
This article has multiple issues. Please help improve it or discuss these issues on the talk page. 
 This article needs attention from an expert on the subject. Please add a reason or a talk parameter to this template to explain the issue with the article. Consider associating this request with a WikiProject. (May 2010) 
 
Calculus 



Integral calculus
 Definitions

 Integration by







The multiple integral is a generalization of the definite integral to functions of more than one real variable, for example, f(x, y) or f(x, y, z). Integrals of a function of two variables over a region in R^{2} are called double integrals.
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 xaxis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. (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 abovementioned region between the surface and the plane.) If there are more variables, a multiple integral will yield hypervolumes of multidimensional functions.
Multiple integration of a function in n variables: f(x_{1}, x_{2}, ..., x_{n}) over a domain D is most commonly represented by nested integral signs in the reverse order of execution (the leftmost integral sign is computed last), followed by the function and integrand arguments in proper order (the integral with respect to the rightmost argument is computed last). The domain of integration is either represented symbolically for every argument over each integral sign, or is abbreviated by a variable at the rightmost integral sign:
 $\backslash int\; \backslash cdots\; \backslash int\_\backslash mathbf\{D\}\backslash ;f(x\_1,x\_2,\backslash ldots,x\_n)\; \backslash ;dx\_1\; \backslash !\backslash cdots\; dx\_n$
Since the concept of an antiderivative is only defined for functions of a single real variable, the usual definition of the indefinite integral does not immediately extend to the multiple integral.
Mathematical definition
For n > 1, consider a socalled "halfopen" ndimensional hyperrectangular domain T, defined as:
 $T=\backslash left\; [\; a\_1,\; b\_1\; \backslash right\; )\; \backslash times\; \backslash left\; [\; a\_2,\; b\_2\; \backslash right\; )\; \backslash times\; \backslash cdots\; \backslash times\; \backslash left\; [\; a\_n,\; b\_n\; \backslash right\; )\; \backslash subseteq\; \backslash mathbf\{R\}^n.$
Partition each interval [a_{j}, b_{j}) into a finite family I_{j} of nonoverlapping subintervals i_{jα}, with each subinterval closed at the left end, and open at the right end.
Then the finite family of subrectangles C given by
 $C=I\_1\backslash times\; I\_2\backslash times\; \backslash cdots\; \backslash times\; I\_n$
is a partition of T; that is, the subrectangles C_{k} are nonoverlapping and their union is T.
Let f : T → R be a function defined on T. Consider a partition C of T as defined above, such that C is a family of m subrectangles C_{m} and
 $T=C\_1\backslash cup\; C\_2\backslash cup\; \backslash cdots\; \backslash cup\; C\_m$
We can approximate the total nthdimensional volume bounded below by T and above by f with the following Riemann sum:
 $\backslash sum\_\{k=1\}^m\; f(P\_k)\backslash ,\; \backslash operatorname\{m\}(C\_k)$
where P_{k} is a point in C_{k} and m(C_{k}) is the product of the lengths of the intervals whose Cartesian product is C_{k}, otherwise known as the measure of C_{k}.
The diameter of a subrectangle C_{k} is the largest of the lengths of the intervals whose Cartesian product is C_{k}. The diameter of a given partition of T is defined as the largest of the diameters of the subrectangles in the partition. Intuitively, as the diameter of the partition C is restricted smaller and smaller, the number of subrectangles m gets larger, and the measure m(C_{k}) of each subrectangle grows smaller. The function f is said to be Riemann integrable if the limit
 $S=\backslash lim\_\{\backslash delta\; \backslash to\; 0\}\; \backslash sum\_\{k=1\}^m\; f(P\_k)\backslash ,\; \backslash operatorname\{m\}\backslash ,\; (C\_k)$
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
 $\backslash int\; \backslash cdots\; \backslash int\_T\backslash ;f(x\_1,x\_2,\backslash ldots,x\_n)\; \backslash ;dx\_1\; \backslash !\backslash cdots\; dx\_n$
Frequently this notation is abbreviated as
 $\backslash int\_T\backslash !f(\backslash mathbf\{x\})\backslash ,d\backslash mathbf\{x\}.$
where x represents the ntuple (x_{1}, ... x_{n}) and dx is the ndimensional volume differential.
The Riemann integral of a function defined over an arbitrary bounded ndimensional set can be defined by extending that function to a function defined over a halfopen 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 properties common to those of integrals of functions of one variable (linearity, commutativity, monotonicity, and so on.). One important property of multiple integrals is that the value of an integral is independent of the order of integrands under certain conditions. This property is popularly known as Fubini's theorem.
