Introduction
Just as the definite integral of a positive function of one variable represents theMathematical definition
For , consider a so-called "half-open" -dimensional hyperrectangular domain , defined as: : partition (set theory)">Partition Partition may refer to: Computing Hardware * Disk partitioning, the division of a hard disk drive * Memory partition, a subdivision of a computer's memory, usually for use by a single job Software * Partition (database), the division of a ...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 asParticular cases
In the case of the integral :Methods of integration
The resolution of problems with multiple integrals consists, in most cases, of finding a way to reduce the multiple integral to anIntegrating constant functions
When the integrand is a constant function , the integral is equal to the product of and the measure of the domain of integration. If and the domain is a subregion of , the integral gives the area of the region, while if the domain is a subregion of , the integral gives the volume of the region.Example. Let and :D = \left\ in which case :\int_3^6 \int_2^4 \ 2 \ dx\, dy =2\int_3^6 \int_2^4 \ 1 \ dx\, dy= 2\cdot\operatorname(D) = 2 \cdot (2 \cdot 3) = 12, since by definition we have: :\int_3^6 \int_2^4 \ 1 \ dx\, dy=\operatorname(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 isExample 1. Consider the function integrated over the domain :T=\left \, a disc withradius In classical geometry, a radius ( : radii) of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length. The name comes from the latin ''radius'', meaning ray but also the ...1 centered at the origin with the boundary included. Using the linearity property, the integral can be decomposed into three pieces: :\iint_T \left(2\sin x - 3y^3 + 5\right) \, dx \, dy = \iint_T 2 \sin x \, dx \, dy - \iint_T 3y^3 \, dx \, dy + \iint_T 5 \, dx \, dy The function is an odd function in the variable and the disc is symmetric with respect to the -axis, so the value of the first integral is 0. Similarly, the function is an odd function of , and is symmetric with respect to the -axis, and so the only contribution to the final result is that of the third integral. Therefore the original integral is equal to the area of the disk times 5, or 5.
Example 2. Consider the function and as integration region the ball with radius 2 centered at the origin, :T = \left \. The "ball" is symmetric about all three axes, but it is sufficient to integrate with respect to -axis to show that the integral is 0, because the function is an odd function of that variable.
Normal domains on
This method is applicable to any domain for which: * the projection of onto either the -axis or the -axis is bounded by the two values, and * 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, and . 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.-axis
If the domain is normal with respect to the -axis, and is a continuous function; then and (both of which are defined on the interval ) are the two functions that determine . Then, by Fubini's theorem: :-axis
If is normal with respect to the -axis and is a continuous function; then and (both of which are defined on the interval ) are the two functions that determine . Again, by Fubini's theorem: :Normal domains on
If is a domain that is normal with respect to the -plane and determined by the functions and , then :Change of variables
The limits of integration are often not easily interchangeable (without normality or with complex formulae to integrate). One makes aExample 1a. The function is ; if one adopts the substitution , therefore , one obtains the new function .* Similarly for the domain because it is delimited by the original variables that were transformed before ( and in example). * the differentials and 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 , two in ); however, more general substitutions can be made using the same principle.
Polar coordinates
In if the domain has a circular symmetry and the function has some particular characteristics one can apply the ''transformation to polar coordinates'' (see the example in the picture) which means that the generic points 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: :Example 2a. The function is and applying the transformation one obtains :f(x, y) = f(\rho \cos \varphi,\rho \sin \varphi) = \rho \cos \varphi + \rho \sin \varphi = \rho(\cos \varphi + \sin \varphi ).
Example 2b. The function is , in this case one has: :The transformation of the domain is made by defining the radius' crown length and the amplitude of the described angle to define the intervals starting from .f(x, y) = \rho^2 \left(\cos^2 \varphi + \sin^2 \varphi\right) = \rho^2 using the Pythagorean trigonometric identity (very useful to simplify this operation).
