Definitions

# Multiple integral

The multiple integral is a type of definite integral extended to functions of more than one real variable, for example, f(x, y) or f(x, y, z).

## Introduction

Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function and the plane which contains its domain. (Note that the same volume can be obtained via the triple integral — the integral of a function in three variables — of the constant function f(x, y, z) = 1 over the above-mentioned region between the surface and the plane.) If there are more variables, a multiple integral will yield hypervolumes of multi-dimensional functions.

Multiple integration of a function in n variables: f(x1, x2, …, xn) over a domain D is most commonly represented by nesting integral signs in the reverse order of execution (the leftmost integral sign is computed last) proceeded by the function and integrand arguments in proper order (the rightmost argument is computed last). The domain of integration is either represented symbolically for every integrand over each integral sign, or is often abbreviated by a variable at the rightmost integral sign:


int ldots int_mathbf{D};f(x_1,x_2,ldots,x_n) ;mathbf{d}x_1 !ldotsmathbf{d}x_n

Since it is impossible to calculate the antiderivative of a function of more than one variable, indefinite multiple integrals do not exist. Therefore all multiple integrals are definite integrals.

## Examples

For example, the volume of the parallelepiped of sides 4 × 6 × 5 may be obtained in two ways:

• By the double integral

$iint_D 5 dx, dy$
of the function f(x, y) = 5 calculated in the region D in the xy-plane which is the base of the parallelepiped.

• By the triple integral

$iiint_mathrm\left\{parallelepiped\right\} 1 , dx, dy, dz$
of the constant function 1 calculated on the parallelepiped itself.

## Mathematical definition

Let n be an integer greater than 1. Consider a so-called half-open n-dimensional rectangle (from here on simply called rectangle). For a plane, n = 2, and the multiple integral is just a double integral.

$T=\left[a_1, b_1\right)times \left[a_2, b_2\right)timescdots times \left[a_n, b_n\right)subset mathbb R^n.$

Divide each interval [ai, bi) into a finite number of non-overlapping subintervals, with each subinterval closed at the left end, and open at the right end. Denote such a subinterval by Ii. Then, the family of subrectangles of the form

$C=I_1times I_2times cdotstimes I_n$

is a partition of T that is, the subrectangles C are non-overlapping and their union is T. The diameter of a subrectangle C is, by definition, the largest of the lengths of the intervals whose product is C, and the diameter of a given partition of T is defined as the largest of the diameters of the subrectangles in the partition.

Let f : TR be a function defined on a rectangle T. Consider a partition

$T=C_1cup C_2cup cdots cup C_m$

of T defined as above, where m is a positive integer. A Riemann sum is a sum of the form

$sum_\left\{k=1\right\}^m f\left(P_k\right), operatorname\left\{m\right\}\left(C_k\right)$

where for each k the point Pk is in Ck and m(Ck) is the product of the lengths of the intervals whose Cartesian product is Ck.

The function f is said to be Riemann integrable if the limit

$S = lim_\left\{delta to 0\right\} sum_\left\{k=1\right\}^m f\left(P_k\right), operatorname\left\{m\right\}, \left(C_k\right)$

exists, where the limit is taken over all possible partitions of T of diameter at most δ. If f is Riemann integrable, S is called the Riemann integral of f over T and is denoted

$int_T!f\left(x\right),dx.$

The Riemann integral of a function defined over an arbitrary bounded n-dimensional set can be defined by extending that function to a function defined over a half-open rectangle whose values are zero outside the domain of the original function. Then, the integral of the original function over the original domain is defined to be the integral of the extended function over its rectangular domain, if it exists.

In what follows the Riemann integral in n dimensions will be called multiple integral.

### Properties

Multiple integrals have many of the same properties of integrals of functions of one variable (linearity, additivity, monotonicity, etc.). Moreover, just as in one variable, one can use the multiple integral to find the average of a function over a given set. More specifically, given a set DRn and an integrable function f over D, the average value of f over its domain is given by

$bar\left\{f\right\} = frac\left\{1\right\}\left\{m\left(D\right)\right\} int_D f\left(x\right), dx,$

where m(D) is the measure of D.

