The first step in any problem like this is to draw a picture. Then you can find the limits a bit easier... Here's a plot: http://www.wolframalpha.com/input/?i=plot{y%3Dx%2B2%2C+x%3D4-y^2}
(Note: You rewrote $f(y)$ incorrectly... it should change into $f(x) = x + 2$)
This problem is really just trying to emphasize how integrating w.r.t. y or x can make a problem a lot harder or easier.
Basically, when integrating w.r.t. x, you're going to have to split the region in two: Region one ($R_1$) is from x=-5 to x=0, and region two ($R_2$) from x=0 to x=4.
This is because your "top" curve changes at x=0. On $R_1$, your area is bounded above by y=x+2, below by $y=-\sqrt{4-x}$. On $R_2$, you are bounded above by $y=\sqrt{4-x}$, below by $y=-\sqrt{4-x}$.
Thus, your area is:
$\int_{-5}^0 [(x+2)-(-\sqrt{4-x})]dx + \int_0^4[(\sqrt{4-x})-(-\sqrt{4-x})]dx$
Simplifies to:
$\int_{-5}^0 [(x+2)+\sqrt{4-x})]dx + \int_0^42\sqrt{4-x}dx$
EDIT: To determine analytically, without graph.
It is possible, but I wouldn't recommend it, as it could (and typically does) involve more work. There is also more of a chance of making a mistake.
First, find all intersections of the functions. In this case, there are three functions: $f_1(x)=x+2$, $f_2(x)=\sqrt{4-x}$, $f_3(x)=-\sqrt{4-x}$. Now find the intersections of these graphs, which are: $(-5, -3),(0, 2),(4, 0)$ This splits the x axis into five regions: $(-\infty, -3), (-3, 0), (0, 4), (4, \infty)$ (Note: the above are not points, but rather ranges of x. I have made the endpoints excluded really because it doesn't matter so much...)
Now, determine the top and bottom curves for each region. This means you must determine, for example, the function with the greatest y value for the region $(-\infty, -3)$. Do this for all ranges. This basically involves a lot of manipulating inequalities.
Now you know your upper and lower limits for each of those regions. Now you need to know what regions you need to find the area of. Do this by a similar process as above for the y-axis.
Now you know your limits of integration as well as your integrands.
I think you will agree that drawing a graph is a better approach... :P