# Thread: How should i approach this integral question?

1. ## How should i approach this integral question?

Hmm I'm having trouble with that Square root (2x - x^2) , how should i deal with that?

Thanks

2. 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 $\sqrt{2x-x^2}$, you want to graph those. Of course, y= x is the straight line from (0,0) to (1,1).

It's $y= \sqrt{2x- x^2}$ that is the problem? Square both sides to get $y^2= 2x- x^2$ so $x^2- 2x+ y^2= 0$. Now complete the square in x by adding 1 to both sides: $x^2- 2x+ 1+ y^2= 1$ but $x^2- 2x+ 1= (x-1)^2$ so this is $(x-1)^2+ y^2= 1$. 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 $(x-1)^2+ y^2= 1$ on the left up to x= y on the right.

Solve $(x-1)^2+ y^2= 1$ 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.