Double integrals over general regions Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it's easier. $$\int\int_{D}{ydA}, \text{$D$ is bounded by $y=x-2, x=y^2$}$$
I'm having trouble with setting up the integral for both type I and type 2 and choosing which one would be easier. I'm sure I'll be able to integrate it with no problem, but setting them up is where I would need some help. 
 A: If you draw region $D$ you will have this:

To find cutoffs you solve the system
$x=y+2$ and $x=y^2$
so
$y^2-y-2=0$
then the points are $(1,-1)$ and $(4,2)$.
Now, to setting up the integral over type $1$ region we need to partition $D$ in two subregions, so
$$\int {\int_D {ydA = \int_0^1 {\int_{ - \sqrt x }^{\sqrt x } {ydydx + } } } } \int_1^4 {\int_{x - 2}^{\sqrt x } {ydydx} }$$

To setting up the integral over type $2$ region
$$\int_{ - 1}^2 {\int_{{y^2}}^{y + 2} {ydxdy} }$$
It seems that second integral is easier than the first one.
A: Plotting both $y=x-2$ and $x=y^2$ in wolfram alpha:

Based upon the points of intersection, you can see that the region D exists for $(x,y)$ with $x\in[1,4]$ and $y\in[-1,2]$.
A: A type 2 region will work. Observe the graph if done as a type 2:

$$\int_a^b\int_{h_1(y)}^{h_2(y)}y\,dx\,dy$$
First you need to find where the two intersect:
$$y+2=y^2$$
$$y^2-y-2=0$$
$$y=-1;y=2$$
$$(1,-1),(4,2)$$
It is clear that the bounds are as follows:
$$\begin{align}
h_1(y)&=y^2 \\
h_2(y)&=y+2 \\
a&=-1 \\
b&=2
\end{align}$$
A: I found the answer. $\int_{0}^1\int_{-\sqrt{x}}^\sqrt{x}{ydydx}+\int_{1}^4\int_{x-2}^\sqrt{x}{ydydx}=\int_{-1}^2\int_{y^2}^{x+2}{ydxdy}$
It's easier to to integrate $\int_{-1}^2\int_{y^2}^{x+2}{ydxdy}$ and after integrating it, the answer comes out to be $\frac94$
