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)