About the order of integration for double integrals I have to compute $\int\int (2x-y) \,dx \, dy $  
on the domain $\{ (x,y) \in R^2 : 1\leq x\leq 4, 0\leq y\leq \sqrt{x} \}$
So mi first try is to do: $\int_0^{\sqrt{x}}\int_{1}^{4} (2x-y)\, dx \, dy = 15 \sqrt{x}-\frac{3}{2}x$
But if i do this like: $\int_1^4 \int_0^{\sqrt{x}} (2x-y) \,dy\, dx = \frac{421}{20}$
In the first case the result still depends on $x$, but in the second case i get a value.
I know i did not changed the order of integration, but i just intepreted in diferent ways what my homework ask.
Im a bit confused here, my question is: what is the correct order of integration?
 A: Only the second integral that you have given is correct. To see this, you need to really think about what you are doing here. The region of integration is the area under the curve $y = \sqrt{x}$ from $x=1$ to $x=4$ (sketch a graph to make this clear). So for each value of $x$ from $1$ to $4$, you need to integrate from $y=0$ to $y=\sqrt{x}$. This corresponds to the integral you have written.
The first integral really doesn't address the problem. As you note, it results in a function of $x$ --- your range of integration is not any geometric area, and you are not computing what you might think you are.
A: You can interpret the double integral as the volume of the solid above the given domain, with height $z$ at point $(x,y)$. where $z=2x-y$.  Therefore the answer must be a specific number, and your first answer is definitely wrong.
To reverse the order of integration and get
$$I=\int_?^?\int_?^? 2x-y\,dx\,dy$$
you should begin by drawing a diagram of the domain.  Please do this for yourself - I am no good at posting diagrams.  To get the limits for the outer integral we need the limits over the whole region for $y$.  From the diagram, $y$ goes from $0$ to $2$, so at this stage we have
$$\int_0^2\int_?^? 2x-y\,dx\,dy\ .$$
Now we need the limits for $x$ on the inner integral.  To do this you need to look ahead: when we come to evaluate the integral with respect to $x$, we will be taking $y$ as a constant.  So we need the $x$ values not over the whole region but for a specific value of $y$.  To visualise this, draw on the diagram a line $y=\hbox{constant}$: what are the $x$ values within the region along this line?
At this point you will see from the diagram that there is a difficulty: the answer to the question I have just asked is different, depending on the value of $y$.  If $1\le y\le2$ then the $x$ values go from the curve $y=\sqrt x$ to the right hand edge of the domain, but if $0\le y\le1$ then the $x$ values go from the left hand edge to the right hand edge.  (Please draw an example of each case on your diagram.)
We therefore have to split up the integral into two parts:
$$\int_0^1\int_?^? 2x-y\,dx\,dy+\int_1^2\int_?^? 2x-y\,dx\,dy\ .$$
From what we have said above, $x$ in the second case goes from $x=y^2$ to $x=4$, and in the first case from $x=1$ to $x=4$.  So we have
$$\int_0^1\int_1^4 2x-y\,dx\,dy+\int_1^2\int_{y^2}^4 2x-y\,dx\,dy$$
and now you can evaluate this.
