# Multiple Integrals

## Double Integrals Over Rectangles

For a rectangular region [latex]S[/latex] defined by [latex]x[/latex] in [latex][a,b][/latex] and [latex]y[/latex] in [latex][c,d][/latex], the double integral of a function [latex]f(x,y)[/latex] in this region is given as [latex]\int_c^d(\int_a^b f(x,y) dx) dy[/latex].### Learning Objectives

Use double integrals to find the volume of rectangular regions in the xy-plane### Key Takeaways

#### Key Points

- The multiple integral is a type of definite integral extended to functions of more than one real variable —for example, [latex]f(x, y)[/latex] or [latex]f(x, y, z)[/latex]. Integrals of a function of two variables over a region in [latex]R^2[/latex] are called double integrals.
- 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 [latex]z = f(x, y))[/latex] and the plane which contains its domain.
- If there are more variables than 3, a multiple integral will yield hypervolumes of multi-dimensional functions.

#### Key Terms

**Fubini's theorem**: a result which gives conditions under which it is possible to compute a double integral using iterated integrals**hypervolume**: a volume in more than three dimensions

*definite integral*extended to functions of more than one real variable—for example, [latex]f(x, y)[/latex] or [latex]f(x, y, z)[/latex]. Integrals of a function of two variables over a region in [latex]R^2[/latex] are called double 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 [latex]x[/latex]-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 [latex]z = f(x, y))[/latex] and the plane which contains its domain. The same volume can be obtained via the triple integral—the integral of a function in three variables—of the constant function [latex]f(x, y, z) = 1[/latex] 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.

### Double Integrals Over Rectangles

Double integrals over rectangular regions are straightforward to compute in many cases. For a rectangular region [latex]S[/latex] defined by [latex]x[/latex] in [latex][a,b][/latex] and [latex]y[/latex] in [latex][c,d][/latex], the double integral of a function [latex]f(x,y)[/latex] in this region is given as: [latex]\begin{align}\int\!\!\!\int_S f(x,y) dxdy &= \int_a^b\left(\int_c^d f(x,y) dy\right) dx \\ &= \int_c^d\left(\int_a^b f(x,y) dx\right) dy\end{align}[/latex]. Here, we exchanged the order of the integration, assuming that [latex]f(x,y)[/latex] satisfies the conditions to apply Fubini's theorem.### Example

Let us assume that we wish to integrate a multivariable function [latex]f[/latex] over a region [latex]A[/latex]: [latex-display]A = \left \{ (x,y) \in \mathbf{R}^2: 11 \le x \le 14 \; \ 7 \le y \le 10 \right \}[/latex-display] [latex-display]f(x,y) = x^2 + 4y[/latex-display] Formulating the double integral, we first evaluate the inner integral with respect to [latex]x[/latex]: [latex-display]\begin{align} \int_{11}^{14} (x^2 + 4y) \ dx & = \left (\frac{1}{3}x^3 + 4yx \right)\Big |_{x=11}^{x=14} \\ & = \frac{1}{3}(14)^3 + 4y(14) - \frac{1}{3}(11)^3 - 4y(11) \\ &= 471 + 12y \end{align}[/latex-display] We then integrate the result with respect to [latex]y[/latex]*:*[latex-display]\begin{align} \int_7^{10} (471 + 12y) \ dy & = (471y + 6y^2)\big |_{y=7}^{y=10} \\ & = 471(10)+ 6(10)^2 - 471(7) - 6(7)^2 \\ &= 1719 \end{align}[/latex-display] We could have computed the double integral starting from the integration over [latex]y[/latex]. Confirm yourself that the result is the same.

## Iterated Integrals

An iterated integral is the result of applying integrals to a function of more than one variable.### Learning Objectives

Use iterated integrals to integrate a function with more than one variable### Key Takeaways

#### Key Points

- The function [latex]f(x,y)[/latex], if [latex]y[/latex] is considered a given parameter, can be integrated with respect to [latex]x[/latex] as follows: [latex]\int f(x,y)dx[/latex].
- The result is a function of [latex]y[/latex] and therefore its integral can be considered again. If this is done, the result is the iterated integral [latex]\int\left(\int f(x,y)\,dx\right)\,dy[/latex].
- It is key to note that this is different, in principle, to the multiple integral [latex]\iint f(x,y)\,dx\,dy[/latex].