Particular cases
In the case of T ⊆ R^{2}, the integral
 $\backslash ell\; =\; \backslash iint\_T\; f(x,y)\backslash ,\; dx\backslash ,\; dy$
is the double integral of f on T, and if T ⊆ R^{3} the integral
 $\backslash ell\; =\; \backslash iiint\_T\; f(x,y,z)\backslash ,\; dx\backslash ,\; dy\backslash ,\; dz$
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 a notational convention which is convenient when computing a multiple integral as an iterated integral, as shown later in this article.
Methods of integration
The resolution of problems with multiple integrals consists, in most of cases, of finding a way to reduce the multiple integral to an iterated integral, a series of integrals of one variable, each being directly solvable. Sometimes, it is possible to obtain the result of the integration by direct examination without any calculations.
Integrating constant functions
When the integrand is a constant function c, the integral is equal to the product of c and the measure of the domain of integration. If c = 1 and the domain is a subregion of R^{2}, the integral gives the area of the region, while if the domain is a subregion of R^{3}, the integral gives the volume of the region.
Example. Let f(x, y) = 2 and
 $D\; =\; \backslash \{\; (x,y)\; \backslash in\; \backslash mathbf\{R\}^2\; \backslash :\; \backslash \; 2\; \backslash le\; x\; \backslash le\; 4\; \backslash ;\; \backslash \; 3\; \backslash le\; y\; \backslash le\; 6\; \backslash \}$
in which case
 $\backslash int\_3^6\; \backslash int\_2^4\; \backslash \; 2\; \backslash \; dx\backslash ,\; dy\; =2\backslash int\_3^6\; \backslash int\_2^4\; \backslash \; 1\; \backslash \; dx\backslash ,\; dy=\; 2\backslash cdot\backslash mbox\{area\}(D)\; =\; (2\; \backslash cdot\; 3)\; \backslash cdot\; 2\; =\; 12$,
since by definition we have:
 $\backslash int\_3^6\; \backslash int\_2^4\; \backslash \; 1\; \backslash \; dx\backslash ,\; dy=\backslash mbox\{area\}(D).$
Use of symmetry
When the domain of integration is symmetric about the origin with respect to at least one of the variables of integration and the integrand is odd with respect to this variable, the integral is equal to zero, as the integrals over the two halves of the domain have the same absolute value but opposite signs. When the integrand is even with respect to this variable, the integral is equal to twice the integral over one half of the domain, as the integrals over the two halves of the domain are equal.
Example 1. Consider the function $f(x,y)\; =\; 2\; \backslash sin(x)3y^3+5$ integrated over the domain
 $T=\backslash left\; \backslash \{\; (\; x,y)\; \backslash in\; \backslash mathbf\{R\}^2\; \backslash :\; \backslash \; x^2+y^2\backslash le\; 1\; \backslash right\; \backslash \},$
a disc with radius 1 centered at the origin with the boundary included.
Using the linearity property, the integral can be decomposed into three pieces:
 $\backslash iint\_T\; (2\backslash sin\; x\; \; 3y^3\; +\; 5)\; \backslash ,\; dx\; \backslash ,\; dy\; =\; \backslash iint\_T\; 2\; \backslash sin\; x\; \backslash ,\; dx\; \backslash ,\; dy\; \; \backslash iint\_T\; 3y^3\; \backslash ,\; dx\; \backslash ,\; dy\; +\; \backslash iint\_T\; 5\; \backslash ,\; dx\; \backslash ,\; dy$
2sin(x) and 3y^{3} 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 integrals 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(y^{2} + z^{2}) and as integration region the sphere with radius 2 centered at the origin,
 $T\; =\; \backslash left\; \backslash \{\; (\; x,y,\; z)\; \backslash in\; \backslash mathbf\{R\}^3\; \backslash :\; \backslash \; x^2+y^2+z^2\; \backslash le\; 4\; \backslash right\; \backslash \}.$
The "ball" is symmetric about all three axes, but it is sufficient to integrate with respect to xaxis to show that the integral is 0, because the function is an odd function of that variable.
Normal domains on R^{2}
This method is applicable to any domain D for which:
 the projection of D onto either the xaxis or the yaxis is bounded by the two values, a and b
 any line perpendicular to this axis that passes between these two values intersects the domain in an interval whose endpoints are given by the graphs of two functions, α and β.
