Multiple integral

Integral as area between two curves.

The multiple integral is a kind of definite integral extended to functions of more than one real variable (i.e. $f(x,y)\,\!$ or $f(x,y,z)\,\!$).

Multiple integrals are not the same as iterated integrals

It is easy 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"; see Fubini's theorem) 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

$\int _{0}^{1}\int _{0}^{1}f(x,y)\,dy\,dx\neq \int _{0}^{1}\int _{0}^{1}f(x,y)\,dx\,dy.$

For an elementary example (doable by the methods of first-year calculus), see examples of Fubini's theorem. 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.

Multiple integrals

If, conceptually, the definite integral for functions of one variable represents the area of the region on the graph between the curve defined by the function and the x-axis, that for functions of two variables (double integral) represents the measure of the space between the surface defined by the function and the plane which contains its domain, so it describes not an area but a volume of a particular solid called cylindroid; one obtains the same value if considering the triple integrals (functions of three variables) calculated with the constant f(x, y, z) = 1. If the number of variables is higher, one will calculate "hypervolumes" (volumes of solid of more than three dimensions) that cannot be graphed.

Multiple integral as volume under a surface.

In that example, the volume of the parallelepiped of sides 4x6x5 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 "bi-dimensional interval" D (region contained in the xy-plane)
• by the triple integral $\iiint _{\mathrm {parallelepiped} }1\,dx\,dy\,dz$ of the constant function 1 calculated on the "three-dimensional" interval coinciding with the same parallelepiped; in that case the volume is considered as the "sum" of all the infinitesimal elements that compose the domain.

Because it is impossible to calculate the antiderivative of a function of more than one variable, indefinite multiple integrals do not exist so they are all definite integrals.

Some practical applications

Graph of the direct calculation of E.

These integrals are used in many applications in physics.

In mechanics the moment of inertia is calculated as a volume integral (that is a triple integral) of the density weighed with the square of the distance from the axis:

$I_{z}=\int _{V}^{.}\rho r^{2}\,dV.$

In electromagnetism, Maxwell's equations can be written with multiple integrals to calculate the total magnetic and electric fields. In this example, the electric field produced by a distribution of charges is obtained by a triple integral of a vector function:

${\vec {E}}={\frac {1}{4\pi \epsilon _{0}}}\int {\frac {{\vec {r}}-{\vec {r}}'}{\left\|{\vec {r}}-{\vec {r}}'\right\|^{3}}}\rho ({\vec {r}}')\,\operatorname {d} ^{3}r'.$

Mathematical definition

Say one is given $f:T\longrightarrow \mathbb {R}$ where $T\subseteq \mathbb {R} ^{n}$. Then δ is the biggest diameter of a decomposition D on T. Then

$\ell =\lim _{\delta \to 0}\sum _{i=1}^{n}f(P_{i})\,\operatorname {mis} \,T_{i}\in \mathbb {R}$

is the multiple integral of F on T (mis T_i is the measure of the infinitesimal interval of the T domain).

Theorems

Multiple integrals naturally inherit the properties of integrals of functions of one variable (linearity, additivity, monotonicity, absolute value, the theorem of the average and the theorem of the weighted average).

Theorem of the weighted average for multiple integrals

Moreover, there exists a theorem to determinate the average value of a function of more than one variable in a D region:

${\bar {f}}={\frac {1}{\mathrm {area} \ D}}\iint _{D}f(x,y)\,dx\,dy.$

In technical applications, such as engineering problems, double and triple integrals are mostly used.

Double integral

From the general definition, in the case of $T\subseteq \mathbb {R} ^{2}$ one has that

$\ell =\iint _{T}f(x,y)\,dx\,dy$

is the double integral of F on T.

Triple integral

The extension of the definition to the triple integral is immediate. From the general definition, in the case of $T\subseteq \mathbb {R} ^{3}$ one has that

$\ell =\iiint _{T}f(x,y,z)\,dx\,dy\,dz$

is the triple integral of F on T.

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 of which is directly solvable.

Direct examination

Sometimes it is possible to avoid direct calculation to obtain the result of the integration.

Constants

In case of constant functions the result is immediate; one need only multiply the measure of the domain for the value of the constant c. If n = 1, on R² this gives the volume of the obtained cylindroid, while on R³ it is a hypervolume.

• Example:

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

Let us integrate f on D:

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

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 $f(x,y)=2\ \sin(x)-3\ y^{3}+5$ and $T=x^{2}+y^{2}\leq 1$ is the integration area (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 $f(x,y,z)=x\ e^{y^{2}+z^{2}}$ and - as integration region - the sphere with radius 2 centered in the origin of the axes $T=x^{2}+y^{2}+z^{2}\leq 4$. 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

x-axis

D is a measurable dominion 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:

$\iint _{T}f(x,y)\ dx\,dy=\int _{a}^{b}dx\int _{\alpha (x)}^{\beta (x)}f(x,y)\,dy.$

x-axis

D is a measurable dominion perpendicular to the x 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: D region for integral by reduction's formulas.

Example:

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

$\iint _{D}(x+y)\,dx\,dy.$

This domain is perpendicular to the x and even to the y axis; to apply 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 this: $[a,b]=[0,1]\,\!$ (it has been chosen that the normality is with respect to the x axis for a better visual understanding).

