Double Integration For Bounded Region For the following question:
$\displaystyle \int\!\!\!\int_D x^2y\,dx\,dy$ where $\displaystyle D$ is the bounded region $\displaystyle x= y^2$ & $\displaystyle y = \frac{1}{2}x$
I get the limits of integration to be:
$$\int_0^4\!\!\!\int_{2y}^{y^2} x^2y\,dx\,dy$$
I got the limts for y by substituting $\displaystyle y = \frac{1}{2}x$ in to $x= y^2$
Is this the right approach (probably not)?
I'm sure my graph is ok, brain freeze from here though.
 A: The points of intersection are $(x,y)=(0,0)$ and $(x,y)=(4,2)$.  First you have
$$
\int_\text{?}^\text{?}\cdots\cdots\cdots dy.
$$
The variable $y$ goes from $0$ to $2$.
Then you have
$$
\int_0^2\left(\int_\text{?}^\text{?}\cdots\cdots\,dx\right) \, dy.
$$
For any particular value of $y$, the other variable $x$ goes from $y^2$ up to $2y$.
Notice that $y^2$ is less than $2y$ if $y$ is between $0$ and $2$.
So you've got
$$
\int_0^2\left(\int_{y^2}^{2y}\cdots\cdots\,dx\right) \, dy.
$$
A: For double integration problems, it's always beneficial to draw a diagram of the region of integration when you can:
 
The limits on the inner integral (it's a $dx$ integral, so horizontal in nature) correspond to integrating over the blue line segment in the diagram for a fixed $y$ value (you were almost correct here, but got the order wrong, integrate from the leftmost point ($x=y^2$) to the rightmost point ($x=2y$) ).  The limits on the outer integral correspond to the $y$-range the horizontal line segments "sweep through" (from $y=0$ to $y=2$, here).