xaxis
If the domain D is normal with respect to the xaxis, and f : D → R is a continuous function; then α(x) and β(x) (defined on the interval [a, b]) are the two functions that determine D. Then:
 $\backslash iint\_D\; f(x,y)\backslash \; dx\backslash ,\; dy\; =\; \backslash int\_a^b\; dx\; \backslash int\_\{\; \backslash alpha\; (x)\}^\{\; \backslash beta\; (x)\}\; f(x,y)\backslash ,\; dy.$
yaxis
If D is normal with respect to the yaxis and f : D → R is a continuous function; then α(y) and β(y) (defined on the interval [a, b]) are the two functions that determine D. Then:
 $\backslash iint\_D\; f(x,y)\backslash \; dx\backslash ,\; dy\; =\; \backslash int\_a^b\; dy\; \backslash int\_\{\backslash alpha\; (y)\}^\{\; \backslash beta\; (y)\}\; f(x,y)\backslash ,\; dx.$
Example
Consider the region (please see the graphic in the example):
 $D\; =\; \backslash \{\; (x,y)\; \backslash in\; \backslash mathbf\{R\}^2\; \backslash :\; \backslash \; x\; \backslash ge\; 0,\; y\; \backslash le\; 1,\; y\; \backslash ge\; x^2\; \backslash \}$
Calculate
 $\backslash iint\_D\; (x+y)\; \backslash ,\; dx\; \backslash ,\; dy.$
This domain is normal with respect to both the x and yaxes. To apply the formulae it is required to find the functions that determine D and the intervals over which these are defined. In this case the two functions are:
 $\backslash alpha\; (x)\; =\; x^2\backslash text\{\; and\; \}\backslash beta\; (x)\; =\; 1$
while the interval is given by the intersections of the functions with x = 0, so the interval is [a, b] = [0, 1] (normality has been chosen with respect to the xaxis for a better visual understanding).
It is now possible to apply the formula:
 $\backslash iint\_D\; (x+y)\; \backslash ,\; dx\; \backslash ,\; dy\; =\; \backslash int\_0^1\; dx\; \backslash int\_\{x^2\}^1\; (x+y)\; \backslash ,\; dy\; =\; \backslash int\_0^1\; dx\; \backslash \; \backslash left[xy\; +\; \backslash frac\{y^2\}\{2\}\; \backslash right]^1\_\{x^2\}$
(at first the second integral is calculated considering x as a constant). The remaining operations consist of applying the basic techniques of integration:
 $\backslash int\_0^1\; \backslash left[xy\; +\; \backslash frac\{y^2\}\{2\}\backslash right]^1\_\{x^2\}\; \backslash ,\; dx\; =\; \backslash int\_0^1\; \backslash left(x\; +\; \backslash frac\{1\}\{2\}\; \; x^3\; \; \backslash frac\{x^4\}\{2\}\; \backslash right)\; dx\; =\; \backslash cdots\; =\; \backslash frac\{13\}\{20\}.$
If we choose normality with respect to the yaxis we could calculate
 $\backslash int\_0^1\; dy\; \backslash int\_0^\{\backslash sqrt\{y\}\}\; (x+y)\; \backslash ,\; dx.$
and obtain the same value.
Normal domains on R^{3}
The extension of these formulae to triple integrals should be apparent:
if T is a domain that is normal with respect to the xyplane and determined by the functions α (x,y) and β(x,y), then
 $\backslash iiint\_T\; f(x,y,z)\; \backslash \; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iint\_D\; \backslash int\_\{\backslash alpha\; (x,y)\}^\{\backslash beta\; (x,y)\}\; f(x,y,z)\; \backslash ,\; dz\; dx\; dy$
(this definition is the same for the other five normality cases on R^{3}).
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 1a. The function is $f(x,\; y)\; =\; (x1)^2\; +\backslash sqrt\; y$; if one adopts this substitution $x\text{'}\; =\; x1,\; \backslash \; y\text{'}=\; y$ therefore $x\; =\; x\text{'}\; +\; 1,\; \backslash \; y=y\text{'}$ one obtains the new function $f\_2(x,y)\; =\; (x\text{'})^2\; +\backslash sqrt\; y$.
 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 absolute value of 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 R^{2}, two in R^{3}); however, more general substitutions can be made using the same principle.
