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The multiple integral is a kind of ''definite'' ''' Integral ''' extended to Functions of more than one real Variable (i.e. $f(x,y)\backslash ,\backslash !$ or $f(x,y,z)\backslash ,\backslash !$). MULTIPLE INTEGRALS ARE NOT THE SAME AS ITERATED INTEGRALS It is easy to confuse the concepts of ''multiple integral'' and Iterated Integral , especially since the same notation is often used for either concept. The notation :$\backslash int\_0^1\backslash int\_0^1\; f(x,y)\backslash ,dy\backslash ,dx$ in some cases means an iterated integral rather than a true double integral. In an iterated integral, the outer integral :$\backslash int\_0^1\; \backslash cdots\; \backslash ,\; dx$ is the integral with respect to ''x'' of the following function of ''x'': :$g(x)=\backslash int\_0^1\; f(x,y)\backslash ,dy.$ 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 either of the two iterated integrals (either "''dy dx''" or "''dx dy''"; see Fubini's Theorem ) 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 :$\backslash int\_0^1\backslash int\_0^1\; f(x,y)\backslash ,dy\backslash ,dx$ eq \int_0^1\int_0^1 f(x,y)\,dx\,dy. For an elementary example (doable by the methods of first-year Calculus ), see Examples Of Fubini's Theorem . This is an instance of rearrangement of a Conditionally Convergent integral. The notation :$\backslash int\_\{[0,1]\; imes[0,1]\}\; f(x,y)\backslash ,dx\backslash ,dy$ may be used if one wishes to be emphatic about intending a double integral rather than an iterated integral. MULTIPLE INTEGRALS If, conceptually, the definite integral for functions of one variable represent the Area of the region on the graph between the curve defined by the function and the ''x''-axis, that for functions of two variables ( Double Integral ) represent the measure of the space between the surface defined by the function and the plane which contains its Domain , so they describe not an area but a Volume of a particular solid called Cylindroid ; you obtain the same value if you consider the ''triple integrals'' (functions of three variables) calculated with the constant ''f''(''x'', ''y'', ''z'') = 1. If the number of variables is higher you will calculate " Hypervolume s" (volumes of solid of more than three dimensions) that you cannot graph. In that example the volume of the Parallelepiped of sides 4x6x5 is obtainable by two ways: by the Double Integral $\backslash iint\_D\; 5\; \backslash \; dx\backslash ,\; dy$ of the function ''f''(''x'', ''y'') = 5 calculated in the "bi-dimensional interval" ''D'' (region contained in the ''xy''-plane) by the triple integral $\backslash iiint\_\backslash mathrm\{parallelepiped\}\; 1\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz$ of the constant function 1 calculated on the "three-dimensional" interval coinciding with the same parallelepiped; in that case the volume is considered as the "sum" of all the Infinitesimal elements that compose the domain. Because is impossible to calculate the Antiderivative of a function of more than one variable, ''indefinite'' multiple integrals do not exist so they are all ''definite'' integrals. SOME PRACTICAL APPLICATIONS These integrals are used in a lot of Physics applications. In Mechanics the Moment Of Inertia is calculated as a volume integral (that is a triple integral) of the Density weighed with the square of the distance from the axis: :$I\_z\; =\; \backslash int\_V^.\; ho\; r^2\backslash ,\; dV.$ In Electromagnetism the Maxwell's Equations can be written with multiple integrals to calculate the total magnetic and electric fields. In the example the Electric Field produced by a Distribution Of Charges is obtained by a ''triple integral'' of a vector function:

:and obtain the same value.


Normal domains on R³

Is immediate the extension of these formulas to the triple integrals.

''T'' is a domain perpendicular to the ''xy''-plane respect to the α (''x'',''y'',''z'') and β(''x'',''y'',''z'') functions. Then:

:$\backslash iiint\_T\; f(x,y,z)\; \backslash \; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iint\_D\; dx\backslash ,\; dy\; \backslash int\_\{\backslash alpha\; (x,y,z)\}^\{\backslash beta\; (x,y,z)\}\; f(x,y,z)\; \backslash ,\; dz$

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


Change of variables


Often, because the limits of integration are not easily interchangeable (without normality or with complex formulas to integrate) one makes a ''change of variables'' to rewrite the integral on more a "comfortable" region, that is described in simpler way through formulas. To do that, the function must be changed to the new coordinates.

