Question

Evaluate the triple integral${\int }_0^2{\int }_0^{-\frac{3}{2}x+3}{\int }_0^{4-2x-\frac{4}{3}y}dzdydx$and given physical interpretation of the integral

Short answer

The value of the triple integral is$4cubicunits.$it represents the volume of the rectangular region.

Step-by-step solution

Given information

We are given a triple integral${\int }_0^2{\int }_0^{-\frac{3}{2}x+3}{\int }_0^{4-2x-\frac{4}{3}y}dzdydx$

Evaluate the integral

We get,

$$\begin{gathered} {\int }_0^2{\int }_0^{-\frac{3}{2}x+3}{\int }_0^{4-2x-\frac{4}{3}y}dzdydx \\ ={\int }_0^2{\int }_0^{-\frac{3}{2}x+3}[4-2x-\frac{4}{3}y]dydx \\ ={\int }_0^2{{[4y-2xy-\frac{2}{3}{y}^2]}_0}^{-\frac{3}{2}x+3}dx \\ ={\int }_0^2[-6x+12+3x^2-6x-\frac{3}{2}{x}^2-6+6x]dx \\ ={\int }_0^2[\frac{3}{2}{x}^2-6x+6]dx \\ ={{[\frac{1}{2}{x}^3-3x^2+6x]}_0}^2 \\ =4 \\ =4cubicunits \end{gathered}$$

It represents the volume of the rectangular region