Question

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

${\int }_0^3{\int }_0^3{\int }_0^{3-y}f(x,y,z)dzdydx$

Short answer

The three-dimensional region is,

$\mathrm{ℝ}=\left\{\right(x,y,z)/0\le x\le 3,0\le y\le 3,0\le z\le 3-y\}$

Step-by-step solution

Step 1. Given Information 

We are given,

${\int }_0^3{\int }_0^3{\int }_0^{3-y}f(x,y,z)dzdydx$

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 iterated triple integral ${\int }_0^3{\int }_0^3{\int }_0^{3-y}f(x,y,z)dzdydx$represents the volume of the half cube represents by $\mathrm{ℝ}=\left\{\right(x,y,z)/0\le x\le 3,0\le y\le 3,0\le z\le 3-y\}$

Or the planer equation $x+y+z=3$.

Since from the given triple integral the limits are $0\le x\le 3,0\le y\le 3,0\le z\le 3-y$.