5.4 Triple integrals  (Page 2/8)

 Page 2 / 8

For $a,b,c,d,e,$ and $f$ real numbers, the iterated triple integral can be expressed in six different orderings:

$\begin{array}{cc}\hfill \underset{e}{\overset{f}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{c}{\overset{d}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{a}{\overset{b}{\int }}f\left(x,y,z\right)dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz& =\underset{e}{\overset{f}{\int }}\left(\underset{c}{\overset{d}{\int }}\left(\underset{a}{\overset{b}{\int }}f\left(x,y,z\right)dx\right)dy\right)dz=\underset{c}{\overset{d}{\int }}\left(\underset{e}{\overset{f}{\int }}\left(\underset{a}{\overset{b}{\int }}f\left(x,y,z\right)dx\right)dz\right)dy\hfill \\ & =\underset{a}{\overset{b}{\int }}\left(\underset{e}{\overset{f}{\int }}\left(\underset{c}{\overset{d}{\int }}f\left(x,y,z\right)dy\right)dz\right)dx=\underset{e}{\overset{f}{\int }}\left(\underset{a}{\overset{b}{\int }}\left(\underset{c}{\overset{d}{\int }}f\left(x,y,z\right)dy\right)dx\right)dz\hfill \\ & =\underset{c}{\overset{e}{\int }}\left(\underset{a}{\overset{b}{\int }}\left(\underset{e}{\overset{f}{\int }}f\left(x,y,z\right)dz\right)dx\right)dy=\underset{a}{\overset{b}{\int }}\left(\underset{c}{\overset{e}{\int }}\left(\underset{e}{\overset{f}{\int }}f\left(x,y,z\right)dz\right)dy\right)dx.\hfill \end{array}$

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Evaluating a triple integral

Evaluate the triple integral ${\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}{\int }_{x=-1}^{x=5}\left(x+y{z}^{2}\right)dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz.$

The order of integration is specified in the problem, so integrate with respect to $x$ first, then y , and then $z.$

$\begin{array}{}\\ \\ \\ \\ \phantom{\rule{1em}{0ex}}{\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}{\int }_{x=-1}^{x=5}\left(x+y{z}^{2}\right)dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz\hfill & & & \\ ={\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}\left[{\frac{{x}^{2}}{2}+xy{z}^{2}|}_{x=-1}^{x=5}\right]dy\phantom{\rule{0.2em}{0ex}}dz\hfill & & & \text{Integrate with respect to}\phantom{\rule{0.2em}{0ex}}x.\hfill \\ ={\int }_{z=0}^{z=1}{\int }_{y=2}^{y=4}\left[12+6y{z}^{2}\right]dy\phantom{\rule{0.2em}{0ex}}dz\hfill & & & \text{Evaluate.}\hfill \\ ={\int }_{z=0}^{z=1}\left[{12y+6\frac{{y}^{2}}{2}{z}^{2}|}_{y=2}^{y=4}\right]dz\hfill & & & \text{Integrate with respect to}\phantom{\rule{0.2em}{0ex}}y.\hfill \\ ={\int }_{z=0}^{z=1}\left[24+36{z}^{2}\right]dz\hfill & & & \text{Evaluate.}\hfill \\ ={\left[24z+36\frac{{z}^{3}}{3}\right]}_{z=0}^{z=1}=36.\hfill & & & \text{Integrate with respect to}\phantom{\rule{0.2em}{0ex}}z.\hfill \end{array}$

Evaluating a triple integral

Evaluate the triple integral $\underset{B}{\iiint }{x}^{2}yz\phantom{\rule{0.2em}{0ex}}dV$ where $B=\left\{\left(x,y,z\right)|-2\le x\le 1,0\le y\le 3,1\le z\le 5\right\}$ as shown in the following figure.

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate y first, then x , and then z .

