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Triple Integrals

# Definition and Properties of Triple Integrals

### Definition of Triple Integral

We can introduce the triple integral similar to double integral as a limit of a Riemann sum. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$).

Let the set of numbers $$\left\{ {{x_0},{x_1}, \ldots ,{x_m}} \right\}$$ be a partition of $$\left[ {a,b} \right]$$ into small intervals so that the following relations are valid:

${a = {x_0} \lt }\kern0pt{ {x_0} \lt {x_1} \lt {x_2} \lt \ldots} { \lt {x_i} \lt \ldots } {\lt {x_{m – 1}} \lt {x_m} }={ b.}$

Figure 1.

Similarly, we can construct partitions of the segment $$\left[ {c,d} \right]$$ along the $$y$$-axis and the segment $$\left[ {p,q} \right]$$ along the $$z$$-axis:

${c }={ {y_0} \lt {y_1} \lt {y_2} \lt \ldots}\kern0pt { \lt {y_j} \lt \ldots }\kern0pt {\lt {y_{n – 1}} \lt {y_n} }={ d,}$
${p }={ {z_0} \lt {z_1} \lt {z_2} \lt \ldots}\kern0pt { \lt {z_k} \lt \ldots }\kern0pt {\lt {z_{\ell – 1}} \lt {z_\ell} }={ q.}$

The Riemann sum of the function $$f\left( {x,y,z} \right)$$ over the partition of $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ is defined by

${\sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {\sum\limits_{k = 1}^{\ell} {f\left( {{u_i},{v_j},{w_k}} \right) }}\kern0pt{{ \Delta {x_i}\Delta {y_j}\Delta {z_k}} } } }$

Here $${\left( {{u_i},{v_j},{w_k}} \right)}$$ is some point in the rectangular box $$\left( {{x_{i – 1}},{x_i}} \right)$$ $$\times \left( {{y_{j – 1}},{y_j}} \right)$$ $$\times \left( {{z_{k – 1}},{z_k}} \right),$$ and the differences are

${\Delta {x_i} = {x_i} – {x_{i – 1}},\;\;}\kern-0.3pt {\Delta {y_j} = {y_j} – {y_{j – 1}},\;\;}\kern-0.3pt {\Delta {z_k} = {z_k} – {z_{k – 1}}.}$

The triple integral of a function $$f\left( {x,y,z} \right)$$ in the parallelepiped $$\left[ {a,b} \right] \times \left[ {c,d} \right] \times \left[ {p,q} \right]$$ is defined as a limit of the Riemann sum, such that the maximum values of the differences $$\Delta {x_i},$$ $$\Delta {y_j}$$ and $$\Delta {z_k}$$ approach zero:

$\require{AMSmath.js} {\iiint\limits_{\left[ {a,b} \right] \times \left[ {c,d} \right] \times \left[ {p,q} \right]} {f\left( {x,y,z} \right)dV \text{ = }} }\kern0pt {\lim\limits_{\substack{ \text{max}\,\Delta {x_i} \to 0\\ \text{max}\,\Delta {y_j} \to 0\\ \text{max}\,\Delta {z_k} \to 0}} {\sum\limits_{i = 1}^m {\sum\limits_{j = 1}^n {\sum\limits_{k = 1}^\ell} {f\left( {{u_i},{v_j},{w_k}} \right)\cdot }}}\kern0pt{{{ \Delta {x_i}\Delta {y_j}\Delta {z_k}} } } }$

To define the triple integral over a general region $$U,$$ we choose a rectangular box $$\left[ {a,b} \right]$$ $$\times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ containing the given region $$U.$$ Then we introduce the function $$g\left( {x,y,z} \right)$$ such that

$\begin{cases} {g\left( {x,y,z} \right) }={ f\left( {x,y,z} \right),} \text{ if}\;f \in U \\ {g\left( {x,y,z} \right) }={ 0,} \text{ if}\;f \notin U \end{cases}$

Then the triple integral of the function $$f\left( {x,y,z} \right)$$ over a general region $$U$$ is defined as

${\iiint\limits_U {f\left( {x,y,z} \right)dV} } = {\iiint\limits_{\left[ {a,b} \right] \times \left[ {c,d} \right] \times \left[ {p,q} \right]} {g\left( {x,y,z} \right)dV} .}$

### Properties of Triple Integrals

Let $$f\left( {x,y,z} \right)$$ and $$g\left( {x,y,z} \right)$$ be functions which are integrable in the region $$U.$$ Then the following properties are valid:

