double integral over the given shaded domain in the Figure Compute the double integral of   f(x, y) = 8 x^3 y  over the given shaded domain in the following Figure :

I am confused about the limits, 
my work so far for the limits is x from 0 to 4 and y from 0 to 2 
 A: The important thing to realise about an iterated integral is that there are implied brackets:
$$\int\!\!\int f(x,y)\,dx\,dy=\int\left(\int f(x,y)\,dx\right)dy\ .$$
To evaluate the integral we have to do the part in brackets first.  This means integrating with respect to $x$, taking $y$ to be a fixed value.  Therefore the limits for $x$ on the inner integral might not be constants, but might be expressions in $y$.  On the other hand, the integral with respect to $y$ is performed after $x$ has been eliminated from the calculation, so the limits for $y$ will be constants.
Some people differ on this, but I recommend establishing the limits from the outside in.  So, start with $y$.  As pointed out already, by the time we come to deal with $y$ we will have removed $x$ from the expression.  Therefore we need the minimum and maximum values of $y$ over the whole region.  I hope it is clear in your problem that this means $y=0$ and $y=2$.  So we have
$$\int_0^2\left(\int_?^? f(x,y)\,dx\right)dy\ .$$
Now the situation regarding $x$ is different.  Remember that we will integrate with respect to $x$, taking $y$ as a constant.  To find the relevant $x$-values, draw on your diagram a line $y=\hbox{constant}$.  (Please do it yourself: I am no good at posting diagrams online.)  That is, a horizontal straight line passing through the shaded region.  The limits for $x$ are the minimum and maximum $x$-values within the shaded region on this line, that is, $x=y$ and $x=4$.  So the integral is
$$\int_0^2\left(\int_y^4 f(x,y)\,dx\right)dy\ .$$
A: Note that $f$ doesn't enter into it.
It's easiest in this case to let $y$ range between $0$ and $2$. Then for a fixed $y$, $x$ ranges between $y$ and $4$.
You could do it the other way around, but you would have to break it into two cases. First let $x$ range between $0$ and $2$; for a fixed $x$ in this range, $y$ ranges between $0$ and $x$. Then let $x$ range between $2$ and $4$; for a fixed $x$ in this range, $y$ ranges between $0$ and $2$.
The first method would give you
$$\int_{y=0}^2 \int_{x=y}^4 8x^3y \,dx\, dy$$
The second method would give you
$$\int_{x=0}^2 \int_{y=0}^x 8x^3y \,dy\, dx + \int_{x=2}^4 \int_{y=0}^2 8x^3y \,dy\, dx$$
They should be equal, but the first looks like less work.
