Hmm I'm having trouble with that Square root (2x - x^2) , how should i deal with that?
Thanks
First graph the region of integration. Since x is between 0 and 1, draw vertical lines at x= 0 and x= 1. Since y is between x and $\displaystyle \sqrt{2x-x^2}$, you want to graph those. Of course, y= x is the straight line from (0,0) to (1,1).
It's $\displaystyle y= \sqrt{2x- x^2}$ that is the problem? Square both sides to get $\displaystyle y^2= 2x- x^2$ so $\displaystyle x^2- 2x+ y^2= 0$. Now complete the square in x by adding 1 to both sides: $\displaystyle x^2- 2x+ 1+ y^2= 1$ but $\displaystyle x^2- 2x+ 1= (x-1)^2$ so this is $\displaystyle (x-1)^2+ y^2= 1$. The graph of that is circle with center at (1, 0) and radius 1. Because y was equal to the positive square root, we have the upper half circle and it should be easy to see that that intersects y= x at (0,0) and (0,1) and is above it.
Reversing the integral, y ranges from 0 to 1 and, for each y, from the left side of the graph $\displaystyle (x-1)^2+ y^2= 1$ on the left up to x= y on the right.
Solve $\displaystyle (x-1)^2+ y^2= 1$ for x. That is a quadratic and will give two solutions. Take the one that is less than 1 for y between 0 and 1.