1. $${\iiint\limits_U {\left[ {f\left( {x,y,z} \right) + g\left( {x,y,z} \right)} \right]dV} }$$ $$= {\iiint\limits_U {f\left( {x,y,z} \right)dV} }$$ $$+{ \iiint\limits_U {g\left( {x,y,z} \right)dV} ;}$$
2. $${\iiint\limits_U {\left[ {f\left( {x,y,z} \right) – g\left( {x,y,z} \right)} \right]dV} }$$ $$= {\iiint\limits_U {f\left( {x,y,z} \right)dV} }$$ $$-{ \iiint\limits_U {g\left( {x,y,z} \right)dV} ;}$$
3. $${\iiint\limits_U {kf\left( {x,y,z} \right)dV} }$$ $$={ k\iiint\limits_U {f\left( {x,y,z} \right)dV},}$$ where $$k$$ is a constant;
4. If $${f\left( {x,y,z} \right)} \le {g\left( {x,y,z} \right)}$$ at any point of the region $$U,$$ then $${\iiint\limits_U {f\left( {x,y,z} \right)dV} }\le{ \iiint\limits_U {g\left( {x,y,z} \right)dV} ;}$$
5. If the region $$U$$ is a union of two non-overlapping regions $${U_1}$$ and $${U_2},$$ then
$${\iiint\limits_U {f\left( {x,y,z} \right)dV} } = {\iiint\limits_{{U_1}} {f\left( {x,y,z} \right)dV} } + {\iiint\limits_{{U_2}} {f\left( {x,y,z} \right)dV} ;}$$
6. Let $$m$$ be the minimum and $$M$$ be the maximum value of a continuous function $$f\left( {x,y,z} \right)$$ in the region $$U.$$ Then the following estimate is valid for the triple integral:
$${m \cdot V }\le {\iiint\limits_U {f\left( {x,y,z} \right)dV} }\le{ M \cdot V,}$$
where $$V$$ is the volume of the integration region $$U.$$
7. The Mean Value Theorem for Triple Integrals
If a function $$f\left( {x,y,z} \right)$$ is continuous in the region $$U,$$ then there exists a point $${M_0} \in U$$ such that
$${\iiint\limits_U {f\left( {x,y,z} \right)dV} }={ f\left( {{M_0}} \right) \cdot V,}$$
where $$V$$ is the volume of the region $$U.$$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Evaluate the maximum value of the triple integral

${ \iiint\limits_U {\frac{{dxdydz}}{{\sqrt {100 – {x^2} – {y^2} – {z^2}} }}} ,}$

where $$U$$ is the ball with the radius $$R = 6$$ centered at the origin.

### ✓Example 2

Evaluate the maximum and minimum values of the triple integral

$\iiint\limits_U {\frac{{dV}}{{\ln \left( {e + x + y + z} \right)}}} ,$

where the region $$U$$ is the parallelepiped:

${U }={ \left\{ {\left( {x,y,z} \right)|\;0 \le x \le 1,\;}\right.}\kern0pt{\left.{0 \le y \le 2,\;0 \le z \le 3} \right\}.}$

### Example 1.

Evaluate the maximum value of the triple integral

${ \iiint\limits_U {\frac{{dxdydz}}{{\sqrt {100 – {x^2} – {y^2} – {z^2}} }}} ,}$

where $$U$$ is the ball with the radius $$R = 6$$ centered at the origin.

#### Solution.

The equation of the ball is given by

${{x^2} + {y^2} + {z^2} }\le{ 36.}$

Using the property $$6,$$ we can write:

$I \le M \cdot V,$

where the volume $$V$$ of the ball is

${V = \frac{4}{3}\pi {R^3} }={ \frac{4}{3}\pi \cdot {6^3} }={ 288\pi .}$

The maximum value $$M$$ of the integrand is

${M = \frac{1}{{\sqrt {100 – 36} }} }={ \frac{1}{8}.}$

From here we can get the maximum value of the triple integral:

${I \le \frac{1}{8} \cdot 288\pi }={ 36\pi .}$

### Example 2.

Evaluate the maximum and minimum values of the triple integral

$\iiint\limits_U {\frac{{dV}}{{\ln \left( {e + x + y + z} \right)}}} ,$

where the region $$U$$ is the parallelepiped:

${U }={ \left\{ {\left( {x,y,z} \right)|\;0 \le x \le 1,\;}\right.}\kern0pt{\left.{0 \le y \le 2,\;0 \le z \le 3} \right\}.}$

#### Solution.

First we calculate the volume of the region of integration $$U:$$
$V = 1 \cdot 2 \cdot 3 = 6.$
The estimate of the integral is defined by the inequality
$m \cdot V \le I \le M \cdot V.$
Here the minimum value $$m$$ of the integrand is
${m }={ \frac{1}{{\ln \left( {e + 1 + 2 + 3} \right)}} } = {\frac{1}{{\ln \left( {e + 6} \right)}}.}$
Accordingly, the maximum value $$M$$ is
$M = \frac{1}{{\ln e}} = 1.$
Thus, the estimate of the integral is
${\frac{6}{{\ln \left( {e + 6} \right)}} \le I }\le{ 6.}$