What is Triple Integral?

In calculus, the triple integral is a type of multiple integral that comes under the definite integral category. It is defined as the multiple integral of the function having three variables over the region $R^3$.

In calculus, if any function in three variables, $f(x,y,z)$, then the triple integral of the function over the three-dimensional region W is written as ${\displaystyle {\iiint }_{w}f\left(x,y,z\right)}\text{}dxdydz$.

Examples:

Important Points to Remember

To learn about triple integral, it is extremely important to have a proper understanding of definite and indefinite integral.

Indefinite Integral

In calculus, the integration of an unbounded function, i.e., the function that has no lower and upper limit is known as indefinite integral. The result of the indefinite integral is always a function as it has no limits.

Definite Integral

In calculus, the integration of a bounded function, i.e., the function that has both limits of integration is known as a definite integral. The result of the definite integral is a specific number due to the bounds of the function.

Difference between Double Integral and Triple Integral

How to Solve Triple Integrals?

Triple Riemann Sum

Triple Riemann sum is one of the ways to solve the triple integration problems. 

Let us suppose a continuous function $f(x,y,z)$is enclosed in box B$(a\le x\le b,c\le y\le d,ande\le z\le f)$, and that it is required to find the triple integral of the function over box B.

To find this, divide the interval $a\le x\le b$ into m sub-intervals,$c\le y\le d$ into n sub-intervals, and$e\le z\le f$ into r sub-intervals. Thus the whole box is divided into the number of small mnr sub-boxes and let the volume of each small box be $\delta V$.

Next, choose a point in each small box that represents it. For any box ijk, let the point be $(x_{ijk,}{y}_{ijk,}{z}_{ijk})$. Assuming the density of the box ijk to be constant, and since it is represented by $f(x_{ijk,}{y}_{ijk,}{z}_{ijk})$, then the mass of the box is given by:

$${\displaystyle \sum _{i=1}^{i=m}{\displaystyle \sum _{j=1}^{j=n}{\displaystyle \sum _{k=1}^{k=r}f(x_{ijk,}{y}_{ijk,}}}}{z}_{ijk})\delta V$$

This is known as the Riemann sum.

But this sum gives us the approximate value of the total mass of the box. To get the exact value of the total mass, we need to increase the sub-boxes. By increasing the number of boxes, the sum approaches a particular value which is called the value of the triple integral.

$\underset{m\to \infty ,n\to \infty ,k\to \infty }{\lim }{\displaystyle {\sum }_{i=1}^{i=m}{\displaystyle {\sum }_{j=1}^{j=n}{\displaystyle {\sum }_{k=1}^{k=r}f(x_{ijk,}}}}{y}_{ijk,}{z}_{ijk})\delta V={\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{y=c}^{y=d}{\displaystyle {\int }_{z=e}^{z=f}f(x,y,z)dzdxdy}}}$

Thus triple integral is the actual mass of box B.

Note:  It is not easy to solve the integral of complicated functions using the Riemann sum. So, to evaluate the triple integral problems easily, we will have to apply the iterated and Fubini’s theorem.

Iterated Integral Theorem

Suppose any region D is defined as:

$a_1\le x\le a_2$, $b_1(x)\le y\le b_2(x)$ and $c_1(x,y)\le z\le c_2(x,y)$

Then the triple integration of $f(x,y,z)$over the region D is calculated by applying the following formula:

Fubini’s Theorem

It states that if there is a continuous function $f(x,y,z)$that is defined over any rectangular box B such that$a\le x\le b$,$c\le y\le d$, and$e\le z\le f$, then to integrate $f(x,y,z)$over the region B, there are six different orders which are given below:

${\displaystyle {\iiint }_{\left\{B\right\}}f(x,y,z)dV={\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{y=c}^{y=d}{\displaystyle {\int }_{z=e}^{z=f}f(x,y,z)\text{dzdydx}}}}}$

$={\displaystyle {\int }_{y=c}^{y=d}{\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{z=e}^{z=f}f(x,y,z)\text{dzdxdy}}}}$

$={\displaystyle {\int }_{z=e}^{z=f}{\displaystyle {\int }_{y=c}^{y=d}{\displaystyle {\int }_{x=a}^{x=b}f(x,y,z)dxdydz}}}$

$={\displaystyle {\int }_{z=e}^{z=f}{\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{y=c}^{y=d}f(x,y,z)dydxdz}}}$

$={\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{z=e}^{z=f}{\displaystyle {\int }_{y=c}^{y=d}f(x,y,z)dydzdx}}}$

  In each case, the result will be the same.

Cylindrical Coordinate System

The three-dimensional coordinate system (also known as an extension of polar coordinates of two dimensional) in which we specify the position of the particular point $Q(x,y,z)$with the help of three factors namely radial distance (r), polar angle (θ), and z, is known as a cylindrical coordinate system.

Following is the relation between the Cartesian $(x,y,z)$and the cylindrical coordinates (r, θ, z):

$x=r\cos (\theta )$

$y=r\sin (\theta )$

z=z

How to Convert Triple Integral into Cylindrical Coordinate System?

