Multiple integral

Multiple integral
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Multiple_integral".

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It has been suggested that Volume integral be merged into this article or section. (Discuss)

Integral as area between two curves.

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).

Double integral as volume under a surface. The rectangular region at the bottom of the body is the domain of integration, while the surface is the graph of the two-variable function to be integrated.

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, …, x_n) 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:

$\iint \ldots \int_\mathbf{D}\;f(x_1,x_2,\ldots,x_n) \;\mathbf{d}x_1 \mathbf{d}x_2\!\ldots\mathbf{d}x_n$

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

$\iiint_\mathrm{parallelepiped} 1 \, dx\, dy\, dz$

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.

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

Divide each interval a_i, b_i) 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 I_i. Then, the family of subrectangles of the form

$C=I_1\times I_2\times \cdots\times I_n$

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 : T → R be a function defined on a rectangle T. Consider a partition

$T=C_1\cup C_2\cup \cdots \cup C_m$

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

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

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

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

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

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

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

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 D ⊆ Rⁿ and an integrable function f over D, the average value of f over its domain is given by

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

where m(D) is the measure of D.

Particular cases

In the case of T ⊆ R², the integral

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

is the double integral of f on T, and if T ⊆ R³ the integral

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

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 this gives the area of the region, while in R³ it is the volume of the region.

For example:

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

Let us integrate f over D:

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

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:

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

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

See also: Order of integration (calculus)

x-axis

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

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

y-axis

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

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

Example

Example: D region for integral by reduction's formulas.

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

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

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:

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

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

It's now possible to apply the formulas:

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

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

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

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

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

and obtain the same value.

Example of a normal domain in R3 (xy-plane).

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:

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

(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 $f(x, y) = (x-1)^2 +\sqrt y$ ;

if one adopts this substitution $x' = x-1, \ y'= y \, \!$ therefore $x = x' + 1, \ y=y' \,\!$

one obtains the new function $f_2(x,y) = (x')^2 +\sqrt y$ .

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

See also: Polar coordinate system

Transformation from cartesian to 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:

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

Example (2-a):

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

and applying the transformation one obtains

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

Example (2-b):

The function is $f(x,y) = x^2 + y^2\,\!$

In this case one has:

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

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 of a domain transformation from cartesian to polar.

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 $D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ y \ge 0 \}$ , 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:

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

The Jacobian determinant of that transformation is the following:

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

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:

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

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 $f(x,y) = x\,\!$ 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:

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

finally let's apply the integration formula:

$\iint_D x \, dx\, dy = \iint_T \rho \cos \phi \ \rho \, d\rho\, d\phi.$

Once the intervals are known, you have

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

Cylindrical coordinates

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:

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

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 (3-a):

The region is $D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ 0 \le z \le 5 \}$ (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: $T = \{ 2 \le \rho \le 3, \ 0 \le \phi \le \pi, \ 0 \le z \le 5 \}$ (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:

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

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 $f(x,y,z) = x^2 + y^2 + z\,\!$ and as integration domain this cylinder: $D = \{ x^2 + y^2 \le 9, \ -5 \le z \le 5 \}$ .

The transformation of D in cylindrical coordinates is the following:

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

while the function becomes

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

Finally you can apply the integration's formula:

$\iiint_D (x^2 + y^2 +z) \, dx\, dy\, dz = \iiint_T ( \rho^2 + z) \rho \, d\rho\, d\phi\, dz;$

developing the formula you have

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

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

Spherical coordinates

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:

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

Note that points on z axis do not have a precise characterization in spherical coordinates, so $φ$ 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: $T = \{ 0 \le \rho \le 4, \ 0 \le \phi \le \pi, \ 0 \le \theta \le 2 \pi \}.$

The Jacobian determinant of this transformation is the following:

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

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

Finally you obtain the final integration formula:

$\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \sin \theta \cos \phi, \rho \sin \theta \sin \phi, \rho \cos \theta) \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi.$

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 R³ (please see example 4-b); in other cases it can be better to use cylindrical coordinates (please see example 4-c).

$\iiint_T f(a,b,c) \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi.$

Note that the extra $ρ 2$ and $sinφ$ 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 $f(x,y,z) = x^2 + y^2 + z^2\,\!$ is the function to integrate.

Its transformation is very easy:

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

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

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

Let's therefore apply the integration's formula:

$\iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz = \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi,$

and, developing, we get

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

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

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 $f(x,y,z) = x^2 + y^2\,\!$ 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:

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

However, applying the transformation, we get

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

Applying the formula for integration we would obtain:

$\iiint_T \rho^2 \sin^2 \theta \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi = \iiint_T \rho^4 \sin^3 \theta \, d\rho\, d\theta\, d\phi$

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

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

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 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

$\int_0^{2 \pi} d\phi \int_0^{3a} \rho^3 d\rho \int_{- \sqrt{9a^2 - \rho^2} }^{\sqrt{9 a^2 - \rho^2} }\, dz = 2 \pi \int_0^{3a} 2 \rho^3 \sqrt{9 a^2 - \rho^2} \, d\rho.$

Now let's apply the transformation

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

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

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

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

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

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:

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

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

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:

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

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:

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

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.

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

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

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

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

Example of an improper domain.

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

See also: Order of integration (calculus)

Fubini's theorem states that if

$\int_{A\times B} |f(x,y)|\,d(x,y)<\infty,$

that is, the integral is absolutely convergent, then the multiple integral will give the same result as the iterated integral,

$\int_{A\times B} f(x,y)\,d(x,y)=\int_A\left(\int_B f(x,y)\,dy\right)\,dx=\int_B\left(\int_A f(x,y)\,dx\right)\,dy.$

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

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

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:

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

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

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

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

The notation

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

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:

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

References

Robert A. Adams - Calculus: A Complete Course (5th Edition) ISBN 0201791315.

External links

Eric W. Weisstein, Multiple Integral at MathWorld.
L.D. Kudryavtsev (2001), "Multiple integral", in Hazewinkel, Michiel, Encyclopaedia of Mathematics, Kluwer Academic Publishers, ISBN 978-1556080104
Mathematical Assistant on Web online evaluation of double integrals in Cartesian coordinates and polar coordinates (includes intermediate steps in the solution, powered by Maxima (software))

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