:Example (1-a):
::The function is $f(x,\; y)\; =\; (x-1)^2\; +\backslash sqrt\; y$;
::if you adopt this substitution $x\text{'}\; =\; x-1,\; \backslash \; y=\; y\text{'}\; \backslash ,\; \backslash !$ therefore $x\; =\; x\text{'}\; +\; 1,\; \backslash \; y=y\text{'}\; \backslash ,\backslash !$
::you obtain new function $f(x,y)\_2\; =\; x\text{'}\; ^2\; +\backslash sqrt\; y$.

• Ditto for the dominion because it's delimitated by the original variables that we had transformed before (''x'' and ''y'' in example).


• the Differentials (''dx'' and ''dy'' as an example) depend on Determining Of The Jacobian Matrix containing the partial derivatives of the transformations regarding the variable new (look at as example differentiates them of the transformation in polar coordinates).

There exist three main "kinds" of changes of variable (one in R², two in R³), however is possible to hand with this method so as to to operate the substitution that more thinks good.


Polar coordinates

See Also: coordinates (mathematics)



In R² if the domain as a circular "symmetry" (if it describes a Circular Crown ) and the function has some "particular" characteristics you can apply the ''passate 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 to change the "shape" of the domain and simplify the operations.

The fundamental relation to make the transformation is the follow:

:$f(x,y)\; ightarrow\; f(\; ho\; \backslash \; \backslash cos\; \backslash phi,\; ho\; \backslash \; \backslash sin\; \backslash phi\; ).$

Example (2-a):

:The function is $f(x,y)\; =\; x\; +\; y\backslash ,\backslash !$

:and applying the transformation you get

::$f(\; ho,\; \backslash phi)\; =\; ho\; \backslash cos\; \backslash phi\; +\; ho\; \backslash sin\; \backslash phi\; =\; ho\; \backslash \; (\backslash cos\; \backslash phi\; +\; \backslash sin\; \backslash phi\; ).$

Example (2-b):
:The function is $f(x,y)\; =\; x^2\; +\; y^2\backslash ,\backslash !$
:In this case you get

::$f(\; ho,\; \backslash phi)\; =\; ho^2\; (\backslash cos\; \backslash phi^2\; +\; \backslash sin\; \backslash phi^2)\; =\; ho^2\backslash ,\backslash !$

:using the Pythagorean Trigonometric Identity (very useful to simplify this operations).

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

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

Example (2-d):
:The domain is $D\; =\; \backslash \{\; x^2\; +\; y^2\; \backslash le\; 9,\; \backslash \; x^2\; +\; y^2\; \backslash ge\; 4,\; \backslash \; y\; \backslash ge\; 0\; \backslash \}$, that is the circular crown in the semiplane of positive ''y'' (please see the picture in the example); you note that φ describes a Plane Angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following Rectangle :

::$T\; =\; \backslash \{\; 2\; \backslash le\; ho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi\; \backslash \}$.

The Jacobian Determinant of that transformation is the following:

:
\begin{vmatrix}
\cos \phi & - ho \sin \phi \
\sin \phi & ho \cos \phi
\end{vmatrix} = ho


which has been got 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 ''d''ρ ''d''φ.

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

:$\backslash iint\_D\; f(x,y)\; \backslash \; dx\backslash ,\; dy\; =\; \backslash iint\_T\; f(\; ho\; \backslash cos\; \backslash phi,\; ho\; \backslash sin\; \backslash phi)\; ho\; \backslash ,\; d\; ho\backslash ,\; d\; \backslash phi.$

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

Example (2-e):
:The function is $f(x,y)\; =\; x\backslash ,\backslash !$ 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:

::$f(x,y)\; =\; x\; \backslash longrightarrow\; f(\; ho,\backslash phi)\; =\; ho\; \backslash \; \backslash cos\; \backslash phi.$

:finally let's apply the integration formula:

::$\backslash iint\_D\; x\; \backslash ,\; dx\backslash ,\; dy\; =\; \backslash iint\_T\; ho\; \backslash cos\; \backslash phi\; \backslash \; ho\; \backslash ,\; d\; ho\backslash ,\; d\backslash phi.$

:Once known the intervals you have

::


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:
$f(x,y,z)\; ightarrow\; f(\; ho\; \backslash \; \backslash cos\; \backslash phi,\; ho\; \backslash \; \backslash sin\; \backslash phi,\; z)$

The domain's transformation is easy because graphically only the shape of the base varies while the three-dimensional development follows that of the starting region.

