Triple Integrals (original) (raw)
Last Updated : 1 Oct, 2025
A Triple Integral is a type of multiple integral that involves a function of three variables. Also called a volume integral, it is used to calculate quantities (like volume or mass) over a three-dimensional region.
- It extends the features of integration to three variables (x, y, z).
- The integration is performed successively with respect to each variable.
- The order of integration can change, but the final result remains the same.
Expressed in general form as:
In the above general form:
- f(x, y, z) is the function being integrated.
- v is the three-dimensional region of integration.
- dv represents a small volume element, which can be expressed in different coordinate systems:
Representation of Triple Integrals
Triple integrals can be represented as below:
\iiint kV dV = \int_0^z\int_0^y\int_0^x kV dx \ dy \ dz
**Where x, y and z are three different variables on which integration is performed
**Also Check
How to Solve Triple Integrals?
Below are the steps that can be followed for solving Triple Integral Problems :
**Step 1: The first step for solving problems involving triple integral is to identify what are the variable and their corresponding differentials used in the triple integral equation.
Here, k is a constant.
\iiint k dV = \int_0^z\int_0^y\int_0^x kV dx dy dz
So, here variables involved for the above triple integrals are x, y, z represented as dx, dy, dz.
**Step 2: Now supposedly the variables are given values, supposedly each of x, y and z is given a lower value of zero and an upper value of 6. So, the equation will look like:
\iiint k dV = \int_0^6\int_0^6\int_0^6 k dx dy dz
When values are allotted to variables we will evaluate each of the integral and then put values for each of the variable.
**Step 3: In this step, we will evaluate the variable to arrive at a final solution for the integral.
\iiint k dV = \int_0^6\left(\int_0^6\left(\int_0^6 k dx\right) dy\right) dz'
\Rightarrow \iiint k dV = \int_0^6\left(\int_0^6 k[x]_0^6 dy\right) dz
\Rightarrow \iiint k dV = 6k \int_0^6 [y]_0^6 dz
\Rightarrow \iiint k dV = 36k\int_0^6 dz
\Rightarrow \iiint k dV = 216 k
Properties of Triple Integration
Some of the properties of Triple Integration are:
- Linearity
- Additivity
- Monotonicity
- Divergence Theorem
**Note: Divergece Theorem is not a property of Triple Integration in literal sense, but as it involves the triple integral or volume integral. Thus we can consider this as property.
Linearity
This property denotes triple integrals to give the same result under the same limits when addition/subtraction is performed collectively on the triple integral and when addition/subtraction is performed on individual units of triple integral variable terms.
\iiint R [f(x,y,z) \pm g(x,y,z)] dV = \iiint R f(x,y,z) dV \pm \iiint R g(x,y,z) dV
Additivity
This property denotes triple integrals to give the same result under the same limits when evaluated as a single unit or split into multiple units. Here, R is a region given as a union of S and T, where S and T are disjoint partitions of R, i.e., S and T units have nothing in common.
If R = S ⋃ T, S ⋂ T = Φ then,
\iiint R f(x,y,z) dz dy dx = \iiint S f(x,y,z)dz dy dx + \iiint T f(x,y) dz dy dx
Monotonicity
This property denotes triple integrals to be fetching the same results for the same limits, irrespective the variable is evaluated as a part of the triple integral or it is outside the triple integral evaluation.
If f(x, y, z) ≥ g(x, y), then \iiint R f(x,y) dz dy dx ≥ \iiint R g(x,y)dy~dx
\iiint R k f(x, y, z) dz dy dx = k \iiint R f(x, y, z) dz dy dx
Divergence Theorem
The theorem mentions the normal component of a vector point function supposedly takes it as F over a closed surface, say 'S', is the volume integral of the divergence of 'F' taken over volume 'V' enclosed by the closed surface S.
It is denoted as below.
\iiint_V ▽\vec F. dV = \iint_s \vec F. \vec n. dS
Application of Triple Integrals
Triple integration can be used in numerous ways to calculate the volume of three-dimensional figures. Below are the applications of integrals:
- They find their application for three-dimensional figures, where an integral is performed to evaluate the volume. Triple integrals are also referred to as volume integrals.