Sometimes, it is easier to solve the complicated integral in terms of the cylindrical coordinate system instead of solving it in terms of the cartesian system.

By using the formula for the change of variables method, one can easily convert a triple integral from the Cartesian system into a cylindrical system. The formula of change of variable method is given below:

$${\displaystyle {\iiint }_{D}f(x,y,z)dV={\displaystyle {\iiint }_{D}f(r\cos (\theta ),r\sin (\theta ),z)\left|J\right|}}drd\theta dz$$

where J is a jacobian and is equal to:

$\left|J\right|=\det \left(\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial z}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial z}\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial z}\end{array}\right)$

$=\det \left(\begin{array}{ccc}\cos \theta & -r\sin \theta & 0\\ \sin \theta & r\cos \theta & 0\\ 0& 0& 1\end{array}\right)$

$=r{\cos }^2\theta +r{\sin }^2\theta$

$=r$

Therefore, $dV=rdrd\theta dz$

In a cylindrical system, we assume that $r\ge 0,0\le \theta \le 2\pi ,-\infty \le z\le \infty$.

Note: 

The thing that we have to remember is that the above formula helps us to solve the integral problem in those cases when the region is a cylindrical surface.

Spherical Coordinate System

The spherical coordinate system is the three-dimensional system in which we specify the position of the particular point Q(x,y,z) with the help of three factors namely radial distance (r), polar angle (θ), and azimuthal angle ($\varphi$).

The relation between the Cartesian and the spherical system:

How to Convert Triple Integral into Spherical Coordinate System?

In mathematics, with the help of the formula of change of variables method, the triple integral which is in the Cartesian system can be easily converted into a spherical system. The thing that we have to remember is that this system is helpful in those cases when the region of integration is a spherical surface. The following is the formula of the change of variable method for the spherical system:

Where J is the Jacobian and it is equal to: 

$\left|J\right|=\det \left(\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }& \frac{\partial x}{\partial \varphi }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }& \frac{\partial y}{\partial \varphi }\\ \frac{\partial z}{\partial r}& \frac{\partial z}{\partial \theta }& \frac{\partial z}{\partial \varphi }\end{array}\right)$

$=\det \left(\begin{array}{ccc}sin\varphi \cos \theta & -r\sin \varphi \sin \theta & r\cos \varphi \cos \theta \\ sin\varphi \sin \theta & r\sin \varphi \cos \theta & r\cos \varphi \sin \theta \\ \cos \varphi & 0& -r\sin \varphi \end{array}\right)$

$=r^2\sin \varphi$

 Common Mistake

• The common mistake that we do while solving an integral problem is the wrong use of the methods of integration. Before solving any problem of integral it is recommended to revise all the methods of integral.
• The second thing that you have to take care of while solving a definite integral problem is the substitution of the limits. After solving the indefinite integral of the function, the upper and the lower limit of the integral should be substituted properly.

Key Concept

• Fubini’s theorem is used to evaluate the triple integration of a continuous function over a rectangular box.
• The iterated theorem is another important theorem that is used to evaluate the triple integration problems.  

Formulas

(1) Triple integral:

$\underset{m\to \infty ,n\to \infty ,k\to \infty }{\lim }{\displaystyle {\sum }_{i=1}^{i=m}{\displaystyle {\sum }_{j=1}^{j=n}{\displaystyle {\sum }_{k=1}^{k=r}f(x_{ijk,}}}}{y}_{ijk,}{z}_{ijk})\delta V={\displaystyle {\int }_{x=a}^{x=b}{\displaystyle {\int }_{y=c}^{y=d}{\displaystyle {\int }_{z=e}^{z=f}f(x,y,z)dzdxdy}}}$

(2) In cylindrical system:

${\displaystyle {\iiint }_{D}f(x,y,z)dV={\displaystyle {\iiint }_{D}f(r\cos (\theta ),r\sin (\theta ),z)r}}drd\theta dz$

(3) In spherical system:

${\displaystyle {\iiint }_{D}f(x,y,z)dxdydz={\displaystyle {\iiint }_{D}f(r\sin (\varphi )cos(\theta ),rsin(\varphi )sin(\theta ),r\cos (\varphi ))r^2\sin (\varphi )}}drd\theta d\varphi$

Context and Application

This topic is significant in the professional exams for both undergraduate and graduate courses like

• Bachelor of Science in Mathematics.
• Bachelor of Science in Non-medical.
• Bachelor of Technology in Mechanical Engineering.
• Master of Science in Mathematics.

Practice Problems

Example 1: How to write double and triple integral in latex?

Answer: In latex coding, use $i\mathrm{int}$to write double integral and $ii\mathrm{int}$ for triple integral.

Example 2: Write the formula with the help of which you can evaluate the average value of the function for any integrable function $f(x,y,z)$ over a region B.

