Definition and Properties of Triple Integrals

Solved Problems

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:

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

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

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