Q. 34 - Calculus

Question

Evaluate the triple integrals over the specified rectangular solid region.

$\begin{array}{r}{\iiint }_{\mathcal{R}} \ln \left(xy{z}^2\right)dV\text{, where }\\ \mathcal{R}=\left\{\right(x,y,z)\mid 1\le x\le 3,1\le y\le e,\text{ and }1\le z\le 2\}\end{array}$

Step-by-Step Solution

Answer

${\iiint }_{\mathcal{R}} \ln \left(xy{z}^2\right)dV =(e-1)[3\mathrm{In}3+8\mathrm{In}2-6]+2$

Step 1. Given information.

We have been given the triple integral:

$\begin{array}{r}{\iiint }_{\mathcal{R}} \ln \left(xy{z}^2\right)dV\text{, where }\\ \mathcal{R}=\left\{\right(x,y,z)\mid 1\le x\le 3,1\le y\le e,\text{ and }1\le z\le 2\}\end{array}$

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

Step 2. Evaluate.

By Fubini's theorem of triple integral :

$\begin{array}{r}{\iiint }_R \mathrm{In}\left(xy{z}^2\right)dv={\int }_1^3 {\int }_1^e {\int }_1^2 \mathrm{In}\left(xy{z}^2\right)dzdydx \\ ={\int }_1^3 {\int }_1^e {\int }_1^2 (\mathrm{In}x+\mathrm{In}y+2\mathrm{In}z)dzdydx \\ ={\int }_1^3 {\int }_1^e \left[{\int }_1^2 (\mathrm{In}x+\mathrm{In}y+2\mathrm{In}z)dz\right]dydx\\ ={\int }_1^3 {\int }_1^e \left[\mathrm{In}x{\int }_1^2 dz+\mathrm{In}y{\int }_1^2 dz+2{\int }_1^2 \mathrm{In}zdz\right]dydx\\ ={\int }_1^3 {\int }_1^e \left[\mathrm{In}x(z{)}_1^2+\mathrm{In}y(z{)}_1^2+2(z\mathrm{In}z-z{)}_1^2\right]dydx\\ ={\int }_1^2 {\int }_1^e [\mathrm{In}x(2-1)+\mathrm{In}y(2-1)+2(2\mathrm{In}2-2-1\ln 1+1\left)\right]dydx\end{array}$

Step 3. Integrate with respect to y.

Integrate with respect to y

$\begin{array}{r}={\int }_1^2 \left[{\int }_1^e (\mathrm{In}x+\mathrm{In}y+2(2\mathrm{In}2-1\left)\right]dy\right]dx\\ ={\int }_1^2 \left[\mathrm{In}x{\int }_1^e dy+{\int }_1^e \mathrm{In}ydy+2(2\mathrm{In}2-1){\int }_1^e dy\right]dx\\ ={\int }_1^2 \left[\mathrm{In}x(y{)}_1^e+(y\mathrm{In}y-y{)}_1^e+2(2\mathrm{In}2-1)(y{)}_1^e\right]dx \\ ={\int }_1^2 [\mathrm{In}x(e-1)+(e\mathrm{In}e-e+1)+2(2\mathrm{In}2-1\left)\right(e-1\left)\right]dx\end{array}$

Since $\ln e=1, \ln 1=0$

$\begin{array}{r}=(e-1){\int }_1^3 \mathrm{In}xdx+{\int }_1^3 dx+2(2\mathrm{In}2-1)(e-1){\int }_1^3 dx\\ =(e-1)[x\mathrm{In}x-x{]}_1^3+(x{)}_1^3+2(2\mathrm{In}2-1)(e-1)[x{]}_1^3 \\ =(e-1)[3\mathrm{In}3-3+1]+(3-1)+2(2\mathrm{In}2-1)(e-1)(3-1)\\ =(e-1)[3\mathrm{In}3-2]+2+4(2\mathrm{In}2-1)(e-1) \\ =(e-1)[3\mathrm{In}3-2+8\mathrm{In}2-4]+2 \\ =(e-1)[3\mathrm{In}3+8\mathrm{In}2-6]+2 \end{array}$