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:

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

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 \(U.\)

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Solved Problems

Click a problem to see the solution.

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:

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

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

\[{\frac{6}{{\ln \left( {e + 6} \right)}} \le I }\le{ 6.}\]