Multiple integral

[1] If there are more variables, a multiple integral will yield hypervolumes of multidimensional functions.

For n > 1, consider a so-called "half-open" n-dimensional hyperrectangular domain T, defined as Partition each interval [aj, bj) into a finite family Ij of non-overlapping subintervals ijα, with each subinterval closed at the left end, and open at the right end.

[3] If f is Riemann integrable, S is called the Riemann integral of f over T and is denoted Frequently this notation is abbreviated as where x represents the n-tuple (x1, ..., xn) and dnx is the n-dimensional volume differential.

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

Consider the function f(x,y) = 2 sin(x) − 3y3 + 5 integrated over the domain a disc with radius 1 centered at the origin with the boundary included.

Therefore the original integral is equal to the area of the disk times 5, or 5π.Example 2.

Consider the function f(x, y, z) = x exp(y2 + z2) and as integration region the ball with radius 2 centered at the origin,

If the domain D is normal with respect to the x-axis, and f : D → R is a continuous function; then α(x) and β(x) (both of which are defined on the interval [a, b]) are the two functions that determine D. Then, by Fubini's theorem:[5] If D is normal with respect to the y-axis and f : D → R is a continuous function; then α(y) and β(y) (both of which are defined on the interval [a, b]) are the two functions that determine D. Again, by Fubini's theorem: If T is a domain that is normal with respect to the xy-plane and determined by the functions α(x, y) and β(x, y), then This definition is the same for the other five normality cases on R3.

One makes a change of variables to rewrite the integral in a more "comfortable" region, which can be described in simpler formulae.

using the Pythagorean trigonometric identity (can be 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.

The domain is D = {x2 + y2 ≤ 9, x2 + y2 ≥ 4, y ≥ 0}, that is the circular crown in the positive y half-plane (please see the picture in the example); φ describes a plane angle while ρ varies from 2 to 3.

Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates: φ is valid in the [0, 2π] interval while ρ, which is a measure of a length, can only have positive values.Example 2e.

From the previous analysis of D we know the intervals of ρ (from 2 to 3) and of φ (from 0 to π).

Now we change the function: Finally let's apply the integration formula: Once the intervals are known, you have In R3 the integration on domains with a circular base can be made by the passage to cylindrical coordinates; the transformation of the function is made by the following relation: 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.

The region is D = {x2 + y2 ≤ 9, x2 + y2 ≥ 4, 0 ≤ z ≤ 5} (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: (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 to polar coordinates: therefore, they become ρ dρ dφ 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 transformation of D in cylindrical coordinates is the following: while the function becomes Finally one can apply the integration formula: developing the formula you have 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.

The domain D is the ball with center at the origin and radius 3a, and f(x, y, z) = x2 + y2 is the function to integrate.

Looking at the domain, it seems convenient to adopt the passage to spherical coordinates, in fact, the intervals of the variables that delimit the new T region are: However, applying the transformation, we get Applying the formula for integration we obtain: which can be solved by turning it into an iterated integral.

Alternatively, this problem can be solved by using the passage to cylindrical coordinates.

Thanks to the passage to 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.

For example, doing the previous calculation with order reversed gives the same result: Consider the region (please see the graphic in the example): Calculate This domain is normal with respect to both the x- and y-axes.

The remaining operations consist of applying the basic techniques of integration: If we choose normality with respect to the y-axis we could calculate and obtain the same value.

Using the methods previously described, it is possible to calculate the volumes of some common solids.

This is in agreement with the formula for the volume of a prism 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.

By the Fichtenholz–Lichtenstein theorem, if f is bounded on [0, 1] × [0, 1] and both iterated integrals exist, then they are equal.

Quite generally, just as in one variable, one can use the multiple integral to find the average of a function over a given set.

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: The gravitational potential associated with a mass distribution given by a mass measure dm on three-dimensional Euclidean space R3 is[11] If there is a continuous function ρ(x) representing the density of the distribution at x, so that dm(x) = ρ(x)d3x, where d3x is the Euclidean volume element, then the gravitational potential is In electromagnetism, Maxwell's equations can be written using multiple integrals to calculate the total magnetic and electric fields.

[12] In the following example, the electric field produced by a distribution of charges given by the volume charge density ρ( r→ ) is obtained by a triple integral of a vector function: This can also be written as an integral with respect to a signed measure representing the charge distribution.