You could do this:

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 semicirclebelowthe 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?

--Chris