Question

Решите задачу 5,7 пожалуйста

Answer 1

$1)\; \; D:\; \left \{ {{x^2=}2-y \atop {x+y=0}} \right.\; \; \left \{ {{y=2-x^2} \atop {y=-x}} \right.\; \; \to \; \; 2-x^2=-x\; ,\; \; x^2-x-2=0\; ,\\\\x_1=-1\; ,\; x_2=2\; \; ,\; \; y(-1)=1\; ,\; y(2)=-2\\\\y=2-x^2\; \; \to \; \; x^2=2-y\; \; ,\; \; x=\pm \sqrt{2-y}\\\\x+y=0\; \; \to \; \; x=-y\\\\\iint \limits _{D}\, f(x,y)\, dx\, dy=\int\limits^2_{-1}\, dx\int\limits^{2-x^2}_{-x}\, f(x,y)\, dy=\\\\\\=\int\limits^{1}_{-2}\, dy\int\limits^{\sqrt{2-y}}_{-y}\, f(x,y)\, dx+\int\limits^2_1\, dy \int\limits^{\sqrt{2-y}}_{-\sqrt{2-y}}\, dx$

$2)\; \; D:\; \left \{ {{y=1+x^2} \atop {y=5\; ,\; x\geq 0}} \right.\\\\1+x^2=5\; \; \to \; \; x^2=4\; \; ,\; \; x=\pm 2\\\\\iint \limits _{D}3x\, dx\, dy=\int\limits^2_1\, 3x\, dx\int\limits^5_{1+x^2}\, dy=\int\limits^2_1\, 3x(5-1-x^2)dx=\int\limits^2_1\, (12x-3x^3)\, dx=\\\\=(6x^2-\frac{3x^4}{4})\Big |_1^2=24-12-(6-\frac{3}{4})=6+\frac{3}{4}=\frac{27}{4}$