Chapter 6.2, Problem 37E

To determine

To calculate: The volume of the described solid (tetrahedron).

Answer to Problem 37E

The volume of the tetrahedron is $\underset{\_}{10{\text{cm}}^3}$.

Explanation of Solution

Given information:

The tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges.

The length of the perpendicular edges are 3 cm, 4 cm, and 5 cm.

Consider that the cross-section at height z is a triangle similar to the triangle at its base.

The dimensions of the tetrahedron as shown in Figure 1.

Refer to Figure 1.

The proportionality factor for the cross section as follows:

$y=\frac{5-z}{5}$

The region lies between $a=0$ and $b=5$.

Calculation:

The expression to find the volume of the tetrahedron as shown below.

$V={\displaystyle {\int }_a^{b}A\left(z\right)dz}$ (1)

Find the area of the tetrahedron as shown below.

$\begin{array}{c}A\left(z\right)=\frac{1}{2}bh\\ =\frac{1}{2}\times \left(3\times \frac{5-z}{5}\right)\times \left(4\times \frac{5-z}{5}\right)\\ =6{\left(\frac{5-z}{5}\right)}^2\\ =6{\left(1-\frac{z}{5}\right)}^2\end{array}$

Substitute 0 for a, 5 for b, and $6{\left(1-\frac{z}{5}\right)}^2$ for $A\left(z\right)$ in Equation (1).

$V={\displaystyle {\int }_0^{5}6{\left(1-\frac{z}{5}\right)}^{2}dz}$ (2)

Consider $u=1-\frac{z}{5}$ (3)

Differentiate both sides of the Equation (3).

$\begin{array}{l}du=-\frac{1}{5}dz\\ dz=-5du\end{array}$

Find the upper limit value of u using Equation (3).

Substitute 0 for z in Equation (3).

$\begin{array}{c}u=1-\frac{0}{5}\\ =1\end{array}$

Find the lower limit value of u using Equation (3).

Substitute 5 for z in Equation (3).

$\begin{array}{c}u=1-\frac{5}{5}\\ =0\end{array}$

Substitute u for $1-\frac{z}{5}$ and $-5du$ for dz in Equation (2).

$\begin{array}{c}V=6{\displaystyle {\int }_1^0{u}^2\left(-5du\right)}\\ =-30{\left(\frac{u^3}{3}\right)}_1^0\\ =-30\left(0-\frac{1}{3}\right)\\ =10{\text{cm}}^3\end{array}$

Therefore, the volume of the tetrahedron is $\underset{\_}{10{\text{cm}}^3}$.