I need to change integration order of
$\int_{1}^{2}\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$
The region is bounded between $1\leq x\leq2$ and $0\leq y\leq1$
The upper limit, $y=\sqrt{2x-x^2}$ in terms of x is:
$x=1+\sqrt{1-y^2}$ or $x=1-\sqrt{1-y^2}$
How to determine which of them is the correct upper limit, by the graph, or by algebraic way?
It depends of your choice; indeed the outer integral limit has to be always fixed, and for varying the inner limits. If you want, you can integrate with $dy $ being the the outer, this is, $dx dy$ but you need to keep the outer integral from 0 to 1, and the inner function as a function of the outer one, since when you integrate the inner in respect to $x$ in this case, $x$ will vanish and you will be with just a function in terms of $y$.