Example (3-a):

:The region is $D\; =\; \backslash \{\; x^2\; +\; y^2\; \backslash le\; 9,\; \backslash \; x^2\; +\; y^2\; \backslash ge\; 4,\; \backslash \; 0\; \backslash le\; z\; \backslash le\; 5\; \backslash \}$ (that is the "tube" whose base is the circular crown of the 2-d example and whose height is 5); if you apply the transformation you'll get this region: $T\; =\; \backslash \{\; 2\; \backslash le\; ho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi,\; \backslash \; 0\; \backslash le\; z\; \backslash le\; 5\; \backslash \}$ (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 for the passage in cylindrical coordinates:

:$\backslash iiint\_D\; f(x,y,z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; f(\; ho\; \backslash cos\; \backslash phi,\; ho\; \backslash sin\; \backslash phi,\; z)\; ho\; \backslash ,\; d\; ho\backslash ,\; d\backslash phi\backslash ,\; dz.$

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

::$T\; =\; \backslash \{\; 0\; \backslash le\; ho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; -5\; \backslash le\; z\; \backslash le\; 5\; \backslash \}.$

:while the function becomes

::$f(\; ho\; \backslash \; \backslash cos\; \backslash phi,\; ho\; \backslash \; \backslash sin\; \backslash phi,\; z)\; =\; ho^2\; +\; z\backslash ,\backslash !$


:Finally you can apply the integration's formula:

::$\backslash iiint\_D\; (x^2\; +\; y^2\; +z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; (\; ho^2\; +\; z)\; ho\; \backslash ,\; d\; ho\backslash ,\; d\backslash phi\backslash ,\; dz;$

:developing the formula you have

::

::


Spherical coordinates


In R³ some domains have a spherical symmetry, so it's possible to determinate the coordinates of every point of the integration's 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:
$f(x,y,z)\; \backslash longrightarrow\; f(\; ho\; \backslash sin\; heta\; \backslash cos\; \backslash phi,\; ho\; \backslash sin\; heta\; \backslash sin\; \backslash phi,\; ho\; \backslash cos\; heta)\backslash ,\backslash !$

Please note that points on ''z'' axis don't have a precise characterization in spherical coordinates, so ''θ'' can varies from 0 to π .

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

Example (4-a):

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

:The Jacobian determinant of this transformation is the following:

::
\begin{vmatrix}
\sin heta \cos \phi & ho \cos heta \cos \phi & - ho \sin heta \sin \phi \
\sin heta \sin \phi & ho \cos heta \sin \phi & ho \sin heta \cos \phi \
\cos heta & - ho \sin heta & 0
\end{vmatrix} = ho^2 \sin heta


:The ''dx dy dz'' differentials therefore are transformed to ρ² sin(θ) ''d''ρ ''d''θ ''d''φ.

:Finally you obtain the final integration formula:

::$\backslash iiint\_D\; f(x,y,z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; f(\; ho\; \backslash sin\; heta\; \backslash cos\; \backslash phi,\; ho\; \backslash sin\; heta\; \backslash sin\; \backslash phi,\; ho\; \backslash cos\; heta)\; ho^2\; \backslash sin\; heta\; \backslash ,\; d\; ho\backslash ,\; d\; heta\backslash ,\; d\backslash phi.$

:It's better to use this method in case of spherical domains and in case of functions that you can easily simplify by the first fundamental relation of trigonometry extended in '''R'''³ (please see the 4-b example); in the other cases it's better to use the passage in cylindrical coordinates (please see the 4-c example).

Example (4-b):



:Its transformation is very easy:

::$f(\; ho\; \backslash sin\; heta\; \backslash cos\; \backslash phi,\; ho\; \backslash sin\; heta\; \backslash sin\; \backslash phi,\; ho\; \backslash cos\; heta)\; =\; ho^2,\backslash ,$

:while we know the intervals of the transformed region ''T'' from ''D'':


::$(0\; \backslash le\; ho\; \backslash le\; 4,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; 0\; \backslash le\; heta\; \backslash le\; \backslash pi).\backslash ,$

:Let's apply therefore the integration's formula:

::$\backslash iiint\_D\; (x^2\; +\; y^2\; +z^2)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; ho^2\; \backslash \; ho^2\; \backslash sin\; heta\; \backslash ,\; d\; ho\backslash ,\; d\; heta\backslash ,\; d\backslash phi,$

:and developing we get

::

::

Example (4-c):
:The domain ''D'' is the ball with center in the origin and radius ''3a'' ($D\; =\; x^2\; +\; y^2\; +\; z^2\; \backslash le\; 9a^2\; \backslash ,\backslash !$) and $f(x,y,z)\; =\; x^2\; +\; y^2\backslash ,\backslash !$ is the function to integrate.

