Q. 38 - Calculus

Question

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

${\int }_0^2{\int }_0^3{\int }_{4x/3}^{4}f(x,y,z)dzdxdy$

Step-by-Step Solution

Answer

The three-dimensional region is,

$\mathrm{ℝ}=\left\{(x,y,z)/0\le x\le 3,0\le y\le 2,\frac{4x}{3}\le z\le 4\right\}$

Step 1. Given Information

We are given, 

${\int }_0^2{\int }_0^3{\int }_{4x/3}^{4}f(x,y,z)dzdxdy$

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 ${\int }_0^2{\int }_0^3{\int }_{\frac{4x}{3}}^{4}f(x,y,z)dzdxdy$ represents the volume of rectangular solid represents by $\mathrm{ℝ}=\left\{(x,y,z)/0\le x\le 3,0\le y\le 2,\frac{4x}{3}\le z\le 4\right\}$.

Since by the triple integral the limits are $0\le x\le 3,0\le y\le 2,\frac{4x}{3}\le z\le 4$.