If f is a continuous and positive function, exchange the order of integration appropriately. $$\int_{1}^2\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$$ Note that $x \in [1,2]$ and $y\in [2-x,\sqrt{2x-x^2}]$
if $y = 2-x \implies x=2-y$ so our new lower limit is $ 2-y $, however I don't know how to calculate the new upper limit, can someone help me
Hints that should take you there: First sketch the curves.
So the region is the part above the green line and below the blue line When changing to $dxdy$, we now want to go from left function to right function instead of higher function to lower function, so you need the green line as a function of $x=something in y$ and then the blue line as well. When you go to solve for $x$ in $y=\sqrt{2x-x^2}$ you will have to complete the square, then when you take the square root you will get a $\pm$. Make sure you pick the right positive or negative to get the right half blue curve, not the left.
after that, your y bounds are easy, they go from 0 to 1.