Polar coordinates
In R^{2} 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:
 $f(x,y)\; \backslash rightarrow\; f(\backslash rho\; \backslash cos\; \backslash phi,\backslash rho\; \backslash sin\; \backslash phi\; ).$
Example 2a. The function is $f(x,y)\; =\; x\; +\; y$ and applying the transformation one obtains
 $f(\backslash rho,\; \backslash phi)\; =\; \backslash rho\; \backslash cos\; \backslash phi\; +\; \backslash rho\; \backslash sin\; \backslash phi\; =\; \backslash rho(\backslash cos\; \backslash phi\; +\; \backslash sin\; \backslash phi\; ).$
Example 2b. The function is $f(x,y)\; =\; x^2\; +\; y^2$, in this case one has:
 $f(\backslash rho,\; \backslash phi)\; =\; \backslash rho^2\; (\backslash cos^2\; \backslash phi\; +\; \backslash sin^2\; \backslash phi)\; =\; \backslash rho^2$
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 2c. The domain is $D\; =\; \backslash \{\; x^2\; +\; y^2\; \backslash le\; 4\; \backslash \}$, 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 2d. 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 positive y halfplane (please see the picture in the example); φ describes a plane angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following rectangle:
 $T\; =\; \backslash \{\; 2\; \backslash le\; \backslash rho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi\; \backslash \}.$
The Jacobian determinant of that transformation is the following:
 $\backslash frac\{\backslash partial\; (x,y)\}\{\backslash partial\; (\backslash rho,\; \backslash phi)\}\; =$
\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:
 $\backslash iint\_D\; f(x,y)\; \backslash \; dx\backslash ,\; dy\; =\; \backslash iint\_T\; f(\backslash rho\; \backslash cos\; \backslash phi,\; \backslash rho\; \backslash sin\; \backslash phi)\; \backslash rho\; \backslash ,\; d\; \backslash rho\backslash ,\; d\; \backslash phi.$
φ is valid in the [0, 2π] interval while ρ, which is a measure of a length, can only have positive values.
Example 2e. The function is f(x, y) = x and the domain is the same as in Example 2d. 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:
 $f(x,y)\; =\; x\; \backslash longrightarrow\; f(\backslash rho,\backslash phi)\; =\; \backslash rho\; \backslash cos\; \backslash phi.$
finally let's apply the integration formula:
 $\backslash iint\_D\; x\; \backslash ,\; dx\backslash ,\; dy\; =\; \backslash iint\_T\; \backslash rho\; \backslash cos\; \backslash phi\; \backslash rho\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash phi.$
Once the intervals are known, you have
 $\backslash int\_0^\backslash pi\; \backslash int\_2^3\; \backslash rho^2\; \backslash cos\; \backslash phi\; \backslash ,\; d\; \backslash rho\; \backslash ,\; d\; \backslash phi\; =\; \backslash int\_0^\backslash pi\; \backslash cos\; \backslash phi\; \backslash \; d\; \backslash phi\; \backslash left[\; \backslash frac\{\backslash rho^3\}\{3\}\; \backslash right]\_2^3\; =\; \backslash left[\; \backslash sin\; \backslash phi\; \backslash right]\_0^\backslash pi\; \backslash \; \backslash left(9\; \; \backslash frac\{8\}\{3\}\; \backslash right)\; =\; 0.$
Cylindrical coordinates
In R^{3} 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:
$f(x,y,z)\; \backslash rightarrow\; f(\backslash rho\; \backslash cos\; \backslash phi,\; \backslash rho\; \backslash sin\; \backslash phi,\; z)$
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 3a. 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 Example 2d and whose height is 5); if the transformation is applied, this region is obtained: $T\; =\; \backslash \{\; 2\; \backslash le\; \backslash rho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\backslash pi,\; \backslash \; 0\; \backslash le\; z\; \backslash le\; 5\; \backslash \}$ (that is the parallelepiped whose base is similar to the rectangle in Example 2d 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:
 $\backslash iiint\_D\; f(x,y,z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; f(\backslash rho\; \backslash cos\; \backslash phi,\; \backslash rho\; \backslash sin\; \backslash phi,\; z)\; \backslash rho\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash phi\backslash ,\; dz.$
This method is convenient in case of cylindrical or conical domains or in regions where it is easy to individuate the z interval and even transform the circular base and the function.
Example 3b. The function is $f(x,y,z)\; =\; x^2\; +\; y^2\; +\; z$ 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:
 $T\; =\; \backslash \{\; 0\; \backslash le\; \backslash rho\; \backslash le\; 3,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; 5\; \backslash le\; z\; \backslash le\; 5\; \backslash \}.$
while the function becomes
 $f(\backslash rho\; \backslash cos\; \backslash phi,\; \backslash rho\; \backslash sin\; \backslash phi,\; z)\; =\; \backslash rho^2\; +\; z$
Finally one can apply the integration formula:
 $\backslash iiint\_D\; (x^2\; +\; y^2\; +z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; (\; \backslash rho^2\; +\; z)\; \backslash rho\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash phi\backslash ,\; dz;$
developing the formula you have
 $\backslash int\_\{5\}^5\; dz\; \backslash int\_0^\{2\; \backslash pi\}\; d\backslash phi\; \backslash int\_0^3\; (\; \backslash rho^3\; +\; \backslash rho\; z\; )\backslash ,\; d\backslash rho\; =\; 2\; \backslash pi\; \backslash int\_\{5\}^5\; \backslash left[\; \backslash frac\{\backslash rho^4\}\{4\}\; +\; \backslash frac\{\backslash rho^2\; z\}\{2\}\; \backslash right]\_0^3\; \backslash ,\; dz\; =\; 2\; \backslash pi\; \backslash int\_\{5\}^5\; \backslash left(\; \backslash frac\{81\}\{4\}\; +\; \backslash frac\{9\}\{2\}\; z\backslash right)\backslash ,\; dz\; =\; \backslash cdots\; =\; 405\; \backslash pi.$
Spherical coordinates
In R^{3} 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:
$f(x,y,z)\; \backslash longrightarrow\; f(\backslash rho\; \backslash cos\; \backslash theta\; \backslash sin\; \backslash phi,\; \backslash rho\; \backslash sin\; \backslash theta\; \backslash sin\; \backslash phi,\; \backslash rho\; \backslash cos\; \backslash phi)$
Points on z axis do not have a precise characterization in spherical coordinates, so θ can vary between 0 to 2π .
