# 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

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