Example 2c. The domain is , that is a circumference of radius 2; it's evident that the covered angle is the circle angle, so varies from 0 to 2, 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 , that is the circular crown in the positive half-plane (please see the picture in the example); describes a plane angle while varies from 2 to 3. Therefore the transformed domain will be the following rectangle: :T = \. TheJacobian determinant In vector calculus, the Jacobian matrix (, ) of a vector-valued function of several variables is the matrix of all its first-order partial derivatives. When this matrix is square, that is, when the function takes the same number of variables ...of that transformation is the following: :\frac = \begin \cos \varphi & - \rho \sin \varphi \\ \sin \varphi & \rho \cos \varphi \end = \rho which has been obtained by inserting the partial derivatives of , in the first column respect to and in the second respect to , so the differentials in this transformation become . 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 \varphi, \rho \sin \varphi) \rho \, d \rho\, d \varphi. is valid in the interval while , which is a measure of a length, can only have positive values.
Example 2e. The function is and the domain is the same as in Example 2d. From the previous analysis of we know the intervals of (from 2 to 3) and of (from 0 to ). Now we change the function: :f(x,y) = x \longrightarrow f(\rho,\varphi) = \rho \cos \varphi. finally let's apply the integration formula: :\iint_D x \, dx\, dy = \iint_T \rho \cos \varphi \rho \, d\rho\, d\varphi. Once the intervals are known, you have :\int_0^\pi \int_2^3 \rho^2 \cos \varphi \, d \rho \, d \varphi = \int_0^\pi \cos \varphi \ d \varphi \left \frac \right2^3 = \Big \sin \varphi \Big0^\pi \ \left(9 - \frac \right) = 0.
Cylindrical coordinates
In the integration on domains with a circular base can be made by the ''passage toExample 3a. The region is (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 = \ (that is, the parallelepiped whose base is similar to the rectangle in Example 2d and whose height is 5). Because the component is unvaried during the transformation, the differentials vary as in the passage to polar coordinates: therefore, they become . 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 \varphi, \rho \sin \varphi, z) \rho \, d\rho\, d\varphi\, 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 and as integration domain thiscylinder A cylinder (from ) has traditionally been a three-dimensional solid, one of the most basic of curvilinear geometric shapes. In elementary geometry, it is considered a prism with a circle as its base. A cylinder may also be defined as an infin ...: . The transformation of in cylindrical coordinates is the following: :T = \. while the function becomes :f(\rho \cos \varphi, \rho \sin \varphi, z) = \rho^2 + z Finally one can apply the integration formula: :\iiint_D \left(x^2 + y^2 +z\right) \, dx\, dy\, dz = \iiint_T \left( \rho^2 + z\right) \rho \, d\rho\, d\varphi\, dz; developing the formula you have :\int_^5 dz \int_0^ d\varphi \int_0^3 \left( \rho^3 + \rho z \right)\, d\rho = 2 \pi \int_^5 \left \frac + \frac \right0^3 \, dz = 2 \pi \int_^5 \left( \frac + \frac z\right)\, dz = \cdots = 405 \pi.
Spherical coordinates
In 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 toExample 4a. The domain is (sphere with radius 4 and center at the origin); applying the transformation you get the region :It is 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 to (see Example 4b); in other cases it can be better to use cylindrical coordinates (see Example 4c). :T = \. The Jacobian determinant of this transformation is the following: :\frac = \begin \cos \theta \sin \varphi & - \rho \sin \theta \sin \varphi & \rho \cos \theta \cos \varphi \\ \sin \theta \sin \varphi & \rho \cos \theta \sin \varphi & \rho \sin \theta \cos \varphi \\ \cos \varphi & 0 & - \rho \sin \varphi \end = \rho^2 \sin \varphi The differentials therefore are transformed to . This yields the final integration formula: :\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) \rho^2 \sin \varphi \, d\rho\, d\theta\, d\varphi.
Example 4b. is the same region as in Example 4a and is the function to integrate. Its transformation is very easy: :f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2, while we know the intervals of the transformed region from : :T=\. We therefore apply the integration formula: :\iiint_D \left(x^2 + y^2 + z^2\right) \, dx\, dy\, dz = \iiint_T \rho^2 \, \rho^2 \sin \theta \, d\rho\, d\theta\, d\varphi, and, developing, we get :\iiint_T \rho^4 \sin \theta \, d\rho\, d\theta\, d\varphi = \int_0^ \sin \varphi \,d\varphi \int_0^4 \rho^4 d \rho \int_0^ d\theta = 2 \pi \int_0^ \sin \varphi \left \frac \right0^4 \, d \varphi = 2 \pi \left \frac \right0^4 \Big \cos \varphi \Big0^ = \frac.
