Finding volumes — when to use double integrals and triple integrals?

Question

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?.

Answer 1

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.

Answer 2

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.

Answer 3

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.