The better integration domain for this passage is obviously the sphere.
Example 4a. 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: $T\; =\; \backslash \{\; 0\; \backslash le\; \backslash rho\; \backslash le\; 4,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi,\; \backslash \; 0\; \backslash le\; \backslash theta\; \backslash le\; 2\; \backslash pi\; \backslash \}.$
The Jacobian determinant of this transformation is the following:
 $\backslash frac\{\backslash partial\; (x,y,z)\}\{\backslash partial\; (\backslash rho,\; \backslash theta,\; \backslash phi)\}\; =$
\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 ρ^{2} sin(φ) dρ dθ dφ.
Finally you obtain the final integration formula:
 $\backslash iiint\_D\; f(x,y,z)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; f(\backslash rho\; \backslash sin\; \backslash phi\; \backslash cos\; \backslash theta,\; \backslash rho\; \backslash sin\; \backslash phi\; \backslash sin\; \backslash theta,\; \backslash rho\; \backslash cos\; \backslash phi)\; \backslash rho^2\; \backslash sin\; \backslash phi\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi.$
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 R^{3} (please see Example 4b); in other cases it can be better to use cylindrical coordinates (please see Example 4c).
 $\backslash iiint\_T\; f(a,b,c)\; \backslash rho^2\; \backslash sin\; \backslash phi\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi.$
The extra $\backslash rho^2$ and $\backslash sin\; \backslash phi$ come from the Jacobian.
In the following examples the roles of φ and θ have been reversed.
Example 4b. D is the same region as in Example 4a and $f(x,y,z)\; =\; x^2\; +\; y^2\; +\; z^2$ is the function to integrate. Its transformation is very easy:
 $f(\backslash rho\; \backslash sin\; \backslash phi\; \backslash cos\; \backslash theta,\; \backslash rho\; \backslash sin\; \backslash phi\; \backslash sin\; \backslash theta,\; \backslash rho\; \backslash cos\; \backslash phi)\; =\; \backslash rho^2,$
while we know the intervals of the transformed region T from D:
 $(0\; \backslash le\; \backslash rho\; \backslash le\; 4,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi,\; \backslash \; 0\; \backslash le\; \backslash theta\; \backslash le\; 2\; \backslash pi).$
Let's therefore apply the integration's formula:
 $\backslash iiint\_D\; (x^2\; +\; y^2\; +z^2)\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz\; =\; \backslash iiint\_T\; \backslash rho^2\; \backslash \; \backslash rho^2\; \backslash sin\; \backslash theta\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi,$
and, developing, we get
 $\backslash iiint\_T\; \backslash rho^4\; \backslash sin\; \backslash theta\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi\; =\; \backslash int\_0^\{\backslash pi\}\; \backslash sin\; \backslash phi\; \backslash ,d\backslash phi\; \backslash int\_0^4\; \backslash rho^4\; d\; \backslash rho\; \backslash int\_0^\{2\; \backslash pi\}\; d\backslash theta\; =\; 2\; \backslash pi\; \backslash int\_0^\{\backslash pi\}\; \backslash sin\; \backslash phi\; \backslash left[\; \backslash frac\{\backslash rho^5\}\{5\}\; \backslash right]\_0^4\; \backslash ,\; d\; \backslash phi\; =\; 2\; \backslash pi\; \backslash left[\; \backslash frac\{\backslash rho^5\}\{5\}\; \backslash right]\_0^4\; \backslash left[\; \backslash cos\; \backslash phi\; \backslash right]\_0^\{\backslash pi\}\; =\; \backslash frac\{4096\; \backslash pi\}\{5\}.$
Example 4c. The domain D is the ball with center in the origin and radius 3a,
 $D\; =\; \backslash left\; \backslash \{\; x^2\; +\; y^2\; +\; z^2\; \backslash le\; 9a^2\; \backslash right\; \backslash \}$
and $f(x,y,z)\; =\; x^2\; +\; y^2$ 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:
 $0\; \backslash le\; \backslash rho\; \backslash le\; 3a,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; 0\; \backslash le\; \backslash theta\; \backslash le\; \backslash pi.$
However, applying the transformation, we get
 $f(x,y,z)\; =\; x^2\; +\; y^2\; \backslash longrightarrow\; \backslash rho^2\; \backslash sin^2\; \backslash theta\; \backslash cos^2\; \backslash phi\; +\; \backslash rho^2\; \backslash sin^2\; \backslash theta\; \backslash sin^2\; \backslash phi\; =\; \backslash rho^2\; \backslash sin^2\; \backslash theta$.
