Answer:
[tex]\frac{1}{3}[/tex]
Step-by-step explanation:
Let's first find the intersection points.
[tex]y = x(2-x)\\x = y(2-y)\\y = y(2-y)(2-y(2-y))\\x = x(2-x)(2-x(2-x))[/tex]
So intersection points for x = 0, 1 and for y = 0, 1.
Now, let's write x = y(2-y) instead of y.
[tex]x - 1 = 2y - y^2 - 1\\x - 1 = -(y-1)^2\\1-x = (y-1)^2\\y = 1 - \sqrt{1-x}[/tex]
[tex]\int_D f(x,y)dA=\int\limits^1_0[x(2-x)-(1-\sqrt{1-x})]dx=\\\\= (x^2-\frac{x^3}{3} -x-\frac{2}{3}(1-x)^{\frac{3}{2} } )|^1_0=-\frac{1}{3} +\frac{2}{3}= \frac{1}{3}[/tex]
