Why do I need two integrals to invert the integration order of $\int_{-4}^{3} \int_{x^2-9}^{-x+3} 3dydx$? 

(the first pic read "Domain of integration" and the second one reads "Therefore, the inversion of the order of integration gives")
Why do I need 2 integrals?
Also what I normally do in these kinds of exercises is that I try to make y get all the constants because it's the outer integral. So I get $-9 \le y \le 7$. x is trickier because I have $x^2-9 \le y \le -x+3$.
If I isolate x on each side separately I get $x \le \pm \sqrt{y+9}$ and $x \le 3-y$ which doesn't make much sense. Is this supposed to happen?
 A: The original double integral corresponds to dividing the region into vertical strips. When you do this, the strip at a point $x$ extends from $y=x^2-9$ at the bottom to $y=-x+3$ at the top; that is why the limits on the inner integral, the one with respect to $y$, are $x^2-9$ and $-x+3$. The vertical strips themselves run from $x=-4$ on the left to $x=3$ on the right; this explains the limits on the outer integral, the one with respect to $x$.
Reversing the order of integration corresponds to dividing the region into horizontal strips. Unfortunately, the ends of the horizontal strips are determined differently for strips above the $x$-axis and strips below the $x$-axis. A horizontal strip above the $x$-axis runs from $x=-\sqrt{y+9}$ on the left to $x=-y+3$ on the right. These limits are valid, however, only for strips on or above the $x$-axis, i.e., when $0\le y\le 7$. We can use them to set up a double integral for the area of the part of the region lying above the $x$-axis: the strips run from $y=0$ to $y=7$, so the outer integral, the one with respect to $y$, will have $0$ and $7$ as its limits, and at each of these values of $y$ the strip will run from $-\sqrt{y+9}$ on the left to $-y+3$ on the right, and the inner integral, the one with respect to $x$ will have the corresponding limits:
$$\int_0^7\int_{-\sqrt{y+9}}^{-y+3}3\,dxdy\;.\tag{1}$$
Below the $x$-axis the horizontal strip at $y$ still starts at $-\sqrt{y+9}$ on the left, but it now runs to $\sqrt{y+9}$ on the right, so the inner integral must now be $\int_{-\sqrt{y+9}}^{\sqrt{y+9}}3\,dx$. The values of $y$ for which this is the case run from $-9$ at the bottom up to $0$, so the area of the part of the region below the $x$-axis is given by the double integral
$$\int_{-9}^0\int_{-\sqrt{y+9}}^{\sqrt{y+9}}3\,dxdy\;,\tag{2}$$
and the area of the whole region is the sum of $(1)$ and $(2)$.
This problem is a good illustration of why it generally pays to look at the region before deciding which way to slice it into strips, i.e., whether to integrate first with respect to $y$ or first with respect to $x$.
A: Your domain is given by $-4\leq x\leq 3, x^2-9\leq y -x+3$. If you want to change your integral you have to express your integral as $a\leq y\leq b, h(y)\leq x\leq g(y)$. As you said, $a=-9, b=7$. What about h(y) and g(y)? you look at your region of integration and for each $y$, you see the bounds for $x$. You can see that the left bound would be
$h(y)=-\sqrt{y+9}$, but what about the right bound? you see that the bound on the right consists of a straight line and a parabolla, namely, you have $$g(y)=\left\{\begin{array}{cc}
-y+3& \hbox{if }0\leq y\leq 7\\
\sqrt{y+9}&\hbox{if }-9\leq y\leq 0
\end{array}\right.$$
So, you can see it as a really just one integral, but when you compute it you need two. 
$$\int_{-9}^7\int_{h(y)}^{g(y)}3dxdy=\int_{0}^7\int_{-\sqrt{y+9}}^{-y+3}3dxdy+\int_{-9}^0\int_{-\sqrt{y+9}}^{\sqrt{y+9}}3dxdy$$
About your method
Your method also works, but you need to be careful, with inequalities. You have $x^2-9\leq y\leq -x+3$. 
The right inequality $y\leq -x+3$, is the same as $x\leq -y+3$.
Now, your left inequality $x^2-9\leq y$, which is equivalent to $x^2\leq y+9$, is solved when $-\sqrt{y+9}\leq x\leq \sqrt{y+9}$.
SO, YOU HAVE TWO CONSTRAINTS:
$x\leq -y+3$ AND $-\sqrt{y+9}\leq x\leq \sqrt{y+9}$.
This can be expressed as 
$$-\sqrt{y+9}\leq x\leq \min(-y+3, \sqrt{y+9})$$
Now, 
$$\min(-y+3, \sqrt{y+9})=\left\{\begin{array}{cc}
-y+3& \hbox{if }0\leq y\leq 7\\
\sqrt{y+9}&\hbox{if }-9\leq y\leq 0
\end{array}\right.$$
And you get the bounds of your integral.
A: When you're inverting the interal you're actually integrating first w.r.t. $dx$. Notice that in that case if you don't split the region into two parts , in your case , the lower half of $X$ axis and upper half of $X$ axis then while figuring out the upper and lower limit of $x$ you won't get the limits because the region is not closed with a single curve and symmetric, there's a straight line involved.