:Looking at the domain seems convenient adopting the passage in spherical coordinates, in fact are immediate the intervals of the variables that delimit the new ''T'' region:

::$0\; \backslash le\; ho\; \backslash le\; 3a,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; 0\; \backslash le\; heta\; \backslash le\; \backslash pi.\backslash ,$

:However applying the transformation we get

::$f(x,y,z)\; =\; x^2\; +\; y^2\; \backslash longrightarrow\; ho^2\; \backslash sin^2\; heta\; \backslash cos^2\; \backslash phi\; +\; ho^2\; \backslash sin^2\; heta\; \backslash sin^2\; \backslash phi\; =\; ho^2\; \backslash sin^2\; heta$.

:Applying the formula for integration it would be obtained:

::$\backslash iiint\_T\; ho^2\; \backslash sin^2\; heta\; ho^2\; \backslash sin\; heta\; \backslash ,\; d\; ho\backslash ,\; d\; heta\backslash ,\; d\backslash phi\; =\; \backslash iiint\_T\; ho^4\; \backslash sin^3\; heta\; \backslash ,\; d\; ho\backslash ,\; d\; heta\backslash ,\; d\backslash phi$

:very hard to solve. This problem will be solved by using the passage in cylindrical coordinates. The new ''T'' intervals are

::$0\; \backslash le\; ho\; \backslash le\; 3a,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; -\; \backslash sqrt\{9a^2\; -\; ho^2\}\; \backslash le\; z\; \backslash le\; \backslash sqrt\{9a^2\; -\; ho^2\};$

:the ''z'' interval has been obtained by dividing the ball in two semispheres simply by solving the Inequality from the formula of ''D'' (and then directly transforming ''x² + y²'' in ''ρ²''). The new function is simply ''ρ²''. Therefore let's apply the integration formula

::$\backslash iiint\_T\; ho^2\; ho\; \backslash \; d\; ho\; d\; \backslash phi\; dz$.

:Then we get

::$\backslash int\_0^\{2\; \backslash pi\}\; d\backslash phi\; \backslash int\_0^\{3a\}\; ho^3\; d\; ho\; \backslash int\_\{-\; \backslash sqrt\{9a^2\; -\; ho^2\}\; \}^\{\backslash sqrt\{9\; a^2\; -\; ho^2\}\; \}\backslash ,\; dz\; =\; 2\; \backslash pi\; \backslash int\_0^\{3a\}\; 2\; ho^3\; \backslash sqrt\{9\; a^2\; -\; ho^2\}\; \backslash ,\; d\; ho.$

:Now let's apply the transformation

::

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

::$-\; 2\; \backslash pi\; \backslash int\_\{9\; a^2\}^\{0\}\; ho^2\; \backslash sqrt\{t\}\backslash ,\; dt$

:because $ho^2\; =\; 9\; a^2\; -\; t\backslash ,\backslash !$, we get

::$-2\; \backslash pi\; \backslash int\_\{9\; a^2\}^0\; (9\; a^2\; -\; t)\; \backslash sqrt\{t\}\backslash ,\; dt,$

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

::

::

: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 - CALCULATIONS OF VOLUME


Thanks to the methods previously described are possible to demonstrate the value of the volume of some solid ones.

• Cylinder : Considering like dominion the circular base of radius ''R'' and like function the constant of the height ''h'', applies the passage in polar coordinates directly:

::

• height = $\backslash pi\; R^2\; \backslash cdot\; h$


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

::

• Tetrahedron (triangular Pyramid to base 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''.

:: $\backslash mathrm\{Volume\}\; =\; \backslash int\_0^\backslash ell\; dx\; \backslash int\_0^\{\backslash ell-x\; \}\backslash ,\; dy\; \backslash int\_0^\{\backslash ell-x-y\; \}\backslash ,\; dz\; =\; \backslash int\_0^\backslash ell\; dx\; \backslash int\_0^\{\backslash ell-x\; \}\; (\backslash ell\; -\; x\; -\; y)\backslash ,\; dy$

::

::

: ''Verification'': Volume = base area × height/3 =


MULTIPLE IMPROPER INTEGRAL

In case of unbounded domains or unbounded integrands near the frontier of the dominion you have the double Improper Integral or the triple improper integral.


SEE ALSO

• Integral


• Main analysis' theorems that uses multiple integrals:


• --- Divergence Theorem


• --- Stokes Theorem


• --- Green's Theorem

BIBLIOGRAPHY

''Sources verified on: Robert A. Adams - Calcolo differenziale 2, Funzioni di più variabili ISBN 8840810242''