#### Key Terms

**Fubini's theorem**: a result which gives conditions under which it is possible to compute a double integral using iterated integrals

### Example

For the iterated integral [latex]\int\left(\int (x+y) \, dx\right) \, dy[/latex], the integral [latex]\int (x+y) \, dx = \frac{x^2}{2} + yx[/latex] is computed first. The result is then used to compute the integral with respect to [latex]y[/latex]: [latex-display]\displaystyle{\int \left(\frac{x^2}{2} + yx \right) \, dy = \frac{yx^2}{2} + \frac{xy^2}{2}}[/latex-display] It should be noted, however, that this example omits the constants of integration. After the first integration with respect to [latex]x[/latex], we would rigorously need to introduce a "constant" function of [latex]y[/latex]. That is, If we were to differentiate this function with respect to [latex]x[/latex], any terms containing only [latex]y[/latex] would vanish, leaving the original integral. Similarly for the second integral, we would introduce a "constant" function of [latex]x[/latex], because we have integrated with respect to [latex]y[/latex]. In this way, indefinite integration does not make much sense for functions of several variables. While the antiderivatives of single variable functions differ at most by a constant, the antiderivatives of multivariable functions differ by unknown single-variable terms, which could have a drastic effect on the behavior of the function.## Double Integrals Over General Regions

Double integrals can be evaluated over the integral domain of any general shape.### Learning Objectives

Use double integrals to integrate over general regions### Key Takeaways

#### Key Points

- If the domain [latex]D[/latex] is normal with respect to the [latex]x[/latex]- axis, and [latex]f:D \to R[/latex] is a continuous function, then [latex]\alpha(x)[/latex] and [latex]\beta(x)[/latex] (defined on the interval [latex][a, b][/latex]) are the two functions that determine [latex]D[/latex]: [latex]\iint_D f(x,y)\ dx\, dy = \int_a^b dx \int_{ \alpha (x)}^{ \beta (x)} f(x,y)\, dy[/latex]
- Applying this general method, the projection of [latex]D[/latex] onto either the [latex]x[/latex]-axis or the [latex]y[/latex]-axis should be bounded by the two values, [latex]a[/latex] and [latex]b[/latex].
- For a domain [latex]D = \{ (x,y) \in \mathbf{R}^2 \: \ x \ge 0, y \le 1, y \ge x^2 \}[/latex], we can write the integral over [latex]D[/latex] as[latex]\iint_D (x+y) \, dx \, dy = \int_0^1 dx \int_{x^2}^1 (x+y) \, dy[/latex].

#### Key Terms

**domain**: the set of all possible mathematical entities (points) where a given function is defined

- the projection of [latex]D[/latex] onto either the [latex]x[/latex]-axis or the [latex]y[/latex]-axis is bounded by the two values, [latex]a[/latex] and [latex]b[/latex]
*.* - 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, [latex]\alpha[/latex] and [latex]\beta[/latex].

### Example

Consider the following region: [latex-display]D = \{ (x,y) \in \mathbf{R}^2 \: \ x \ge 0, y \le 1, y \ge x^2 \}[/latex-display] Calculate [latex]\iint_D (x+y) \, dx \, dy[/latex]. This domain is normal with respect to both the [latex]x[/latex]- and [latex]y[/latex]-axes. To apply the formulae, you must first find the functions that determine [latex]D[/latex] and the intervals over which these are defined. In this case the two functions are [latex]\alpha (x) = x^2[/latex] and [latex]\beta (x) = 1[/latex], while the interval is given by the intersections of the functions with [latex]x=0[/latex], so the interval is [latex][a,b] = [0,1][/latex] (normality has been chosen with respect to the [latex]x[/latex]-axis for a better visual understanding). It is now possible to apply the formula: [latex]\begin{align}\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}\end{align}[/latex] (At first the second integral is calculated considering [latex]x[/latex] as a constant). The remaining operations consist of applying the basic techniques of integration: [latex]\begin{align}\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 \\ &= \frac{13}{20}\end{align}[/latex]## Double Integrals in Polar Coordinates

When domain has a cylindrical symmetry and the function has several specific characteristics, apply the transformation to polar coordinates.### Learning Objectives

Solve double integrals in polar coordinates### Key Takeaways

#### Key Points

- The fundamental relation to make the transformation is the following: [latex]f(x,y) \rightarrow f(\rho \cos \varphi,\rho \sin \varphi )[/latex].
- To switch the integral from Cartesian to polar coordinates, the [latex]dx \,\, dy[/latex] differentials in this transformation become [latex]\rho \,\, d\rho \,\,d\varphi[/latex].
- Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates: [latex]\iint_D f(x,y) \ dx\,\, dy = \iint_T f(\rho \cos \varphi, \rho \sin \varphi) \rho \,\, d \rho\,\, d \varphi[/latex].