Answer: The formula of the average value of the function for any integrable function$f(x,y,z)$over the region B is:

Example 3: Evaluate the following:

Solution:  To observe the limits of the function properly, find that the region of integration is a rectangular box. So to evaluate the given integral, we have six ways. The result will always the same. First way:

${\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{y=1}^{y=2}{\displaystyle {\int }_{x=2}^{x=3}1dxdydz}}}$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{y=1}^{y=2}{\displaystyle {\int }_{x=2}^{x=3}{\left[x\right]}_{x=2}^{x=3}}}}dydz$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{y=1}^{y=2}\left[3-2\right]}}dydz$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{y=1}^{y=2}dy}}dz$

$={{\displaystyle {\int }_{z=0}^{z=1}\left[y\right]}}_{y=1}^{y=2}dz$

$={\displaystyle {\int }_{z=0}^{z=1}\left[2-1\right]}dz$

$={\displaystyle {\int }_{z=0}^{z=1}d}z$

$={\left[z\right]}_{z=0}^{z=1}$

$=\left[1-0\right]$

=1

Second Way:

${\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{x=2}^{x=3}{\displaystyle {\int }_{y=1}^{y=2}1dydxdz}}}$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{x=2}^{x=3}{\left[y\right]}_{y=1}^{y=2}}}dxdz$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{x=2}^{x=3}\left[2-1\right]}}dxdz$

$={\displaystyle {\int }_{z=0}^{z=1}{\displaystyle {\int }_{x=2}^{x=3}d}}xdz$

$={{\displaystyle {\int }_{z=0}^{z=1}\left[x\right]}}_{x=2}^{x=3}dz$

$={\displaystyle {\int }_{z=0}^{z=1}\left[3-2\right]}dz$

$={\displaystyle {\int }_{z=0}^{z=1}dz}$

$={\left[z\right]}_{z=0}^{z=1}$

$=1-0$

$=1$

Similarly, solve the above problem in four other ways:

3)  Third way: Firstly integrate the function with respect to x, then with respect to z, and lastly with respect to y.

4)  Fourth way: With respect to y first, second with respect to z, and with respect to x in the last.

5)  Fifth way: First with respect to z, then respect to x, and third with respect to y.

6)  Sixth way: Initially integrate with respect to z, secondly with respect to y, and in the end with respect to x.

Example 4: Evaluate:

Solution: In the problem, observe that the limits of x are in terms of y and z and the limits of y are in terms of z and the limits of z are constant numbers. So in such a problem, the order matters. Here, first, have to solve the integral with respect to x, then with respect to y, and lastly with respect to z as it has constant limits.

${\displaystyle {\int }_0^1{\displaystyle {\int }_0^z{\displaystyle {\int }_{y-z}^{y+z}dxdydz}}}={{\displaystyle {\int }_0^1{\displaystyle {\int }_0^z\left[x\right]}}}_{y-z}^{y+z}dydz$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^z\left[y+z-y+z\right]}}dydz$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^z\left[2z\right]}}dydz$

$={{\displaystyle {\int }_0^{1}2z\left[y\right]}}_0^{z}dz$

$={\displaystyle {\int }_0^{1}2z\left[z-0\right]}dz$

$={\displaystyle {\int }_0^{1}2z^2}dz$

$=2{\left(\frac{z^3}{3}\right)}_0^1$

$=\frac{2}{3}\left(1^3-0^3\right)$

$=\frac{2}{3}$

Example 5: Evaluate ${\displaystyle {\iiint }_{D}z(x^2}+y^2)dxdydz$ where the solid region is$D=\{(x,y,z):x^2+y^2\le 1,2\le z\le 3\}$. (Round your answer up to three decimal places)

Solution: The given region is a part of a right circular cylinder. Therefore, the given problem can be easily solved by changing the integral into cylindrical coordinates. In a cylindrical system, the limits change into:

$$D^{\text{'}}=\{(r,\theta ,z):0\le r\le 1,0\le \theta \le 2\pi ,2\le z\le 3\}$$

Now:

${\displaystyle {\iiint }_{D}z(x^2}+y^2)dxdydz={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_2^{3}z(r^2}}}{\cos }^2(\theta )+r^2{\sin }^2(\theta ))rdzd\theta dr$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_2^{3}z(r^2}}}(1))rdzd\theta dr$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_2^{3}z{r}^3}}}dzd\theta dr$

$={{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }{r}^3\left[z^2\right]}}}_2^{3}d\theta dr$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }{r}^3\left[3^2-2^2\right]}}d\theta dr$

$={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }5r^3}}d\theta dr$

$={\displaystyle {\int }_0^{1}5r^3{[\theta ]}_0^{2\pi }}dr$

$={\displaystyle {\int }_0^{1}5r^3[2\pi -0]}dr$

$={\displaystyle {\int }_0^{1}10\pi r^3}dr$

$=10\pi {[\frac{r^4}{4}]}_0^1$

$=10\pi [\frac{1^4}{4}-\frac{0^3}{4}]$

$=\frac{5\pi }{2}$

$\approx 3.927$