Applying the formula for integration we would obtain:
 $\backslash iiint\_T\; \backslash rho^2\; \backslash sin^2\; \backslash theta\; \backslash rho^2\; \backslash sin\; \backslash theta\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi\; =\; \backslash iiint\_T\; \backslash rho^4\; \backslash sin^3\; \backslash theta\; \backslash ,\; d\backslash rho\backslash ,\; d\backslash theta\backslash ,\; d\backslash phi$
which is very hard to solve. This problem will be solved by using the passage in cylindrical coordinates. The new T intervals are
 $0\; \backslash le\; \backslash rho\; \backslash le\; 3a,\; \backslash \; 0\; \backslash le\; \backslash phi\; \backslash le\; 2\; \backslash pi,\; \backslash \; \; \backslash sqrt\{9a^2\; \; \backslash rho^2\}\; \backslash le\; z\; \backslash le\; \backslash sqrt\{9a^2\; \; \backslash rho^2\};$
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 x^{2} + y^{2} in ρ^{2}). The new function is simply ρ^{2}. Applying the integration formula
 $\backslash iiint\_T\; \backslash rho^2\; \backslash rho\; \backslash \; d\; \backslash rho\; d\; \backslash phi\; dz$.
Then we get
 $\backslash begin\{align\}$
\int_0^{2\pi} d\phi \int_0^{3a} \rho^3 d\rho \int_{\sqrt{9a^2  \rho^2}}^{\sqrt{9 a^2  \rho^2}}\, dz &= 2 \pi \int_0^{3a} 2 \rho^3 \sqrt{9 a^2  \rho^2} \, d\rho \\
&= 2 \pi \int_{9 a^2}^0 (9 a^2  t) \sqrt{t}\, dt && t = 9 a^2  \rho^2 \\
&= 2 \pi \int_0^{9 a^2} \left ( 9 a^2 \sqrt{t}  t \sqrt{t} \right ) \, dt \\
&= 2 \pi \left[ \int_0^{9 a^2} 9 a^2 \sqrt{t} \, dt  \int_0^{9 a^2} t \sqrt{t} \, dt\right] \\
&= 2 \pi \left[9 a^2 \frac{2}{3} t^{ \frac{3}{2} }  \frac{2}{5} t^{ \frac{5}{2}} \right]_0^{9 a^2} \\
&= 2 \cdot 27 \pi a^5 \left ( 6  \frac{18}{5} \right ) \\
&= \frac{648 \pi}{5} a^5.
\end{align}
Thanks to the passage in cylindrical coordinates it was possible to reduce the triple integral to an easier onevariable integral.
See also the differential volume entry in nabla in cylindrical and spherical coordinates.
Examples
Double integral
Let us assume that we wish to integrate a multivariable function f over a region A.
 $A\; =\; \backslash left\; \backslash \{\; (x,y)\; \backslash in\; \backslash mathbf\{R\}^2\; \backslash :\; \backslash \; 11\; \backslash le\; x\; \backslash le\; 14\; \backslash ;\; \backslash \; 7\; \backslash le\; y\; \backslash le\; 10\; \backslash right\; \backslash \}\; \backslash text\{\; and\; \}\; f(x,y)\; =\; x^2\; +\; 4y\backslash ,$
From this we formulate the double integral
 $\backslash int\_7^\{10\}\; \backslash int\_\{11\}^\{14\}\; (x^2\; +\; 4y)\; \backslash \; dx\backslash ,\; dy$
The inner integral is performed first, integrating with respect to x and taking y as a constant, as it is not the variable of integration. The result of this integral, which is a function depending only on y, is then integrated with respect to y.
 $\backslash begin\{align\}$
\int_{11}^{14} (x^2 + 4y) \ dx & = \left (\frac{1}{3}x^3 + 4yx \right)\Big _{x=11}^{x=14} \\
& = \frac{1}{3}(14)^3 + 4y(14)  \frac{1}{3}(11)^3  4y(11) \\
&= 471 + 12y
\end{align}
We then integrate the result with respect to y.
