Q. 32 - Calculus

Question

Evaluate the triple integrals over the specified rectangular solid region.

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

Step-by-Step Solution

Answer

${\iiint }_{\mathcal{R}} (x+2y+3z)dV =1720$

Step 1. Given information.

We have been given the triple integral:

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

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

Step 2. Evaluate.

By Fubini's theorem of triple integral :

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

Step 3. Integrate with respect to y.

Integrate with respect to y

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

Integrate with respect to x

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