{x^3}-\frac{1}{x^3})\, dx+\\\\+\frac{1}{3} \int\limits^{\sqrt2}_1\, (x^2+1)(\frac{8}{x^3}-x^3)\, dx=\\\\=\frac{1}{3}\int\limits^{\sqrt{2/3}}_{1/\sqrt3}\, (27x^5-x^2+27x^3-1)\, dx+\frac{1}{3}\int\limits^1_{\sqrt{2/3}}\, (\frac{7}{x}+\frac{7}{x^3})\, dx+\\\\+\frac{1}{3}\int \limits _{1}^{\sqrt2}\, (\frac{8}{x}-x^5+\frac{8}{x^3}-x^3)\, dx=$

$=\frac{1}{3}\cdot \Big (\frac{9x^6}{2}-\frac{x^3}{3}+\frac{27x^4}{4}-x\Big )\Big |_{1/\sqrt3}^{\sqrt{2/3}}+\frac{7}{3}\cdot \Big (ln|x|-\frac{1}{2x^2}\Big )\Big |_{\sqrt{2/3}}^1+\\\\+\frac{1}{3}\cdot \Big (8ln|x|-\frac{x^6}{6}-\frac{4}{x^2}-\frac{x^4}{4}\Big )\Big |_1^{\sqrt2}=\\\\=\frac{1}{3}\cdot \Big (\frac{4}{3}-\frac{2\sqrt2}{9\sqrt3}+3-\sqrt{\frac{2}{3}}-\frac{1}{6}+\frac{1}{9\sqrt3}-\frac{3}{4}+\frac{1}{\sqrt3}\Big )+\frac{7}{3}\cdot \Big (-\frac{1}{2}-ln\sqrt{\frac{2}{3}}+\frac{3}{4}\Big )+$

$+\frac{1}{3}\cdot \Big (8ln\sqrt2-\frac{8}{6}-2-1+\frac{1}{6}+4+\frac{1}{4}\Big )=\\\\=\frac{1}{3}\cdot \Big (\frac{37}{3}-\frac{11}{9}\cdot \sqrt{\frac{2}{3}}+\frac{10}{9\sqrt3}\Big )+\frac{7}{3}\cdot \Big (\frac{1}{4}-ln\sqrt{\frac{2}{3}}\Big )+\frac{1}{3}\cdot \Big (8ln\sqrt2+\frac{1}{12}\Big )=\\\\=\frac{85}{18}+\frac{10-11\sqrt2}{27\sqrt3}-\frac{7}{6}\cdot ln\frac{2}{3}+\frac{8\sqrt2}{3}$