Question

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

${\int }_0^3{\int }_0^{1-y/3}{\int }_0^{2-(2/3)y-2z}f(x,y,z)dxdzdy$

Short answer

The three-dimensional region is given by planer equation,

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

Step-by-step solution

Step 1. Given Information 

We are given,

${\int }_0^3{\int }_0^{1-y/3}{\int }_0^{2-(2/3)y-2z}f(x,y,z)dxdzdy$

Step 2. The three dimensional region. 

By the definition of triple integral ${\int }_{a_1}^{a_1}{\int }_{b_1}^{b_2}{\int }_{c_1}^{c_2}f(x,y,z)dzdydx$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

Given triple integral ${\int }_0^3{\int }_0^{1-\frac{y}{3}}{\int }_0^{2-\frac{2}{3}y-2x}f(x,y,z)dxdzdy$represents the volume of the tetrahedron whose vertices are $(0,0,0)(2,0,0)(0,3,0)(0,0,1)$.

Since from the given integral the limits are $0\le y\le 3,0\le x\le 1-\frac{y}{3},0\le z\le 2-\frac{2}{3}y-2x$

The planer equation becomes

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

It represents the region of the tetrahedron with vertices $(0,0,0)(2,0,0)(0,3,0)(0,0,1)$.