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Calculate the integral of $f(x, y)$ $=$ $3x$ over the region bounded above by $y=x(2-x)$ and below by $x=y(2-y)$.

What I tried to do here was first expand the two equations out to get $y=x^2-x$ and $x=y^2-y$.

I then rearranged the second equation to get $y=1-\sqrt{1-x}$.

But when i tried to graph this region, I ran into a problem as the second equation is invalid for values of $x$ larger than $1$.

But I do feel that the first equation will be the lower bound of the double integral and the second equation will be upper bound of the integral. But because of the above problem, I can not find any intersection points so I am a little confused on how to proceed further.

Any help would be highly appreciated!

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Your problem stems from how you simplified the two equations that define the boundary of the region.

$$\begin{align} y=x(2-x)&\implies y=2x-x^2\color{red}\ne x^2-x \\ x=y(2-y)&\implies x=2y-y^2\color{red}\ne y^2-y \end{align}$$

You were correct that the second original equation can be rearranged as $y=1-\sqrt{1-x}$, so you can set that equal to the first equation and solve the resulting equation to find the points of intersection of the two graphs.

$$2x-x^2=1-\sqrt{1-x}\implies x=0,1$$

Now that you know the points of intersection, you can set up and evaluate the integral, the latter part of which I'll leave for you to do.

$$\int_0^1\int_{1-\sqrt{1-x}}^{2x-x^2}3x\,dydx$$

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