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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!

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If $D$ is the region, then computing the double integral over $D$ of the constant function $1$ will give you a volume, but the volume will be the same as the original area, because the volume of cylinder is the area of the base times the height. If the height is $1$, then... – Arturo Magidin Jun 4 '12 at 19:22
up vote 3 down vote accepted

$\int$ 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.

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Thank you for your reply! You've said "If the function $f(x_i,y_i)=1$", but when that would be the case? – Chris Jun 4 '12 at 19:43
sorry, that should be "if the function $f(x,y)=1$". It's meant to be the same as the condition you set in your problem (when $f=1$). – Robert Mastragostino Jun 4 '12 at 19:53
Yes, that wasn't the problem that bugged me, but I got it now, so, thank you! – Chris Jun 4 '12 at 20:06

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.

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So if $f(x) = 1$, we are in the case that the area is the area of a rectangle or the volume is the volume of a parallelepiped, right? – Chris Jun 4 '12 at 19:49
@Chris That depends on the region of integration - in general it's an extruded prism but not necessarily a rectangle/parallelipiped; for instance, if your region of integration is a circle, then you're looking at the volume of a cylinder of height 1 constructed on that circle. – Steven Stadnicki Jun 4 '12 at 19:55
Yes, thank you. Could you explain me the term "extruded prism" as I am not familiar with the term? – Chris Jun 4 '12 at 20:05
@Chris I think a photo will be worth a thousand words here. Have a look at : this is a prism extruded from a decagon. – Steven Stadnicki Jun 4 '12 at 20:20
OK, that's what I thought, but wasn't 100% sure, thank you! :) – Chris Jun 4 '12 at 20:25

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.

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Thank you, that's good to remember, too. – Chris Jun 4 '12 at 19:45

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