$\begin{array}{cc}\hfill \underset{B}{\iiint }{x}^{2}yz\phantom{\rule{0.2em}{0ex}}dV& =\underset{1}{\overset{5}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{-2}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{3}{\int }}\left[{x}^{2}yz\right]dy\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz=\underset{1}{\overset{5}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{-2}{\overset{1}{\int }}\left[{{x}^{2}\frac{{y}^{2}}{2}z|}_{0}^{3}\right]dx\phantom{\rule{0.2em}{0ex}}dz\hfill \\ & =\underset{1}{\overset{5}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{-2}{\overset{1}{\int }}\frac{9}{2}{x}^{2}z\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz=\underset{1}{\overset{5}{\int }}\left[{\frac{9}{2}\frac{{x}^{3}}{3}z|}_{-2}^{1}\right]dz=\underset{1}{\overset{5}{\int }}\frac{27}{2}z\phantom{\rule{0.2em}{0ex}}dz={\frac{27}{2}\frac{{z}^{2}}{2}|}_{1}^{5}=162.\hfill \end{array}$

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to $x$ first, then $z,$ and then $y.$

$\begin{array}{cc}\hfill \underset{B}{\iiint }{x}^{2}yz\phantom{\rule{0.2em}{0ex}}dV& =\underset{0}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{5}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{-2}{\overset{1}{\int }}\left[{x}^{2}yz\right]dx\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dy=\underset{0}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{5}{\int }}\left[{\frac{{x}^{3}}{3}yz|}_{-2}^{1}\right]dz\phantom{\rule{0.2em}{0ex}}dy\hfill \\ & =\underset{0}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{5}{\int }}3yz\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dy=\underset{0}{\overset{3}{\int }}\left[{3y\frac{{z}^{2}}{2}|}_{1}^{5}\right]dy=\underset{0}{\overset{3}{\int }}36y\phantom{\rule{0.2em}{0ex}}dy={36\frac{{y}^{2}}{2}|}_{0}^{3}=18\left(9-0\right)=162.\hfill \end{array}$

Evaluate the triple integral $\underset{B}{\iiint }z\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}dV$ where $B=\left\{\left(x,y,z\right)|0\le x\le \pi ,\frac{3\pi }{2}\le y\le 2\pi ,1\le z\le 3\right\}.$

$\underset{B}{\iiint }z\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}dV=8$

Triple integrals over a general bounded region

We now expand the definition of the triple integral to compute a triple integral over a more general bounded region $E$ in ${ℝ}^{3}.$ The general bounded regions we will consider are of three types. First, let $D$ be the bounded region that is a projection of $E$ onto the $xy$ -plane. Suppose the region $E$ in ${ℝ}^{3}$ has the form

$E=\left\{\left(x,y,z\right)|\left(x,y\right)\in D,{u}_{1}\left(x,y\right)\le z\le {u}_{2}\left(x,y\right)\right\}.$

For two functions $z={u}_{1}\left(x,y\right)$ and $z={u}_{2}\left(x,y\right),$ such that ${u}_{1}\left(x,y\right)\le {u}_{2}\left(x,y\right)$ for all $\left(x,y\right)$ in $D$ as shown in the following figure. We can describe region E as the space between u 1 ( x , y ) and u 2 ( x , y ) above the projection D of E onto the x y -plane.

Triple integral over a general region

The triple integral of a continuous function $f\left(x,y,z\right)$ over a general three-dimensional region

$E=\left\{\left(x,y,z\right)|\left(x,y\right)\in D,{u}_{1}\left(x,y\right)\le z\le {u}_{2}\left(x,y\right)\right\}$

in ${ℝ}^{3},$ where $D$ is the projection of $E$ onto the $xy$ -plane, is

$\underset{E}{\iiint }f\left(x,y,z\right)dV=\underset{D}{\iint }\left[\underset{{u}_{1}\left(x,y\right)}{\overset{{u}_{2}\left(x,y\right)}{\int }}f\left(x,y,z\right)dz\right]dA.$

Similarly, we can consider a general bounded region $D$ in the $xy$ -plane and two functions $y={u}_{1}\left(x,z\right)$ and $y={u}_{2}\left(x,z\right)$ such that ${u}_{1}\left(x,z\right)\le {u}_{2}\left(x,z\right)$ for all $\left(x,z\right)$ in $D.$ Then we can describe the solid region $E$ in ${ℝ}^{3}$ as

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