Question

Describe the three-dimensional region expressed in each iterated integral in Exercises 35–44.

${\int }_0^2{\int }_0^{1-3x/2}{\int }_{2x+(4/3)y-4}^{0}f(x,y,z)dzdydx$

Short answer

The three-dimensional region is given by planer equation,

$\frac{x}{2}+\frac{y}{3}-\frac{z}{4}=1$

Step-by-step solution

Step 1. Given Information 

We are given,

${\int }_0^2{\int }_0^{1-3x/2}{\int }_{2x+(4/3)y-4}^{0}f(x,y,z)dzdydx$

Step 2. The three dimensional region. 

By the definition of triple integral ${\int }_{a_1}^{a_1}{\int }_{b_4}^{b_2}{\int }_{a_1}^{c_2}f(x,y,z)dzdydx$represent the volume of the solid region$\mathrm{ℝ}=\left\{(x,y,z)\mid a_1\le x\le a_2,b_1\le y\le b_2,c_1\le z\le c_2\right\}$

Using this definition, we get

The given integral ${\int }_0^2{\int }_0^{1-\frac{3x}{2}}{\int }_{2x+\frac{4}{3}y-4}^{0}f(x,y,z)dzdydx$represents the volume of the tetrahedron whose vertices are$(0,0,0)(2,0,0)(0,3,0)(0,0,-4)$

Since from the given integral

$$\begin{gathered} z=2x+\frac{4}{3}y-4 \\ \Rightarrow 6x+4y-3z=12 \\ \frac{x}{2}+\frac{y}{3}-\frac{z}{4}=1 \end{gathered}$$

From this planer equation the vertices of the tetrahedron becomes $(0,0,0)(2,0,0)(0,3,0)(0,0,-4)$.