### Particular cases

In the case of TR2, the integral

$ell = iint_T f\left(x,y\right), dx, dy$

is the double integral of f on T, and if TR3 the integral

$ell = iiint_T f\left(x,y,z\right), dx, dy, dz$

is the triple integral of f on T.

Notice that, by convention, the double integral has two integral signs, and the triple integral has three; this is just notational convenience, and comes handy when computing a multiple integral as an iterated integral (as shown later in the article).

## Methods of integration

The resolution of problems with multiple integrals consists in most of cases in finding the way to reduce the multiple integral to a series of integrals of one variable, each being directly solvable.

### Direct examination

Sometimes, it is possible to obtain the result of the integration without any direct calculations.

#### Constants

In the case of a constant function, the result is straightforward: simply multiply the measure by the constant function c. If c = 1, and is integrated over a subregion of R2 the product gives the area of the region, while in R3 it is the volume of the region.

• For example:

$D = \left\{ \left(x,y\right) in mathbb\left\{R\right\}^2 : 2 le x le 4 ; 3 le y le 6 \right\}$ and $f\left(x,y\right) = 2,!$

Let us integrate f over D:

$int_3^6 int_2^4 2 dx, dy = mbox\left\{area\right\}\left(D\right) cdot 2 = \left(2 cdot 3\right) cdot 2 = 12.$

#### Use of the possible symmetries

In the case of a domain where there are symmetries respecting at least one of the axes and where the function has at least one parity in respect to a variable, the integral becomes null (the sum of opposite and equal values is null).

It is sufficient that - in functions on Rn - the dependent variable is odd with the symmetric axis.

• Example (1):

Given f(x, y) = 2 sin x − 3y3 + 5 and T = x2 + y2 ≤ 1 the integration area (a disc with radius 1 centered in the origin of the axes, boundary included).

Using the property of linearity, the integral can be decomposed in three pieces:

$iint_T \left(2 sin x - 3 y^3 + 5\right) dx , dy = iint_T 2 sin x dx , dy - iint_T 3 y^3 dx dy + iint_T 5 dx dy$

2  sin x and 3y3 are both odd functions and moreover it is evident that the T disc has a symmetry for the x and even the y axis; therefore the only contribution to the final result of the integral is that of the constant function 5 because the other two pieces are null.

• Example (2):

Consider the function f(x, y, z) = x   exp(y2 + z2) and as integration region the sphere with radius 2 centered in the origin of the axes T = x2 + y2 + z2 ≤ 4. The "ball" is symmetric about all three axes, but it is sufficient to integrate with respect to x-axis to show that the integral is 0, because the function is an odd function of that variable.

### Formulae of reduction

Formulae of reduction use the concept of simple domain to make possible the decomposition of the multiple integral as a product of other one-variable integrals. These have to be solved from the right to the left considering the other variables as constants (which is the same procedure as the calculus of partial derivatives).

#### Normal domains on R2

##### x-axis
If D is a measurable domain perpendicular to the x-axis and $f: D longrightarrow mathbb\left\{R\right\}$ is a continuous function; then α(x) and β(x) (defined in the [a,b] interval) are the two functions that determine D. Then:

$iint_T f\left(x,y\right) dx, dy = int_a^b dx int_\left\{ alpha \left(x\right)\right\}^\left\{ beta \left(x\right)\right\} f\left(x,y\right), dy.$
##### y-axis
If D is a measurable domain perpendicular to the y-axis and $f: D longrightarrow mathbb\left\{R\right\}$ is a continuous function; then α(y) and β(y) (defined in the [a,b] interval) are the two functions that determine D. Then:

$iint_T f\left(x,y\right) dx, dy = int_a^b dy int_\left\{ alpha \left(y\right)\right\}^\left\{ beta \left(y\right)\right\} f\left(x,y\right), dx.$
##### Example

Consider this region: $D = \left\{ \left(x,y\right) : x ge 0, y le 1, y ge x^2 \right\}$ (please see the graphic in the example). Calculate

$iint_D \left(x+y\right) , dx , dy.$

This domain is perpendicular to both the x and to the y axes. To apply the formulas you have to find the functions that determine D and its definition's interval.
In this case the two functions are:

$alpha \left(x\right) = x^2,!$ and $beta \left(x\right) = 1,!$

while the interval is given from the intersections of the functions with $x = 0,!$, so the interval is $\left[a,b\right] = \left[0,1\right],!$ (normality has been chosen with respect to the x axis for a better visual understanding).

