Your two curves only intersect once, at $(0,1)$. The "root" at $x=-3$ (which should be $3$) is extraneous, caused by squaring both sides. This means there is no region bounded by the two curves. If there were, the approach would be correct. For example, take $y=1-x$ and $y=1-x^2$. These intersect at $(0,1)$ and $(1,0)$ and the area between them would be $\int_0^1(x^2-x)\;dx$
Added: No, if you plot the two curves I gave there is a lens-shaped area between the two curves. If you want the area between two curves you do $\int y_2-y_1 \;dx$. The area between a curve and the x-axis is the special case $y_2=0$
In your original problem it does make sense to talk of the area bounded by the $x$ axis and the two curves. You could see this plot. $\sqrt{1+x}$ rises above the axis at $x=-1$. The two curves intersect (as you found) at $(0,1)$, then $1-x$ hits the axis at $x=1$. So your integral would be $area=\int_{-1}^0 \sqrt{1+x} \;dx + \int_0^1(1-x)\;dx$