91Ó°ÊÓ

We have been given the triple integral:

$\begin{array}{r}{∭}_{\mathcal{R}} (x+2y+3z)dV\text{, where}\\ \mathcal{R}=\left\{\right(x,y,z)âˆ\pounds 0≤x≤4,1≤y≤5,\text{and}2≤z≤7\}\end{array}$

We have to evaluate this over the specified rectangular solid regions.

By Fubini's theorem of triple integral :

$\begin{array}{r}{∭}_R (x+2y+3z)dv={∫}_0^4 {∫}_1^5 {∫}_2^7 (x+2y+3z)dzdydx\\ ={∫}_0^4 {∫}_1^5 \left[{∫}_2^7 (x+2y+3z)dz\right]dydx\\ ={∫}_0^4 {∫}_1^5 \left[x{∫}_2^7 dz+2y{∫}_2^7 dz+3{∫}_2^7 zdz\right]dydx\\ ={∫}_0^4 {∫}_1^5 \left[x(z{)}_2^7+2y(z{)}_2^7+3{\left(\frac{z^2}{2}\right)}_2^7\right]dydx\\ ={∫}_0^4 {∫}_1^5 \left[x(7−2)+2y(7−2)+3\left(\frac{7^2}{2}−\frac{2^2}{2}\right)\right]dydx\\ ={∫}_0^4 {∫}_1^5 \left[5x+10y+\frac{135}{2}\right]dydx\end{array}$

Integrate with respect to y

$\begin{array}{r}={∫}_0^4 \left[{∫}_1^5 \left(5x+10y+\frac{135}{2}\right)dy\right]dx\\ ={∫}_0^4 \left[5x{∫}_1^5 dy+10{∫}_1^5 ydy+\frac{135}{2}{∫}_1^5 dy\right]dx\\ ={∫}_0^4 \left[5x(y{)}_1^5+10{\left(\frac{y^2}{2}\right)}_1^5+\frac{135}{2}(y{)}_1^5\right]dx\\ ={∫}_0^4 \left[5x(5−1)+10\left(\frac{5^2}{2}−\frac{1^2}{2}\right)+\frac{135}{2}(5−1)\right]dx\\ ={∫}_0^4 [20x+120+270]dx\\ ={∫}_0^4 [20x+390]dx\end{array}$

Integrate with respect to x

$\begin{array}{r}=20{∫}_0^4 xdx+390{∫}_0^4 dx\\ =20{\left[\frac{x^2}{2}\right]}_0^4+390[x{]}_0^4\\ =20\left[\frac{4^2}{2}\right]+390\left[4\right]\\ =160+1560\\ =1720\end{array}$