It's now possible to apply the formulas:

$iint_D \left(x+y\right) , dx , dy = int_0^1 dx int_\left\{x^2\right\}^1 \left(x+y\right) , dy = int_0^1 dx left\left[xy + frac\left\{y^2\right\}\left\{2\right\} right\right]^1_\left\{x^2\right\}$

(at first the second integral is calculated considering x as a constant). The remaining operations consist of applying the basic techniques of integration:

$int_0^1 left\left[xy + frac\left\{y^2\right\}\left\{2\right\} right\right]^1_\left\{x^2\right\} , dx = int_0^1 left\left(x + frac\left\{1\right\}\left\{2\right\} - x^3 - frac\left\{x^4\right\}\left\{2\right\} right\right) dx = cdots = frac\left\{13\right\}\left\{20\right\}.$

If we choose the normality in respect to the y axis we could calculate

$int_0^1 dy int_0^\left\{sqrt\left\{y\right\}\right\} \left(x+y\right) , dx.$

and obtain the same value.

#### Normal domains on R3

The extension of these formulae to triple integrals should be apparent:

T is a domain perpendicular to the xy-plane respect to the α (x,y,z) and β(x,y,z) functions. Then:

$iiint_T f\left(x,y,z\right) dx, dy, dz = iint_D dx, dy int_\left\{alpha \left(x,y,z\right)\right\}^\left\{beta \left(x,y,z\right)\right\} f\left(x,y,z\right) , dz$

(this definition is the same for the other five normality cases on R3).

### Change of variables

The limits of integration are often not easily interchangeable (without normality or with complex formulae to integrate), one makes a change of variables to rewrite the integral in a more "comfortable" region, which can be described in simpler formulae. To do so, the function must be adapted to the new coordinates.

Example (1-a):
The function is $f\left(x, y\right) = \left(x-1\right)^2 +sqrt y$;
if one adopts this substitution $x\text{'} = x-1, y\text{'}= y , !$ therefore $x = x\text{'} + 1, y=y\text{'} ,!$
one obtains the new function $f_2\left(x,y\right) = \left(x\text{'}\right)^2 +sqrt y$.

• Similarly for the domain because it is delimited by the original variables that were transformed before (x and y in example).
• the differentials dx and dy transform via the determinant of the Jacobian matrix containing the partial derivatives of the transformations regarding the new variable (consider, as an example, the differential transformation in polar coordinates).

There exist three main "kinds" of changes of variable (one in R2, two in R3); however, a suitable substitution can be found using the same principle in a more general way.

#### Polar coordinates

In R2 if the domain has a circular "symmetry" and the function has some "particular" characteristics you can apply the transformation to polar coordinates (see the example in the picture) which means that the generic points P(x,y) in cartesian coordinates switch to their respective points in polar coordinates. That allows one to change the "shape" of the domain and simplify the operations.

The fundamental relation to make the transformation is the following:

$f\left(x,y\right) rightarrow f\left(rho cos phi,rho sin phi \right).$

Example (2-a):

The function is $f\left(x,y\right) = x + y,!$

and applying the transformation one obtains

$f\left(rho, phi\right) = rho cos phi + rho sin phi = rho \left(cos phi + sin phi \right).$

Example (2-b):

The function is $f\left(x,y\right) = x^2 + y^2,!$
In this case one has:

$f\left(rho, phi\right) = rho^2 \left(cos^2 phi + sin^2 phi\right) = rho^2,!$

using the Pythagorean trigonometric identity (very useful to simplify this operation).

The transformation of the domain is made by defining the radius' crown length and the amplitude of the described angle to define the ρ, φ intervals starting from x, y.

Example (2-c):

The domain is $D = x^2 + y^2 le 4,!$, that is a circumference of radius 2; it's evident that the covered angle is the circle angle, so φ varies from 0 to 2π, while the crown radius varies from 0 to 2 (the crown with the inside radius null is just a circle).