#### Key Terms

**Cartesian**: of or pertaining to co-ordinates based on mutually orthogonal axes**Jacobian determinant**: the determinant of the Jacobian matrix

### Change of variable

The polar coordinates [latex]r[/latex] and [latex]\varphi[/latex] can be converted to the Cartesian coordinates [latex]x[/latex] and [latex]y[/latex] by using the trigonometric functions sine and cosine: [latex-display]x = r \cos \varphi \, \\ y = r \sin \varphi \,[/latex-display] The Cartesian coordinates [latex]x[/latex] and [latex]y[/latex] can be converted to polar coordinates [latex]r[/latex] and [latex]\varphi[/latex] with [latex]r \geq 0[/latex] and [latex]\varphi[/latex] in the interval [latex](−\pi, \pi][/latex]: [latex-display]r = \sqrt{x^2 + y^2} \\ \varphi = \tan ^{-1} (\frac yx)[/latex-display] The fundamental relation to make the transformation is as follows: [latex-display]f(x,y) \rightarrow f(\rho \cos \phi,\rho \sin \phi )[/latex-display]### Examples

- Given the function [latex]f(x,y) = x + y[/latex] and applying the transformation, one obtains [latex]f(\rho, \phi) = \rho \cos \phi + \rho \sin \phi = \rho(\cos \phi + \sin \phi )[/latex].
- Given the function [latex]f(x,y) = x^2 + y^2[/latex], one can obtain [latex]f(\rho, \phi) = \rho^2 (\cos^2 \phi + \sin^2 \phi) = \rho^2[/latex] using the Pythagorean trigonometric identity, which is very useful to simplify this operation. Particularly in this case, you can see that the representation of the function f became simpler in polar coordinates. This is the case because the function has a cylindrical symmetry. In general, the best practice is to use the coordinates that match the built-in symmetry of the function.

### Integrals in Polar Coordinates

The Jacobian determinant of that transformation is the following: [latex-display]\displaystyle{\frac{\partial (x,y)}{\partial (\rho, \phi)}} = \begin{vmatrix} \cos \phi & - \rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{vmatrix} = \rho[/latex-display] which has been obtained by inserting the partial derivatives of [latex]x = \rho \cos(\varphi)[/latex], [latex]y = \rho \sin(\varphi)[/latex] in the first column with respect to [latex]\rho[/latex] and in the second column with respect to [latex]\varphi[/latex], so the [latex]dx \, dy[/latex] differentials in this transformation become [latex]\rho \,d \rho \,d\varphi[/latex]. Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates: [latex-display]\iint_D f(x,y)dx \, dy = \iint_T f(\rho \cos \varphi, \rho \sin \varphi)\rho[/latex-display]### Example

Integrate the function [latex]f(x,y) = x[/latex] over the domain: [latex-display]D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ y \ge 0 \}[/latex-display] From [latex]f(x,y) = x \longrightarrow f(\rho,\phi) = \rho \cos \phi[/latex], [latex-display]\begin{align} \iint_D x \, dx\, dy &= \iint_T \rho \cos \phi \rho \, d\rho\, d\phi \\ &= \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 \end{align}[/latex-display]## Triple Integrals in Cylindrical Coordinates

When the function to be integrated has a cylindrical symmetry, it is sensible to integrate using cylindrical coordinates.### Learning Objectives

Evaluate triple integrals in cylindrical coordinates### Key Takeaways

#### Key Points

- Switching from Cartesian to cylindrical coordinates, the transformation of the function is made by the following relation [latex]f(x,y,z) \rightarrow f(\rho \cos \varphi, \rho \sin \varphi, z)[/latex].
- In switching to cylindrical coordinates, the [latex]dx\, dy\, dz[/latex] differentials in the integral become [latex]\rho \, d\rho \,d\varphi \,dz[/latex].
- Therefore, an integral evaluated in Cartesian coordinates can be switched to an integral in cylindrical coordinates as[latex]\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[/latex].