It's now possible to apply the formulas:

$\iint _{D}(x+y)\,dx\,dy=\int _{0}^{1}dx\int _{x^{2}}^{1}(x+y)\,dy=\int _{0}^{1}dx\ \left[xy\ +\ {\frac {y^{2}}{2}}\ \right]_{x^{2}}^{1}$

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

$\int _{0}^{1}\left[xy\ +\ {\frac {y^{2}}{2}}\ \right]_{x^{2}}^{1}\,dx=\int _{0}^{1}\left(x+{\frac {1}{2}}-x^{3}-{\frac {x^{4}}{2}}\right)dx=\cdots ={\frac {13}{20}}.$

If you choose the normality respect the y axis you 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

Often, because the limits of integration are 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 is described in a simpler way through formulae. To do so, the function must be changed 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 dominion because it is delimitated by the original variables that were transformed before (x and y in example).
• the differentials (dx and dy as an example) depend on determining 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, is possible to hand with this method so as to conduct the substitution that appears more suitable.

Polar coordinates

See also: coordinates (mathematics)

Passage from cartesian to polar coordinates.

In R² if the domain has a circular "symmetry" (if it describes a circular crown) and the function has some "particular" characteristics you can apply the passate 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 follow:

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

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:

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

using the Pythagorean trigonometric identity (very useful to simplify this operation).

The transformation of the domain is made by defining the radius' crown's 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 $D=x^{2}+y^{2}\leq 4\,\!$, that is a circumference of radius 2; it's evident that the described angle is the circle angle, so φ varies from 0 to 2π, while the crown's radius varies from 0 to 2 (the crown with the inside radius null is just a circle).

Example (2-d):

The domain is $D=\{x^{2}+y^{2}\leq 9,\ x^{2}+y^{2}\geq 4,\ y\geq 0\}$, that is the circular crown in the semiplane of positive y (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:

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

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, because it 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

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

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's transformation is easy because graphically only the shape of the base varies while the three-dimensional development follows that of the starting region.

Example (3-a):

The region is $D=\{x^{2}+y^{2}\leq 9,\ x^{2}+y^{2}\geq 4,\ 0\leq z\leq 5\}$ (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: $T=\{2\leq \rho \leq 3,\ 0\leq \phi \leq \pi ,\ 0\leq z\leq 5\}$ (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 for the passage in 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: $D=\{x^{2}+y^{2}\leq 9,\ -5\leq z\leq 5\}$.

The transformation of D in cylindrical coordinates is the following:

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

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

$\int _{-5}^{5}dz\int _{0}^{2\pi }d\phi \int _{0}^{3}(\rho ^{3}+\rho z)\,d\rho =2\pi \int _{-5}^{5}\left[{\frac {\rho ^{4}}{4}}+{\frac {\rho ^{2}z}{2}}\right]_{0}^{3}\,dz$

$=2\pi \int _{-5}^{5}\left({\frac {81}{4}}+{\frac {9}{2}}z\right)\,dz=\cdots =2\pi \left({\frac {405}{2}}+225\right).$

Spherical coordinates

Spherical coordinates.

In R³ some domains have a spherical symmetry, so it's possible to determinate the coordinates of every point of the integration's 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:

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

Please note that points on z axis do not have a precise characterization in spherical coordinates, so θ can vary from 0 to π .

The better integration domain for this passage is obviously the sphere.

Example (4-a):

The domain is $D=x^{2}+y^{2}+z^{2}\leq 16$ (sphere with radius 4 and center in the origin); applying the transformation you get this region: $T=\{0\leq \rho \leq 4,\ 0\leq \phi \leq 2\pi ,\ 0\leq \theta \leq \pi \}.$

The Jacobian determinant of this transformation is the following:

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

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

Finally you obtain the final integration formula:

$\iiint _{D}f(x,y,z)\,dx\,dy\,dz=\iiint _{T}f(\rho \sin \theta \cos \phi ,\rho \sin \theta \sin \phi ,\rho \cos \theta )\rho ^{2}\sin \theta \,d\rho \,d\theta \,d\phi .$

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's better to use the passage in cylindrical coordinates (please see example 4-c).

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:

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

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 ($D=x^{2}+y^{2}+z^{2}\leq 9a^{2}\,\!$) 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 obvious:

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

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

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

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

$\int _{0}^{2\pi }d\phi \int _{0}^{3a}\rho ^{3}d\rho \int _{-{\sqrt {9a^{2}-\rho ^{2}}}}^{\sqrt {9a^{2}-\rho ^{2}}}\,dz=2\pi \int _{0}^{3a}2\rho ^{3}{\sqrt {9a^{2}-\rho ^{2}}}\,d\rho .$

Now let's apply the transformation

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

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

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

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

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

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:

$2\pi \left[\int _{0}^{9a^{2}}9a^{2}{\sqrt {t}}\,dt-\int _{0}^{9a^{2}}t{\sqrt {t}}\,dt\right]=2\pi \left[9a^{2}{\frac {2}{3}}t^{\frac {3}{2}}-{\frac {2}{5}}t^{\frac {5}{2}}\right]_{0}^{9a^{2}}$

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

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 - Calculations of volume

Thanks to the methods previously described it is possible to demonstrate the value of the volume of some solid volumes.

• Cylinder: Considering like domain the circular base of radius R and like function the constant of the height h, applies the passage in polar coordinates directly:

$\mathrm {Volume} =\int _{0}^{2\pi }d\phi \int _{0}^{r}h\rho \ d\rho =h2\pi \left[{\frac {\rho ^{2}}{2}}\right]_{0}^{R}=\pi R^{2}h$

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 to base 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 = ${\frac {\ell ^{2}}{2}}\cdot \ell /3={\frac {\ell ^{3}}{6}}.$

File:Dominio improprio.jpg

Example of an improper domain.

Multiple improper integral

In case of unbounded domains or unbounded integrands near the frontier of the dominion you have the double improper integral or the triple improper integral.

See also

• Integral
• Main analysis' theorems that uses multiple integrals:
  • Divergence theorem
  • Stokes theorem
  • Green's theorem

Bibliography

Sources verified on: Robert A. Adams - Calcolo differenziale 2, Funzioni di più variabili ISBN 88-408-1024-2