Then we see that
Completing the square gives us
Now comparing this to the standard form of a circle: , where is the center of the circle and is the radius of the circle, we now see that is centered at and has a radius of one.
Now note that we are not graphing the entire circle. Its just the upper half, since that is hinted by the positive square root. If it was , that would be dealing with the semicircle below the x axis.
Now, if we were to integrate this guy, it would require some extra thinking:
This is the same thing as saying
Now apply the trig substitution
Can you take it from here?