#### Key Terms

**differential**: an infinitesimal change in a variable, or the result of differentiation**cylindrical coordinate**: a three-dimensional coordinate system that specifies point positions by the distance from a chosen reference axis, the direction from the axis relative to a chosen reference direction, and the distance from a chosen reference plane perpendicular to the axis

^{3}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: [latex-display]f(x,y,z) \rightarrow f(\rho \cos \varphi, \rho \sin \varphi, z)[/latex-display] 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. Also in switching to cylindrical coordinates, the [latex]dx\, dy\, dz[/latex] differentials in the integral become [latex]\rho \, d\rho \,d\varphi \,dz[/latex].

### Example 1

The region is: [latex-display]D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ 0 \le z \le 5 \}[/latex-display] If the transformation is applied, this region is obtained: [latex-display]T = \{ 2 \le \rho \le 3, \ 0 \le \varphi \le 2\pi, \ 0 \le z \le 5 \}[/latex-display] because the*z*component is unvaried during the transformation, the [latex]dx\, dy\, dz[/latex] differentials vary as in the passage in polar coordinates: therefore, they become: [latex]\rho \, d\rho \,d\varphi \,dz[/latex]. Finally, it is possible to apply the final formula to cylindrical coordinates: [latex-display]\displaystyle{\iiint Df(x,y,z)dx\,dy\,dz=\iiint Tf(\rho\cos\varphi,\rho\sin\varphi,z)\rho\, d\rho \,d\varphi \,dz}[/latex-display] This method is convenient in case of cylindrical or conical domains or in regions where it is easy to individuate the [latex]z[/latex] interval and even transform the circular base and the function.

### Example 2

The function [latex]f(x,y,z) = x^2 + y^2 + z[/latex] is and as integration domain this cylinder: [latex-display]D = \{ x^2 + y^2 \le 9, \ -5 \le z \le 5 \}[/latex-display] The transformation of [latex]D[/latex] in cylindrical coordinates is the following: [latex-display]T = \{ 0 \le \rho \le 3, \ 0 \le \phi \le 2 \pi, \ -5 \le z \le 5 \}[/latex-display] while the function becomes: [latex-display]f(\rho \cos \varphi, \rho \sin \varphi, z) = \rho^2 + z[/latex-display] Therefore, the integral becomes: [latex]\begin{align}\displaystyle{\iiint_D (x^2 + y^2 +z) \, dx\, dy\, dz} &\displaystyle{= \iiint_T ( \rho^2 + z) \rho \, d\rho\, d\phi\, dz} \\ &= \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\\ &= 405 \pi \end{align}[/latex]## Triple Integrals in Spherical Coordinates

When the function to be integrated has a spherical symmetry, change the variables into spherical coordinates and then perform integration.### Learning Objectives

Evaluate triple integrals in spherical coordinates### Key Takeaways

#### Key Points

- Switching from Cartesian to spherical coordinates, the function is transformed by this relation: [latex]f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)[/latex].
- For the transformation, the [latex]dx\, dy\, dz[/latex] differentials in the integral are transformed to [latex]\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz[/latex].
- Therefore, an integral evaluated in Cartesian coordinates can be switched to an integral in spherical coordinates as [latex]\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.[/latex]

#### Key Terms

**spherical coordinate**: a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith**Jacobian determinant**: the determinant of the Jacobian matrix

### Integrals in Spherical Coordinates

The Jacobian determinant of this transformation is the following: [latex-display]\displaystyle{\frac{\partial (x,y,z)}{\partial (\rho, \theta, \varphi)}} = \begin{vmatrix} \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{vmatrix} = \rho^2 \sin \varphi[/latex-display] The [latex]dx\, dy\, dz[/latex] differentials therefore are transformed to [latex]\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz[/latex]. Finally, you obtain the final integration formula: 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 [latex]R^3[/latex]; in other cases it can be better to use cylindrical coordinates.### Example

Integrate [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex] over the domain [latex]D = x^2 + y^2 + z^2 \le 16[/latex]. In spherical coordinates: [latex-display]f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2[/latex-display] while the intervals of the transformed region [latex]T[/latex] from [latex]D[/latex]: [latex-display]0 \leq \rho \leq 4, 0 \leq \varphi \leq \pi, 0 \leq \theta \leq 2\pi[/latex-display] Therefore: [latex-display]\begin{align} \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, \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi}= \frac{4096 \pi}{5} \end{align}[/latex-display]## Triple Integrals

For [latex]T \subseteq R^3[/latex], the triple integral over [latex]T[/latex] is written as [latex]\iiint_T f(x,y,z)\, dx\, dy\, dz[/latex].### Learning Objectives

Use triple integrals to integrate over three-dimensional regions### Key Takeaways

#### Key Points

- By convention, the triple integral has three integral signs (and a double integral has two integral signs); this is a notational convention which is convenient when computing a multiple integral as an iterated integral.
- If [latex]T[/latex] is a domain that is normal with respect to the [latex]xy[/latex]-plane and determined by the functions [latex]\alpha (x,y)[/latex] and [latex]\beta(x,y)[/latex], then [latex]\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy[/latex].
- To integrate a function with spherical symmetry such as [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex], consider changing integration variable to spherical coordinates.