Example 4c. The domain is the ball with center at the origin and radius , :See also the differential volume entry in nabla in cylindrical and spherical coordinates.D = \left \ and 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 region are obviously: :T=\. However, applying the transformation, we get :f(x,y,z) = x^2 + y^2 \longrightarrow \rho^2 \sin^2 \theta \cos^2 \varphi + \rho^2 \sin^2 \theta \sin^2 \varphi = \rho^2 \sin^2 \theta. Applying the formula for integration we obtain: :\iiint_T \rho^2 \sin^2 \theta \rho^2 \sin \theta \, d\rho\, d\theta\, d\varphi = \iiint_T \rho^4 \sin^3 \theta \, d\rho\, d\theta\, d\varphi which can be solved by turning it into an iterated integral.\iiint_T \rho^4 \sin^3 \theta \, d\rho\, d\theta\, d\varphi = \underbrace_ \,\underbrace_\, \underbrace_ .I = \left.\int_0^\rho^4 d\rho = \frac\right\vert_0^ = \fraca^5 ,II = \int_0^\pi \sin^3\theta \, d\theta = -\int_0^\pi \sin^2\theta \, d(\cos \theta) = \int_0^\pi (\cos^2\theta-1) \, d(\cos \theta) = \left.\frac\^\pi_0 - \left.\cos\theta\^\pi_0 = \frac ,III = \int_0^ d \varphi = 2\pi . Collecting all parts,\iiint_T \rho^4 \sin^3 \theta \, d\rho\, d\theta\, d\varphi = I\cdot II\cdot III = \fraca^5\cdot \frac\cdot 2\pi = \frac\pi a^5 . Alternatively, this problem can be solved by using the passage to cylindrical coordinates. The new intervals are :T=\left\; the interval has been obtained by dividing the ball into twohemisphere Hemisphere refers to: * A half of a sphere As half of the Earth * A hemisphere of Earth ** Northern Hemisphere ** Southern Hemisphere ** Eastern Hemisphere ** Western Hemisphere ** Land and water hemispheres * A half of the (geocentric) celes ...s simply by solving theinequality Inequality may refer to: Economics * Attention inequality, unequal distribution of attention across users, groups of people, issues in etc. in attention economy * Economic inequality, difference in economic well-being between population groups * ...from the formula of (and then directly transforming into ). The new function is simply . Applying the integration formula :\iiint_T \rho^2 \rho \, d \rho \, d \varphi \, dz. Then we get :\begin \int_0^ d\varphi \int_0^ \rho^3 d\rho \int_^\, dz &= 2 \pi \int_0^ 2 \rho^3 \sqrt \, d\rho \\ &= -2 \pi \int_^0 (9 a^2 - t) \sqrt\, dt && t = 9 a^2 - \rho^2 \\ &= 2 \pi \int_0^ \left ( 9 a^2 \sqrt - t \sqrt \right ) \, dt \\ &= 2 \pi \left( \int_0^ 9 a^2 \sqrt \, dt - \int_0^ t \sqrt \, dt\right) \\ &= 2 \pi \left a^2 \frac23 t^ - \frac t^ \right0^ \\ &= 2 \cdot 27 \pi a^5 \left ( 6 - \frac \right ) \\ &= \frac a^5. \end Thanks to the passage to cylindrical coordinates it was possible to reduce the triple integral to an easier one-variable integral.
Examples
Double integral over a rectangle
Let us assume that we wish to integrate a multivariable function over a region : :Double integral over a normal domain
Consider the region (please see the graphic in the example): :Calculating volume
Using the methods previously described, it is possible to calculate the volumes of some common solids. *Multiple improper integral
In case of unbounded domains or functions not bounded near the boundary of the domain, we have to introduce the doubleMultiple integrals and iterated integrals
Triple integral
Main article: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 and an integrable function over , the average value of over its domain is given by :See also
* MainReferences
Further reading
* * *Herman, Edwin “Jed” & Strang, Gilbert (2016): ''Calculus : Volume 3'' : OpenStax, Rice University, Houston, Texas, USA. .External links
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