# Does a double integral calculate an area or a volume?

This takes a little explanation. I realize that double integrals can be used to calculate both an area or a volume but should I assume that in the case of calculating the area I am really calculating the volume and multiplying it by a height of 1 which just gives you the area?

Or are they really two different techniques that depend solely on the context of the assigned problem ?

• Does it matter? It sounds like you have a confident understanding of the issues either way. Commented Sep 16, 2017 at 14:10
• You could ask the same question for a single integral: if $f$ is nonnegative, $\int_a^b f(x) \, dx = \iint_{a<x<b,0<y<f(x)} 1 \, dx \, dy$. Commented Sep 16, 2017 at 14:17
• Funny! I asked the very same thing three years ago. You might be interested in reading some of the answers I received back then math.stackexchange.com/questions/649034/…
– Cure
Commented Sep 16, 2017 at 14:24

Intuitively, integration is defined as a suitable limit of a sum. Depending on what we are adding in the sum this limit can be different things.

Also for a single integral of a function $y=f(x)$ we can have an area as in $$A=\int_a^b f(x)dx$$ or arc length as in $$L=\int_a^b \sqrt{1+{f'(x)}^2}dx$$

• Let us say that we are calculating the volume under a given surface , $z=f(x,y)$ .I have read in text books that the volume under this surface is $\iint f(x,y) \, dx \, dy$. However , i am eager to know how this limit comes using the definition of an integration being a limit of a sum. Can you demonstrate by taking a small cube and finding its volume and then summing over these small volumes by taking the limit . Sorry for bothering. Commented Feb 21, 2021 at 18:23

Under normal circumstances, the triple integral evaluates Volume as it spans three dimensions and double integral evaluates area as it spans two dimensions. A case in point is something like this what @chappers had already indicated.

For example, the volume of a solid between two planes could be written as

$\int_{0}^{2\pi}\int_{0}^{r}\int_{z_{bottom}}^{z_{top}} dV = \int_{0}^{2\pi}\int_{0}^{r}\int_{z_{bottom}}^{z_{top}} rdzdrd\theta = \int_{0}^{2\pi}\int_{0}^{r}\left[z_{top} - z_{bottom}\right]rdrd\theta$

The last one is a double integral but the first two are triple integral all measuring the volume.