etc.
Sorry, but this working does not make a lot of sense to me. I expected to see an somewhere .... I also suggest you put brackets around negative numbers when squaring them (-3^2 is likely to be confused with -9 ....) Note also that the arbitrary constant is not required when finding definite integrals.
mr fantastic is answering the question you asked but it is the wrong question! You appear to have completely misunderstood this problem.
You titled this thread "Integral by interpreting it in terms of area" and you have made no attempt to that. The point of this problem is NOT to do any integration at all!
If , then , or . That's a circle with center at (0, 1) and radius 3. Because the original equation used only the positive square root, so we really have the upper half disk. Finally, because the integral is from x=-3 to 0, while the entire circle goes from x=-3 to x= +3, this integral is really the area of a quarter circle plus the rectangular section between that and the x-axis.
What is the area of a circle of radius 3? What is the area of a quarter of that circle? What is the area of a rectangle of length 3 and height 1? What is the sum of those two areas?