Example (2-d):

The domain is $D = \left\{ x^2 + y^2 le 9, x^2 + y^2 ge 4, y ge 0 \right\}$, that is the circular crown in the positive y half-plane (please see the picture in the example); note that φ describes a plane angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following rectangle:

$T = \left\{ 2 le rho le 3, 0 le phi le pi \right\}$.

The Jacobian determinant of that transformation is the following:

$frac\left\{partial \left(x,y\right)\right\}\left\{partial \left(rho, phi\right)\right\} =$
begin{vmatrix} cos phi & - rho sin phi sin phi & rho cos phi end{vmatrix} = rho

which has been obtained by inserting the partial derivatives of x = ρ cos(φ), y = ρ sin(φ) in the first column respect to ρ and in the second respect to φ, so the dx dy differentials in this transformation becomes ρ dρ dφ.

Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates:

$iint_D f\left(x,y\right) dx, dy = iint_T f\left(rho cos phi, rho sin phi\right) rho , d rho, d phi.$

Please note that φ is valid in the [0, 2π] interval while ρ, which is a measure of a length, can only have positive values.

Example (2-e):

The function is $f\left(x,y\right) = x,!$ and as the domain the same in 2-d example.
From the previous analysis of D we know the intervals of ρ (from 2 to 3) and of φ (from 0 to π). Now let's change the function:

$f\left(x,y\right) = x longrightarrow f\left(rho,phi\right) = rho cos phi.$

finally let's apply the integration formula:

$iint_D x , dx, dy = iint_T rho cos phi rho , drho, dphi.$

Once the intervals are known, you have

$int_0^\left\{pi\right\} int_2^3 rho^2 cos phi d rho d phi = int_0^\left\{pi\right\} cos phi d phi left\left[frac\left\{rho^3\right\}\left\{3\right\} right\right]_2^3 = left\left[sin phi right\right]_0^\left\{pi\right\} left\left(9 - frac\left\{8\right\}\left\{3\right\} right\right) = 0.$

#### Cylindrical coordinates

In R3 the integration on domains with a circular base can be made by the passage in cylindrical coordinates; the transformation of the function is made by the following relation:

$f\left(x,y,z\right) rightarrow f\left(rho cos phi,rho sin phi, z\right)$

The domain transformation can be graphically attained, because only the shape of the base varies, while the height follows the shape of the starting region.

Example (3-a):

The region is $D = \left\{ x^2 + y^2 le 9, x^2 + y^2 ge 4, 0 le z le 5 \right\}$ (that is the "tube" whose base is the circular crown of the 2-d example and whose height is 5); if the transformation is applied, this region is obtained: $T = \left\{ 2 le rho le 3, 0 le phi le pi, 0 le z le 5 \right\}$ (that is the parallelepiped whose base is the rectangle in 2-d example and whose height is 5).

Because the z component is unvaried during the transformation, the dx dy dz differentials vary as in the passage in polar coordinates: therefore, they become ρ dρ dφ dz.

Finally, it is possible to apply the final formula to cylindrical coordinates:

$iiint_D f\left(x,y,z\right) , dx, dy, dz = iiint_T f\left(rho cos phi, rho sin phi, z\right) rho , drho, dphi, dz.$

This method is convenient in case of cylindrical or conical domains or in regions where is easy to individuate the z interval and even transform the circular base and the function.

Example (3-b):

The function is $f\left(x,y,z\right) = x^2 + y^2 + z,!$ and as integration domain this cylinder: $D = \left\{ x^2 + y^2 le 9, -5 le z le 5 \right\}$.
The transformation of D in cylindrical coordinates is the following:

$T = \left\{ 0 le rho le 3, 0 le phi le 2 pi, -5 le z le 5 \right\}.$

while the function becomes

$f\left(rho cos phi,rho sin phi, z\right) = rho^2 + z,!$

Finally you can apply the integration's formula:

$iiint_D \left(x^2 + y^2 +z\right) , dx, dy, dz = iiint_T \left(rho^2 + z\right) rho , drho, dphi, dz;$

developing the formula you have

$int_\left\{-5\right\}^5 dz int_0^\left\{2 pi\right\} dphi int_0^3 \left(rho^3 + rho z \right), drho = 2 pi int_\left\{-5\right\}^5 left\left[frac\left\{rho^4\right\}\left\{4\right\} + frac\left\{rho^2 z\right\}\left\{2\right\} right\right]_0^3 , dz$

