How to find integral over a parallelogram? For $a>0$ exist $B\subset \mathbb R ^{2}$ Parallelogram with vertices $(0,0),(2a,0),(a,a)$ and $(3a,a)$. Write integral $\int_{B}^{} \! f(x,y) \, d(x,y)  $ on two ways as iterated integral, integrated by x first and integrated by y first. Make sketch.
My idea:
I found linear equations.
$y_{AB}=0$
$y_{AC}=x$
$y_{BD}=x-2a$
$y_{CD}=a$ 
I made simply sketch of this parallelogram and got that: $0\leq x\leq 3a$
So I have 3 boundaries: $0\leq x\leq a$,$a\leq x\leq 2a$ and $2a\leq x\leq 3a$.
Is this correct and if yes, how to put this in integral? 
 A: If you integrate with respect to $y$ first you have to split up the integral into the three parts you indicated:
$$\eqalign{\int_Bf(x,y)\>{\rm d}(x,y)=\int_0^a\int_0^x f(x,y)\>dy\>dx+\int_a^{2a}\int_0^a &f(x,y)\>dy\>dx\cr
&+\int_{2a}^{3a}\int_{x-2a}^a f(x,y)\>dy\>dx\ .\cr}$$
If, however, you integrate with respect to $x$ first you do not have to split up the integral:
$$\int_Bf(x,y)\>{\rm d}(x,y)=\int_0^a\int_y^{y+2a} f(x,y)\>dx\>dy\ .$$
If you don't understand these formulas go back to your book where double integrals are first introduced: If the inner integral is with respect to $y$ the outer integral is from the minimal occurring $x$ ($x_{\min}=0$ here) to the maximal occurring $x$ ($x_{\max}=3a$ here). For each $x$ between $x_{\min}$ and $x_{\max}$ we have to extract from the figure  the minimal and the maximal occurring $y$ for said $x$. In the example you provided there are three different "laws" for these minimal and maximal $y$'s. Therefore we have to split up the total integral into three parts.
