Why and when in the computation of an area or volume, into the integral we use function f = 1? Sorry if this is something really easy..
As said in the title, why when we want to compute the area or the volume there is something like this:
$$A(D_1) = \iint 1 \ dx\,dy\text{ or }\iiint 1 \, dx \, dy\ ?$$
And when does this occur? 
Is it a special case?
In which, for example, for the area the one side of the parallelogram is "$1$" ? 
Thank you for your time!
 A: $\int$ is essentially a continuous sum. When you do $\int_A f(x,y) dA$, (intuitively) you're evaluating $f(x_i,y_i)dA$ (the function times a little piece of area) and adding those up to get some total. If the function $f(x,y)=1$, then you're just adding up $1\cdot dA=dA$. That is, you're simply adding up the little areas themselves, which when done over the whole surface will give you the entire area. Similar reasoning can be done for any hypervolume.
A: This goes back to the 'grade-school' definition of an integral as the area under a curve.  When you integrate the function $f(x)$ from $a$ to $b$, you're computing the total area beneath the graph of the function over that integral; when $y=f(x)=1$ then this is just $b-a = 1\cdot(b-a)$, the area of the rectangle bounded by the two horizontal lines $y=0, y=1$ and the two vertical lines $x=a, x=b$.  In other words, it's the length - the one-dimensional area - of the region from $a$ to $b$ that you're integrating over.  The 2-dimensional and higher-dimensional cases are just extensions of this; in each you're computing the area of the region of integration as the volume of a region one dimension higher that's a prism of height 1 in the shape of that area.
A: For nn easy case to keep in mind think of $\int_{a}^b dx=b-a$, as $b-a$ is the length of the interval $(a,b)$ which you're integrating over. 
