am I on the right track here?
Yes. I would write:
$\displaystyle A=\int_1^9 u^{\frac{1}{2}}\,du=\frac{2}{3}\left[u^{\frac{3}{2}} \right]_1^9=\frac{2}{3}(27-1)=\frac{52}{3}$
edit: Don't forget the differential in your definite integral and avoid using a rounded decimal approximation (at least if you have the same professor I did).
That's fine. You can check this sort of thing here integral (-2,6) (x+3)^(1/2) dx - Wolfram|Alpha
I see. So we assume u = (x+3) and go from there. Looks more professional to me. Thanks! On thing I'm not sure about though is, how did you get the 9 and 1?
EDIT: Oh nevermind. I get it now.
Oh that tool is amazing! So I take it the graph shown there is the sketch of the region.