Question

Evaluate the triple integrals over the specified rectangular solid region.

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

Short answer

$182$

Step-by-step solution

Step 1. Given information.

We have been given the triple integral:

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

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

Step 2. Evaluate 

By Fubini's theorem of triple integral :

$\begin{array}{r}{\iiint }_R x^{2}yzdv={\int }_{-2}^1 {\int }_0^3 {\int }_1^5 x^{2}yzdzdydx\\ ={\int }_{-2}^1 {\int }_0^3 \left(x^{2}y{\int }_1^5 zdz\right)dydx\\ ={\int }_{-2}^1 {\int }_0^3 \left(x^{2}y{\left(\frac{z^2}{2}\right)}_1^5\right)dydx\\ ={\int }_{-2}^1 {\int }_0^3 \left(x^{2}y\left(\frac{5^2}{2}-\frac{1^2}{2}\right)dydx\right.\\ ={\int }_{-2}^1 {\int }_0^3 \left(x^{2}y\left(12\right)dydx\right.\end{array}$

Step 3. Integrate with respect to y

Integrate with respect to y

$\begin{array}{r}={\int }_{-2}^1 \left[12x^2{\int }_0^3 ydy\right]dx\\ ={\int }_{-2}^1 \left[12x^2{\int }_0^3 ydy\right]dx\\ ={\int }_{-2}^1 \left[12x^2{\left(\frac{y^2}{2}\right)}_0^3\right]dz\\ ={\int }_{-2}^1 \left[12x^2\left(\frac{9}{2}\right)\right]dx\\ =54{\int }_{-2}^1 x^{2}dx\\ =54{\left(\frac{x^3}{3}\right)}_{-2}^1\\ =54\left(\frac{1^3}{3}-\frac{(-2{)}^3}{3}\right)\\ =182\\ \end{array}$