Triple Integral

A tripe integral returns the sum of a property per volume under a trivariate function. It is essentially a 1-dimensional upgrade to double integrals, since our domains can now be volumes, instead of an area.

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Definition §

Triple Integral

Let $f(x,y,z)$ be a trivariate function with the domain $D\subseteq {\mathbb{R}}^3$. Then the triple integral of $f$ is:

$${\iiint }_{D}f(x,y,z)\text{d}V={\iiint }_{D}f(x,y,z)\text{d}x\text{d}y\text{d}z$$

• $V$ is the infinitesimal volume of a region ($\text{d}x\times \text{d}y\times \text{d}z$).

In definite form:

$$Q={\iiint }_{D}f(x,y,z)\text{d}V={\int }_e^f{\int }_c^d{\int }_a^{b}f(x,y,z)\text{d}x\text{d}y\text{d}z$$

• $D$ is the domain cuboid, and can be represented as $[a,b]\times [c,d]\times [e,f]$, which is equivalent to $a\le x\le b$ and $c\le y\le d$ and $e\le z\le f$.

• If $f=1$, then the triple integral over a domain $D$ simply returns the volume of the domain.

Cuboidal Domain §

The simplest case of triple integrals is when they are evaluated over a cuboidal domain. That is, the domain can be expressed as the cartesian product of three intervals:

$$C=[a,b]\times [c,d]\times [e,f]$$

where $a,b,c,d,e,f$ are constants.

If we have $(x,y,z)\in C$, then this is equivalent to saying $a\le x\le b$ and $c\le y\le d$ and $e\le z\le f$, or $D=\{(x,y,z)\in {\mathbb{R}}^3:a\le x\le b,c\le y\le d,e\le z\le f\}$

General Domain §

General (non-cuboid) three dimensional volumes are a lot harder to integrate than their 2D counterparts.

A domain $D$ is an elementary region in ${\mathbb{R}}^3$ if any one variable can be bounded by functions of the other two, where the domains of those functions can be split into strips.

A general domain can be either:

• $z$-simple
• $y$-simple
• $x$-simple

Z-Simple §

Used when the domain volume can be split into two functions $f_1(x,y)$ and $f_2(x,y)$ that act as bounds for $z$.

The domain is of the form:

$$D=\{(x,y,z)\in {\mathbb{R}}^3:(x,y)\in {\mathbb{R}}^2,f_1(x,y)\le z\le f_2(x,y)\}$$

So our integral becomes:

$${\iiint }_{D}f(x,y,z)\text{d}V={\iint }_R\left({\int }_{f_1(x,y)}^{f_2(x,y)}f(x,y,z)\text{d}z\right)\text{d}A$$

We can then project the both the bounding functions onto the XY plane, and then evaluate it just as we would a standard double integral over a general domain.

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So, depending on whether the projected $R$ is horizontally simple or vertically simple, we have either:

$${\iiint }_{D}f(x,y,z)\text{d}V=\{\begin{array}{ll}{\int }_a^b{\int }_{l(x)}^{u(x)}{\int }_{f_1(x,y)}^{f_2(x,y)}f(x,y,z)\text{d}z\text{d}y\text{d}x,& R\text{ is vertically simple}\\ {\int }_d^c{\int }_{l(y)}^{r(y)}{\int }_{f_1(x,y)}^{f_2(x,y)}f(x,y,z)\text{d}z\text{d}x\text{d}y,& R\text{ is horizontally simple}\end{array}$$

X-Simple, Y-Simple §

This same process can be extended to be applied to the (rarer) $x$-simple and $y$-simple cases:

#todo

Theorems §

Applications §

See Applications Of Double & Triple Integrals