# Thread: I can't evaluate this integral involving a square root, please help

1. ## I can't evaluate this integral involving a square root, please help

I am trying to solve #33 on the attached pdf and I'm stuck on the part with the sigma notation for the square root part. Someone please help.

Any help would be greatly appreciated!
Thanks in advance!

2. I would quickly lose that sigma notation if I were you. It is not very practical in terms of computing most integrals (in my opinion). You could propose some sort of substitution, but I suggest a simple geometrical argument. http://www1.wolframalpha.com/Calcula...image/gif&s=37
That is the region.

You know that sqrt(9-x^2) is the equation for a circle of radius 3. We are integrating it between -3 and 0, and so we lose half of the circle there. We also can not take negative y values, so we lose another quarter of it. So we are left we one quarter of a circle of radius 3.
The area of that is (1/4)*(pi)*(3^2)=9*pi/4.
We are not done! We have 1+sqrt(9-x^2). As you can see from the picture, this 1+... gives an additional rectangle of height one and width 3 (since we are considering between -3 and 0), and so simply add this area of 3.

This gives the answer as 3+(9*pi/4)