Multiple integral

In mathematics (specifically multivariable calculus), a multiple integral is a definite integral of a function of several real variables, for instance, f(x, y) or f(x, y, z).

Integrals of a function of two variables over a region in \(\mathbb{R}^2\) (the real-number plane) are called double integrals, and integrals of a function of three variables over a region in \(\mathbb{R}^3\) (real-number 3D space) are called triple integrals.

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 (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. If there are more variables, a multiple integral will yield hypervolumes of multidimensional functions.

Multiple integration of a function in n variables: f(x₁, x₂, ..., 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:

\(\int \cdots \int_\mathbf{D}\, f(x_1,x_2,\ldots,x_n) \,dx_1 \!\cdots dx_n\)

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.

For n > 1, consider a so-called "half-open" n-dimensional hyperrectangular domain T, defined as

\(T= [ a_1, b_1) \times [ a_2, b_2) \times \cdots \times [ a_n, b_n) \subseteq \R^n\).

Partition each interval [a_j, b_j) into a finite family I_j of non-overlapping 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

\(C=I_1\times I_2\times \cdots \times I_n\)

is a partition of T; that is, the subrectangles C_k are non-overlapping 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

\(T=C_1\cup C_2\cup \cdots \cup C_m\)

We can approximate the total (n + 1)-dimensional volume bounded below by the n-dimensional hyperrectangle T and above by the n-dimensional graph of f with the following Riemann sum:

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

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, also known as the measure of C_k.

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

\(\int \cdots \int_T\, f(x_1,x_2,\ldots,x_n) \,dx_1 \!\cdots dx_n\).

\(\int_T\!f(\mathbf{x})\,d^n\mathbf{x}\).

Condensed: the full section is in Wikipedia.

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.

In the case of \(T \subseteq \R^2\), the integral

\(l = \iint_T f(x,y)\, dx\, dy\)

is the double integral of f on T, and if \(T \subseteq \R^3\) the integral

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

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 a notational convention which is convenient when computing a multiple integral as an iterated integral, as shown later in this article.

The resolution of problems with multiple integrals consists, in most 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. For continuous functions, this is justified by Fubini's theorem. Sometimes, it is possible to obtain the result of the integration by direct examination without any calculations.

The following are some simple methods of integration:

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, the integral gives the area of the region, while if the domain is a subregion of R, the integral gives the volume of the region.

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.

This method is applicable to any domain D for which:

• The projection of D onto either the x-axis or the y-axis 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

Such a domain will be here called a normal domain. Elsewhere in the literature, normal domains are sometimes called type I or type II domains, depending on which axis the domain is fibred over. In all cases, the function to be integrated must be Riemann integrable on the domain, which is true (for instance) if the function is continuous.

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.

• Similarly for the domain because it is delimited by the original variables that were transformed before (x and y in example)
• The area element dx dy transforms 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, two in R); however, more general substitutions can be made using the same principle.

Let us assume that we wish to integrate a multivariable function f over a region A:

\(A = \left \{ (x,y) \in \mathbf{R}^2 \ : \ 11 \le x \le 14 \ ; \ 7 \le y \le 10 \right \} \mbox{ and } f(x,y) = x^2 + 4y\,\).

From this we formulate the iterated integral

\(\int_7^{10} \int_{11}^{14} (x^2 + 4y) \, dx\, dy\).

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.

\(\begin{aligned} \int_{11}^{14} \left(x^2 + 4y\right) \, dx & = \left [\frac13 x^3 + 4yx \right]_{x=11}^{x=14} \\ &= \frac13(14)^3 + 4y(14) - \frac13(11)^3 - 4y(11) \\ &= 471 + 12y \end{aligned}\)

We then integrate the result with respect to y.

\(\begin{aligned} \int_7^{10} (471 + 12y) \ dy & = \Big[471y + 6y^2\Big]_{y=7}^{y=10} \\ &= 471(10)+ 6(10)^2 - 471(7) - 6(7)^2 \\ &= 1719 \end{aligned}\)

In cases where the double integral of the absolute value of the function is finite, the order of integration is interchangeable, that is, integrating with respect to x first and integrating with respect to y first produce the same result. That is Fubini's theorem. For example, doing the previous calculation with order reversed gives the same result:

\(\begin{aligned} \int_{11}^{14} \int_{7}^{10} \, \left(x^2 + 4y\right) \, dy\, dx & = \int_{11}^{14} \Big[x^2 y + 2y^2 \Big]_{y=7}^{y=10} \, dx \\ &= \int_{11}^{14} \, (3x^2 + 102) \, dx \\ &= \Big[x^3 + 102x \Big]_{x=11}^{x=14} \\ &= 1719. \end{aligned}\)

Consider the region (please see the graphic in the example):

\(D = \{ (x,y) \in \mathbf{R}^2 \ : \ x \ge 0, y \le 1, y \ge x^2 \}\) .

Calculate

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

This domain is normal with respect to both the x- and y-axes. To apply the formulae it is required to find the functions that determine D and the intervals over which these functions are defined. In this case the two functions are:

\(\alpha (x) = x^2\text{ and }\beta (x) = 1\)

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 x-axis for a better visual understanding).

It is now possible to apply the formula:

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

(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 normality with respect to the y-axis we could calculate

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

Condensed: the full section is in Wikipedia.

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.

\(\mathrm{Volume} = \int_0^{2\pi} d \varphi\, \int_0^R h \rho \, d \rho = 2 \pi h \left[\frac{\rho^2}{2}\right]_0^R = \pi R^2 h\)

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

\(\mathrm{Volume} = \text{base area} \times \text{height}\).

• Sphere: The volume of a sphere with radius R can be calculated by integrating the constant function 1 over the sphere, using spherical coordinates.

\(\begin{aligned} \text{Volume} &= \iiint_D f(x,y,z) \, dx\, dy\, dz \\ &= \iiint_D 1 \, dV \\ &= \iiint_S \rho^2 \sin \varphi \, d\rho\, d\theta\, d\varphi \\ &= \int_0^{2\pi} \, d \theta \int_0^{ \pi } \sin \varphi\, d \varphi \int_0^R \rho^2\, d \rho \\ &= 2 \pi \int_0^\pi \sin \varphi\, d \varphi \int_0^R \rho^2\, d \rho \\ &= 2 \pi \int_0^\pi \sin \varphi \frac{R^3}{3 }\, d \varphi \\ &= \frac23 \pi R^3 \Big[-\cos \varphi\Big]_0^\pi = \frac43 \pi R^3 \end{aligned}\)

• Tetrahedron (triangular pyramid or 3-simplex): 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.

\(\begin{aligned} \text{Volume} &= \int_0^\ell dx \int_0^{\ell-x}\, dy \int_0^{\ell-x-y }\, dz \\ &= \int_0^\ell dx \int_0^{\ell-x } (\ell - x - y)\, dy \\ &= \int_0^\ell \left( l^2 - 2 \ell x + x^2 - \frac{(\ell-x)^2 }{2}\right)\, dx \\ &= \ell^3 - \ell \ell^2 + \frac{\ell^3}{3 } - \left[\frac{\ell^2 x}{2} - \frac{ \ell x^2}{2} + \frac{x^3}{6 }\right]_0^ \ell \\ &= \frac{\ell^3}{3} - \frac{\ell^3}{6} = \frac{ \ell^3}{6}\end{aligned}\)

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

\(\mathrm{Volume} = \frac13 \times \text{base area} \times \text{height} = \frac13 \times \frac{\ell^2}{2} \times \ell = \frac{ \ell^3}{6}\).

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.