This is not a technical question at all, but I'm quite confused about what should I use to compute volumes in $\mathbb{R}^3$ with integration.

I've read somewhere that a double integral gets the volume swapping across the $x$ and $y$ axis while a triple integral just integrate the whole thing at once, how accurate is this? Can a volume expressed by a double integral be expressed by a triple integral?, And can a triple always be expressed by a double? This one doesn't seem true, but I don't have a good answer to why.

I also found this comments while reading openstudy.com made by someone named KingGeorge one year ago:

For a double integral you have to integrate some function, for a triple integral, you integrate 1.

Does this mean that using an integral to get a volume always should look like $\iiint dxdydz$ without any function?

Geometrically, there are a few things you can be looking at. One, you're finding a 4-volume. That is, the 4-dimensional equivalent of volume. Two, if the volume in the region you're integrating in has a changing density, you could be finding the total mass.

I'm not sure if I understand correctly, but this means that a triple integral does not compute exactly the volume I want but a 4-D equivalent?.


4 Answers 4


You can use both double and triple integrals when calculating a volume. Let me explain you using an example for calculating an area, same applies to volume.

Say you are looking to find the area under a curve $f(x)>0$ over the domain of integration. You must have learnt that : $$A = \int^a_bf(x)dx$$

What you are doing is basically summing infinitely many stripes of length $f(x)$ and base length $dx$.

Now observe that the following are the same : $$\int^a_b\int_0^{f(x)} dtdx = \int^a_bf(x)dx = A$$

So to calculate the area I can use both single and double integral. The second one is simply a shortcut. The first integral sums infinitely many little square of dimension $dt\times dx$ within the specified bounds for $t$ and $x$.

Similarly to find volumes : $$\int\int\int^{f(x,y)}_0dtdxdy = \int\int f(x,y)dxdy$$

The only difference is that the triple integral is a more basic approach in the sense that you really do it small cube by small cube. The double integral is a shortcut.

  • 1
    $\begingroup$ Thank you so much for your comment - I understood the concept on the first read. Thanks for keeping it precise, succinct, and not assuming that the reader knows higher level concepts $\endgroup$
    – user71207
    Commented Feb 23, 2021 at 6:33

Consider a rectangle with sides $a, b, c$. Its volume is $\int_0^a \int_0^b \int_0^c 1$ or $\int_0^a \int_0^b c$ or $\int_0^a bc$. Same logic can be applied to any other shape. Volume is a single integral of area of cross section or a double integral of height.

Note that this is not the only way to split shapes though. For example, you could find the volume of a sphere by integrating over surfaces of all spheres inside it with the same center.


The volume ${\rm vol}(B)$ of some body $B\subset{\mathbb R}^3$ is by its nature a triple integral: $${\rm vol}(B)=\int\nolimits_B 1\ {\rm d}(x,y,z)\ .$$ Fubini's theorem permits to compute this integral recursively in terms of simple, resp. double, integrals over certain intervals or two-dimensional domains. Depending on the way $B$ is defined, some of the "inner integrals" appearing in this way can be for free, as in the following examples:

When $B$ is a rotational body around the $z$-axis with meridian curve $\rho=\rho(z)$ $\>(a\leq z\leq b)$, i.e., $$B:=\{(x,y,z)\>|\> a\leq z\leq b,\ \sqrt{x^2+y^2}\leq\rho(z)\}\ ,$$ then we get two "inner integrals" for free and obtain $${\rm vol}(B)=\pi\int_a^b\rho^2(z)\ dz\ .$$

When $B$ is a "cake" of variable height $h(x,y)$ standing on a domain $B'$ in the $(x,y)$-plane then its volume appears as a double integral $${\rm vol}(B)=\int\nolimits_{B'} h(x,y)\ {\rm d}(x,y)\ .$$ Here we have obtained the innermost integral $\int_0^{h(x,y)}\ dz=h(x,y)$ for free.


Both double and triple integrals can be used to calculate volumes of three dimensional objects. For triple integration, you can reduce the triple integral into a double integral by first calculating the Z component (or any component depending on the "type" of object), and then calculating the double integral over the remaining 2D region. Triple integration is used for functions of three variables and double integrals are used for functions of two variables. For a function of 1 variable, you use the methods of calculus 1 to rotate the objects (solids of revolution) about an axes or line.


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