- Supposedly, we have three-dimensional variables of x, y, and z representing the length, breadth, and height of a three-dimensional cuboid. This can be calculated through the following formula:
\iiint k dV = \int_0^z \int_0^y\int_0^x k dx dy dz
\iiint k f( x ,y ,z) dx dy dz
- Triple integrals can thus be used for all kinds of volume evaluations.
Triple Integrals in Engineering Mathematics
Triple integrals are a fundamental tool in engineering mathematics, used extensively in fields like fluid dynamics, thermodynamics, and electromagnetism. They enable engineers and mathematicians to calculate quantities that are distributed across three-dimensional spaces. Here are the uses of triple integrals in real life:
- Volume Calculations
- Mass and Density Calculations
- Center of Mass
- Moment of Inertia
- Fluid Dynamics and Heat Transfer
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Solved Examples on Triple Integrals
**Example 1. Evaluate the triple integral problem
\iiint k dV =\int_0^{z=12}\int_0^{y=12}\int_0^{x=12} k dx dy dz
**Solution:
\iiint k dV =\int_0^{z=12}\int_0^{y=12}\int_0^{x=12} k dx dy dz
⇒ \iiint k dV =\int_0^{z=12}\int_0^{y=12}⇒ k [x] 0 x=12 dy dz
⇒ \iiint k dV = \int_0^{z=12}\int_0^{y=12} k [x]0 x=12 dy dz
⇒ \iiint k dV = \int_0^{z=12} 12k[y]0 y=12 dz
⇒ \iiint k dV = 144k [z]0 z=12
⇒ \iiint k dV = 1728 k
**Example 2. Evaluate the triple integral problem
\int_0^{z=8}\int_0^{y=6}\int_0^{x=4} k dx dy dz
**Solution:
\int_0^{z=8}\int_0^{y=6}\int_0^{x=4} k dx dy dz
= \int_0^{z=8}\int_0^{y=6} k [x]0 x=4 dy dz
= \int_0^{z=8} 4k[y]0 y=6 dz
= 24k [z]0 z=8
= 192 k
**Example 3. Evaluate the triple integral problem
\int_0^{z=3}\int_0^{y=2}\int_0^{x=4} 4x dx dy dz
**Solution:
\int_0^{z=3}\int_0^{y=2}\int_0^{x=4} 4x dx dy dz
= \int_0^{z=3}\int_0^{y=2} 4k [x2/2]0 x=4 dy dz
= \int_0^{z=3} 4k[8][y]0 y=2 dz
= 64k [z]0 z=3
= 192 k
**Example 4. Evaluate the triple integral problem
\int_0^{z=10}\int_0^{y=12}\int_0^{x=5} k dx dy dz
**Solution:
\int_0^{z=10}\int_0^{y=12}\int_0^{x=5} k dx dy dz
= \int_0^{z=10}\int_0^{y=12} [x]0 5 dy dz
= \int_0^{z=10} [y]0 12 5k dz
= [z]0 10 60k
= 600k
**Example 5. Evaluate the triple integral problem
\int_0^{z=18}\int_0^{y=9}\int_0^{x=3} k dx dy dz
**Solution:
\int_0^{z=18}\int_0^{y=9}\int_0^{x=3} k dx dy dz
= \int_0^{z=18}\int_0^{y=9} [x]0 x=3 dy dz
= \int_0^{z=18} [y]0 y=9 3k dz
= [z] 0 z=18 27k
= 486k
Practice Questions on Triple Integrals
**Q1. Solve: \int_0^{z=10} \int_0^{y=7}\int_0^{x=3} 20 \ dx \ dy \ dz
**Q2. Solve: \int_0^{z=8} \int_0^{y=9} \int_0^{x=2} k \ dx \ dy \ dz
**Q3. Solve: \int_0^{z=2}\int_0^{y=2}\int_0^{x=2} dx \ dy \ dz
**Q4. Solve: \int_0^{z=6} \int_0^{y=8} \int_0^{x=7} k dx dy dz
**Q5. Solve: \int_0^{z=5} \int_0^{y=10} \int_0^{x=5} 40k \ dx \ dy \ dz