# 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.

## 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

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!