 $\backslash begin\{align\}$
\int_7^{10} (471 + 12y) \ dy & = (471y + 6y^2)\big _{y=7}^{y=10} \\
& = 471(10)+ 6(10)^2  471(7)  6(7)^2 \\
&= 1719
\end{align}
Observe that the order of integration is sometimes interchangeable:
 $\backslash begin\{align\}$
\int_{11}^{14} \int_{7}^{10} \ (x^2 + 4y) \ dy\, dx & = \int_{11}^{14} \left (x^2 y + 2y^2 \right)\Big _{y=7}^{y=10} \ dx \\
& = \int_{11}^{14} \ (3x^2 + 102) \ dx \\
& = \left (x^3 + 102x \right)\Big _{x=11}^{x=14} \\
&= 1719
\end{align}
The instances where the order is interchangeable is determined by Fubini's Theorem.
Volumes
The volume of the parallelepiped of sides 4 × 6 × 5 may be obtained in two ways:
 By calculating the double integral of the function f(x, y) = 5 over the region D in the xyplane which is the base of the parallelepiped.
 $\backslash iint\_D\; 5\; \backslash \; dx\backslash ,\; dy$
 By calculating the triple integral of the constant function 1 over the parallelepiped itself
 $\backslash iiint\_\backslash mathrm\{parallelepiped\}\; 1\; \backslash ,\; dx\backslash ,\; dy\backslash ,\; dz$
Computing a volume
Using the methods previously described, it is possible to calculate the volumes of some common solids.
 Cylinder: The volume of a cylinder with height h and circular base of radius R can be calculated by integrating the constant function h over the circular base, using polar coordinates.
 $\backslash mathrm\{Volume\}\; =\; \backslash int\_0^\{2\; \backslash pi\; \}\; d\; \backslash phi\; \backslash int\_0^R\; h\; \backslash rho\; \backslash \; d\; \backslash rho\; =\; h\; 2\; \backslash pi\; \backslash left[\backslash frac\{\backslash rho^2\}\{2\; \}\backslash right]\_0^R\; =\; \backslash pi\; R^2\; h$
This is in agreement with the formula
 $\backslash mathrm\{Volume\}\; =\; \backslash text\{base\; area\}\; \backslash times\; height$.
 Sphere: The volume of a sphere with radius R can be calculated by integrating the constant function 1 over the sphere, using spherical coordinates.
 $\backslash begin\{align\}$
\text{Volume} &= \iiint_D f(x,y,z) \, dx\, dy\, dz \\
&= \iiint_D 1 \, dV \\
&= \iiint_S \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi \\
&= \int_0^{2 \pi }\, d \theta \int_0^{ \pi } \sin \phi\, d \phi \int_0^R \rho^2\, d \rho \\
&= 2 \pi \int_0^{ \pi } \sin \phi\, d \phi \int_0^R \rho^2\, d \rho \\
&= 2 \pi \int_0^{ \pi } \sin \phi \frac{R^3}{3 }\, d \phi \\
&= \frac{2}{3 } \pi R^3 [ \cos \phi]_0^{ \pi } = \frac{4}{3 } \pi R^3.
\end{align}
 Tetrahedron (triangular pyramid or 3simplex): The volume of a tetrahedron with its apex at the origin and edges of length l along the x, y and z axes can be calculated by integrating the constant function 1 over the tetrahedron.
 $\backslash begin\{align\}$
\text{Volume} &= \int_0^\ell dx \int_0^{\ellx }\, dy \int_0^{\ellxy }\, dz \\
&= \int_0^\ell dx \int_0^{\ellx } (\ell  x  y)\, dy \\
&= \int_0^\ell \left[\ell^2  2\ell x + x^2  \frac{ (\ellx)^2 }{2 }\right]\, dx \\
&= \ell^3  \ell \ell^2 + \frac{\ell^3}{3 }  \left[\frac{\ell^2 x}{2 }  \frac{\ell x^2}{2} + \frac{x^3}{6 }\right]_0^\ell \\
&= \frac{\ell^3}{3 }  \frac{\ell^3}{6 } = \frac{\ell^3}{6}
\end{align}
 This is in agreement with the formula
 $\backslash mathrm\{Volume\}\; =\; \backslash frac\{1\}\{3\}\; \backslash times\; \backslash text\{base\; area\}\; \backslash times\; \backslash text\{height\}\; =\; \backslash frac\{1\}\{3\}\; \backslash times\; \backslash frac\{\backslash ell^2\}\{2\}\; \backslash times\; \backslash ell\; =\; \backslash frac\{\backslash ell^3\}\{6\}.$
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
 $\backslash int\_\{A\backslash times\; B\}\; f(x,y)\backslash ,d(x,y)<\backslash infty,$
that is, if the integral is absolutely convergent, then the multiple integral will give the same result as the iterated integral,
 $\backslash int\_\{A\backslash times\; B\}\; f(x,y)\backslash ,d(x,y)=\backslash int\_A\backslash left(\backslash int\_B\; f(x,y)\backslash ,dy\backslash right)\backslash ,dx=\backslash int\_B\backslash left(\backslash int\_A\; f(x,y)\backslash ,dx\backslash right)\backslash ,dy.$
In particular this will occur if f(x, y) 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
 $\backslash int\_0^1\backslash int\_0^1\; f(x,y)\backslash ,dy\backslash ,dx$
means, in some cases, 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 xyplane. 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
 $\backslash int\_0^1\backslash int\_0^1\; f(x,y)\backslash ,dy\backslash ,dx\; \backslash neq\; \backslash int\_0^1\backslash int\_0^1\; f(x,y)\backslash ,dx\backslash ,dy.$
This is an instance of rearrangement of a conditionally convergent integral.