$= 2 pi int_\left\{-5\right\}^5 left\left(frac\left\{81\right\}\left\{4\right\} + frac\left\{9\right\}\left\{2\right\} zright\right), dz = cdots = 855 pi.$

#### Spherical coordinates

In R3 some domains have a spherical symmetry, so it's possible to specify the coordinates of every point of the integration region by two angles and one distance. It's possible to use therefore the passage in spherical coordinates; the function is transformed by this relation:

$f\left(x,y,z\right) longrightarrow f\left(rho cos theta sin phi, rho sin theta sin phi, rho cos phi\right),!$

Note that points on z axis do not have a precise characterization in spherical coordinates, so $phi$ can vary between 0 to π .

The better integration domain for this passage is obviously the sphere.

Example (4-a):

The domain is $D = x^2 + y^2 + z^2 le 16$ (sphere with radius 4 and center in the origin); applying the transformation you get this region: $T = \left\{ 0 le rho le 4, 0 le phi le pi, 0 le theta le 2 pi \right\}.$

The Jacobian determinant of this transformation is the following:

$frac\left\{partial \left(x,y,z\right)\right\}\left\{partial \left(rho, theta, phi\right)\right\} =$
begin{vmatrix} cos theta sin phi & - rho sin theta sin phi & rho cos theta cos phi sin theta sin phi & rho cos theta sin phi & rho sin theta cos phi cos phi & 0 & - rho sin phi end{vmatrix} = rho^2 sin phi

The dx dy dz differentials therefore are transformed to ρ2 sin(φ) dρ dθ dφ.

Finally you obtain the final integration formula:

$iiint_D f\left(x,y,z\right) , dx, dy, dz = iiint_T f\left(rho sin theta cos phi, rho sin theta sin phi, rho cos theta\right) rho^2 sin phi , drho, dtheta, dphi.$

It's better to use this method in case of spherical domains and in case of functions that can be easily simplified, by the first fundamental relation of trigonometry, extended in R3 (please see example 4-b); in other cases it can be better to use cylindrical coordinates (please see example 4-c).

$iiint_T f\left(a,b,c\right) rho^2 sin phi , drho, dtheta, dphi.$

Note that the extra $rho^2$ and $sin phi$ come from the Jacobian.

Note that in the following examples the roles of φ and θ have been reversed.

Example (4-b):

D is the same region of the 4-a example and $f\left(x,y,z\right) = x^2 + y^2 + z^2,!$ is the function to integrate.

Its transformation is very easy:

$f\left(rho sin theta cos phi, rho sin theta sin phi, rho cos theta\right) = rho^2,,$

while we know the intervals of the transformed region T from D:

$\left(0 le rho le 4, 0 le phi le 2 pi, 0 le theta le pi\right).,$

Let's therefore apply the integration's formula:

$iiint_D \left(x^2 + y^2 +z^2\right) , dx, dy, dz = iiint_T rho^2 rho^2 sin theta , drho, dtheta, dphi,$

and, developing, we get

$iiint_T rho^4 sin theta , drho, dtheta, dphi = int_0^\left\{pi\right\} sin theta ,dtheta int_0^4 rho^4 d rho int_0^\left\{2 pi\right\} dphi = 2 pi int_0^\left\{pi\right\} sin theta left\left[frac\left\{rho^5\right\}\left\{5\right\} right\right]_0^4 , d theta$

$= 2 pi left\left[frac\left\{rho^5\right\}\left\{5\right\} right\right]_0^4 left\left[- cos theta right\right]_0^\left\{pi\right\} = 4 pi cdot frac\left\{1024\right\}\left\{5\right\} = frac\left\{4096 pi\right\}\left\{5\right\}.$

Example (4-c):

The domain D is the ball with center in the origin and radius 3a ($D = x^2 + y^2 + z^2 le 9a^2 ,!$) and $f\left(x,y,z\right) = x^2 + y^2,!$ is the function to integrate.

Looking at the domain, it seems convenient to adopt the passage in spherical coordinates, in fact, the intervals of the variables that delimit the new T region are obviously:

$0 le rho le 3a, 0 le phi le 2 pi, 0 le theta le pi.,$

However, applying the transformation, we get

$f\left(x,y,z\right) = x^2 + y^2 longrightarrow rho^2 sin^2 theta cos^2 phi + rho^2 sin^2 theta sin^2 phi = rho^2 sin^2 theta$.

