Changing order of integration for $\int_{1}^{2}\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$

Question

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?

the upper bound is $y^2=2x-x^2\iff (x-1)^2+y^2=1$. Try to draw the bounds to better know all. — Tito Eliatron, Feb 03 '21 at 10:11
I did, but i still confused, the upper limit in terms of x depend on the bounds of y values? — yanphili, Feb 03 '21 at 10:17
$x=1+\sqrt{1-y^2}$ means the right semicircle, while $x=1-\sqrt{1-y^2}$ means the left one. — Tito Eliatron, Feb 03 '21 at 10:26

score 0 · Answer 1 · answered Feb 03 '21 at 10:15

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 $y$.

Changing order of integration for $\int_{1}^{2}\int_{2-x}^{\sqrt{2x-x^2}}f(x,y)dydx$

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