The notation
 $\backslash int\_\; f(x,y)\backslash ,dx\backslash ,dy$
may be used if one wishes to be emphatic about intending a double integral rather than an iterated integral.
Some practical applications
Quite generally, just as in one variable, one can use the multiple integral to find the average of a function over a given set. Given a set D ⊆ R^{n} and an integrable function f over D, the average value of f over its domain is given by
 $\backslash bar\{f\}\; =\; \backslash frac\{1\}\{m(D)\}\; \backslash int\_D\; f(x)\backslash ,\; dx,$
where m(D) is the measure of D.
Additionally, multiple integrals are used in many applications in physics. The examples below also show some variations in the notation.
In mechanics, the moment of inertia is calculated as the volume integral (triple integral) of the density weighed with the square of the distance from the axis:
 $I\_z\; =\; \backslash iiint\_V\; \backslash rho\; r^2\backslash ,\; dV.$
The gravitational potential associated with a mass distribution given by a mass measure dm on threedimensional Euclidean space R^{3} is
 $V(\backslash mathbf\{x\})\; =\; \backslash int\_\{\backslash mathbf\{R\}^3\}\; \backslash frac\{G\}\{\backslash mathbf\{x\}\; \; \backslash mathbf\{y\}\}\backslash ,dm(\backslash mathbf\{y\}).$
If there is a continuous function ρ(x) representing the density of the distribution at x, so that dm(x) = ρ(x)d^{ 3}x, where d^{ 3}x is the Euclidean volume element, then the gravitational potential is
 $V(\backslash mathbf\{x\})\; =\; \backslash int\_\{\backslash mathbf\{R\}^3\}\; \backslash frac\{G\}\{\backslash mathbf\{x\}\backslash mathbf\{y\}\}\backslash ,\backslash rho(\backslash mathbf\{y\})\backslash ,d^3\backslash mathbf\{y\}.$
In electromagnetism, Maxwell's equations can be written using multiple integrals to calculate the total magnetic and electric fields. In the following example, the electric field produced by a distribution of charges given by the volume charge density $\backslash rho\; (\backslash vec\; r)$ is obtained by a triple integral of a vector function:
 $\backslash vec\; E\; =\; \backslash frac\; \{1\}\{4\; \backslash pi\; \backslash epsilon\_0\}\; \backslash iiint\; \backslash frac\; \{\backslash vec\; r\; \; \backslash vec\; r\text{'}\}\{\backslash left\; \backslash \; \backslash vec\; r\; \; \backslash vec\; r\text{'}\; \backslash right\; \backslash ^3\}\; \backslash rho\; (\backslash vec\; r\text{'})\backslash ,\; \backslash operatorname\{d\}^3\; r\text{'}.$
This can also be written as an integral with respect to a signed measure representing the charge distribution.
See also
 Main analysis theorems that relate multiple integrals:
Free software for multidimensional numerical integration
 FuncDesigner frameworks, based on interval analysis, license: BSD (free for any purposes)
 Cuba is a freesoftware library of several multidimensional integration algorithms
 Cubature code for adaptive multidimensional integration
References
 Robert A. Adams  Calculus: A Complete Course (5th Edition) ISBN 0201791315.
 R.K.Jain and S.R.K Iyengar Advanced Engineering Mathematics (Third edition) 2009, Narosa Publishing House ISBN 9788173197307
External links
 MathWorld.

 Maxima (software))
de:Integralrechnung#Mehrdimensionale Integration
This article was sourced from Creative Commons AttributionShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, EGovernment Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a nonprofit organization.