Applying the formula for integration we would obtain:

$iiint_T rho^2 sin^2 theta rho^2 sin theta , drho, dtheta, dphi = iiint_T rho^4 sin^3 theta , drho, dtheta, dphi$

which is very hard to solve. This problem will be solved by using the passage in cylindrical coordinates. The new T intervals are

$0 le rho le 3a, 0 le phi le 2 pi, - sqrt\left\{9a^2 - rho^2\right\} le z le sqrt\left\{9a^2 - rho^2\right\};$

the z interval has been obtained by dividing the ball in two hemispheres simply by solving the inequality from the formula of D (and then directly transforming x2 + y2 in ρ2). The new function is simply ρ2. Applying the integration formula

$iiint_T rho^2 rho d rho d phi dz$.

Then we get

$int_0^\left\{2 pi\right\} dphi int_0^\left\{3a\right\} rho^3 drho int_\left\{- sqrt\left\{9a^2 - rho^2\right\} \right\}^\left\{sqrt\left\{9 a^2 - rho^2\right\} \right\}, dz = 2 pi int_0^\left\{3a\right\} 2 rho^3 sqrt\left\{9 a^2 - rho^2\right\} , drho.$

Now let's apply the transformation

$9 a^2 - rho^2 = t,! longrightarrow dt = -2 rho, drho longrightarrow drho = frac\left\{d t\right\}\left\{- 2 rho\right\},!$

(the new intervals become $0, 3a longrightarrow 9 a^2, 0$). We get

$- 2 pi int_\left\{9 a^2\right\}^\left\{0\right\} rho^2 sqrt\left\{t\right\}, dt$

because $rho^2 = 9 a^2 - t,!$, we get

$-2 pi int_\left\{9 a^2\right\}^0 \left(9 a^2 - t\right) sqrt\left\{t\right\}, dt,$

after inverting the integration's bounds and multiplying the terms between parenthesis, it is possible to decompose the integral in two parts that can be directly solved:

$2 pi left\left[int_0^\left\{9 a^2\right\} 9 a^2 sqrt\left\{t\right\} , dt - int_0^\left\{9 a^2\right\} t sqrt\left\{t\right\} , dtright\right] = 2 pi left\left[9 a^2 frac\left\{2\right\}\left\{3\right\} t^\left\{ frac\left\{3\right\}\left\{2\right\} \right\} - frac\left\{2\right\}\left\{5\right\} t^\left\{ frac\left\{5\right\}\left\{2\right\}\right\} right\right]_0^\left\{9 a^2\right\}$

$= 2 cdot 27 pi a^5 \left(6 - frac\left\{2\right\}\left\{5\right\} \right) = 54 pi frac\left\{28\right\}\left\{5\right\} a^5 = frac\left\{1512 pi\right\}\left\{5\right\} a^5.$

Thanks to the passage in cylindrical coordinates it was possible to reduce the triple integral to an easier one-variable integral.

See also the differential volume entry in nabla in cylindrical and spherical coordinates.

## Example of mathematical applications - Computing a volume

Thanks to the methods previously described it is possible to demonstrate the value of the volume of some solid volumes.

• Cylinder: Consider the domain as the circular base of radius R and the function as a constant of the height h. It is possible to write this in polar coordinates like so:

$mathrm\left\{Volume\right\} = int_0^\left\{2 pi \right\} d phi int_0^R h rho d rho = h 2 pi left\left[frac\left\{rho^2\right\}\left\{2 \right\}right\right]_0^R = pi R^2 h$

Verification: Volume = base area * height = $pi R^2 cdot h$

• Sphere: Is a ready demonstration of applying the passage in spherical coordinates of the integrated constant function 1 on the sphere of the same radius R:

$mathrm\left\{Volume\right\} = int_0^\left\{2 pi \right\}, d phi int_0^\left\{ pi \right\} sin theta, d theta int_0^R rho^2, d rho = 2 pi int_0^\left\{ pi \right\} sin theta frac\left\{R^3\right\}\left\{3 \right\}, d theta = frac\left\{2\right\}\left\{3 \right\} pi R^3 \left[- cos theta\right]_0^\left\{ pi \right\} = frac\left\{4\right\}\left\{3 \right\} pi R^3.$

• Tetrahedron (triangular pyramid or 3-simplex): The volume of the tetrahedron with apex in the origin and chines of length l carefully lay down to you on the three cartesian axes can be calculated through the reduction formulas considering, as an example, normality regarding the plan xy and to axis x and like function constant 1.

$mathrm\left\{Volume\right\} = int_0^ell dx int_0^\left\{ell-x \right\}, dy int_0^\left\{ell-x-y \right\}, dz = int_0^ell dx int_0^\left\{ell-x \right\} \left(ell - x - y\right), dy$

$= int_0^ell \left(ell^2 - 2ell x + x^2 - frac\left\{ \left(ell-x\right)^2 \right\}\left\{2 \right\}\right), dx = ell^3 - ell ell^2 + frac\left\{ell^3\right\}\left\{3 \right\} - left\left[frac\left\{ell^2\right\}\left\{2 \right\} - ell x + frac\left\{x^2\right\}\left\{2 \right\}right\right]_0^ell =$

$= frac\left\{ell^3\right\}\left\{3 \right\} - frac\left\{ell^3\right\}\left\{6 \right\} = frac\left\{ell^3\right\}\left\{6\right\}$

Verification: Volume = base area × height/3 = $frac\left\{ell^2\right\}\left\{2 \right\} cdot ell/3 = frac\left\{ell^3\right\}\left\{6\right\}.$

## Multiple improper integral

In case of unbounded domains or functions not bounded near the boundary of the domain, we have to introduce the double improper integral or the triple improper integral.

## Multiple integrals and iterated integrals

Fubini's theorem states that if
that is, the integral is absolutely convergent, then the multiple integral will give the same result as the iterated integral,
$int_\left\{Atimes B\right\} f\left(x,y\right),d\left(x,y\right)=int_Aleft\left(int_B f\left(x,y\right),dyright\right),dx=int_Bleft\left(int_A f\left(x,y\right),dxright\right),dy.$
In particular this will occur if $|f\left(x,y\right)|$ is a bounded function and A and B are bounded sets.

If the integral is not absolutely convergent, care is needed not to confuse the concepts of multiple integral and iterated integral, especially since the same notation is often used for either concept. The notation

$int_0^1int_0^1 f\left(x,y\right),dy,dx$

means, in some cases, an iterated integral rather than a true double integral. In an iterated integral, the outer integral

$int_0^1 cdots , dx$

is the integral with respect to x of the following function of x:

$g\left(x\right)=int_0^1 f\left(x,y\right),dy.$

A double integral, on the other hand, is defined with respect to area in the xy-plane. If the double integral exists, then it is equal to each of the two iterated integrals (either "dy dx" or "dx dy") and one often computes it by computing either of the iterated integrals. But sometimes the two iterated integrals exist when the double integral does not, and in some such cases the two iterated integrals are different numbers, i.e., one has

$int_0^1int_0^1 f\left(x,y\right),dy,dx neq int_0^1int_0^1 f\left(x,y\right),dx,dy.$

This is an instance of rearrangement of a conditionally convergent integral.

The notation

$int_\left\{\left[0,1\right]times\left[0,1\right]\right\} f\left(x,y\right),dx,dy$

may be used if one wishes to be emphatic about intending a double integral rather than an iterated integral.

## Some practical applications

These integrals are used in many applications in physics.

In mechanics the moment of inertia is calculated as volume integral (that is a triple integral) of the density weighed with the square of the distance from the axis:

$I_z = iiint_V rho r^2, dV.$

In electromagnetism, Maxwell's equations can be written by means of multiple integrals to calculate the total magnetic and electric fields. In the following example, the electric field produced by a distribution of charges is obtained by a triple integral of a vector function:

$vec E = frac \left\{1\right\}\left\{4 pi epsilon_0\right\} iiint frac \left\{vec r - vec r\text{'}\right\}\left\{left | vec r - vec r\text{'} right |^3\right\} rho \left(vec r\text{'}\right), operatorname\left\{d\right\}^3 r\text{'}.$