#### Key Terms

**spherical coordinate**: a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith

*xy*-plane and determined by the functions [latex]\alpha (x,y)[/latex] and [latex]\beta(x,y)[/latex], then: [latex-display]\displaystyle{\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy}[/latex-display]

### Example 1

The volume of the parallelepiped of sides 4 by 6 by 5 may be obtained in two ways:- By calculating the double integral of the function [latex]f(x, y) = 5[/latex] over the region [latex]D[/latex] in the [latex]xy[/latex]-plane which is the base of the parallelepiped: [latex]\iint_D 5 \ dx\, dy[/latex]
- By calculating the triple integral of the constant function 1 over the parallelepiped itself: [latex]\iiint_\mathrm{parallelepiped} 1 \, dx\, dy\, dz[/latex]

### Example 2

Integrate [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex] over the domain [latex]D = \left \{ x^2 + y^2 + z^2 \le 16 \right \}[/latex]. 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 [latex]T[/latex] region are obviously: [latex-display](0 \le \rho \le 4, \ 0 \le \phi \le \pi, \ 0 \le \theta \le 2 \pi)[/latex-display] For the function, we get: [latex-display]f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) = \rho^2[/latex-display] Therefore: [latex-display]\begin{align} \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 \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi} = \frac{4096 \pi}{5} \end{align}[/latex-display]## Change of Variables

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

Use a change a variables to rewrite an integral in a more familiar region### Key Takeaways

#### Key Points

- There exist three main "kinds" of changes of variable (to polar coordinate in [latex]R^2[/latex], and to cylindrical and spherical coordinates in [latex]R^3[/latex]); however, more general substitutions can be made using the same principle.
- When changing integration variables, make sure that the integral domain also changes accordingly.
- Change of variable should be judiciously applied based on the built-in symmetry of the function to be integrated.

#### Key Terms

**polar coordinate**: a two-dimensional coordinate system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction

### 1. Polar coordinates

The function to be integrated transforms as: [latex-display]f(x,y) \rightarrow f(\rho \cos \phi,\rho \sin \phi )[/latex-display] and the integral accordingly changes as: [latex]\displaystyle {\iint_D f(x,y) \ dx\, dy = \iint_T f(\rho \cos \phi, \rho \sin \phi) \rho \, d \rho\, d \phi}[/latex]### 2. Cylindrical coordinates

The function to be integrated transforms as: [latex-display]f(x,y,z) \rightarrow f(\rho \cos \phi, \rho \sin \phi, z)[/latex-display] and the integral accordingly changes as: [latex]\displaystyle {\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}[/latex]### 3. Spherical coordinates

The function to be integrated transforms as: [latex-display]f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \phi, \rho \sin \theta \sin \phi, \rho \cos \phi)[/latex-display] and the integral accordingly changes as: [latex]\displaystyle {\iiint_D f(x,y,z) \, dx\, dy\, dz \\ = \iiint_T f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi}[/latex]## Applications of Multiple Integrals

Multiple integrals are used in many applications in physics and engineering.### Learning Objectives

Apply multiple integrals to real world examples### Key Takeaways

#### Key Points

- Given a set [latex]D \subseteq R^n[/latex] and an integrable function [latex]f[/latex] over [latex]D[/latex], the average value of [latex]f[/latex] over its domain is given by [latex]\bar{f} = \frac{1}{m(D)} \int_D f(x)dx[/latex], where [latex]m(D)[/latex] is the measure of [latex]D[/latex].
- The gravitational potential associated with a mass distribution given by a mass measure [latex]dm[/latex] on three-dimensional Euclidean space [latex]R^3[/latex] is [latex]V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x} - \mathbf{r}|}\,dm(\mathbf{r})[/latex].
- An electric field produced by a distribution of charges given by the volume charge density [latex]\rho (\vec r)[/latex] is obtained by a triple integral of a vector function: [latex]\vec E = \frac {1}{4 \pi \epsilon_0} \iiint \frac {\vec r - \vec r'}{\| \vec r - \vec r' \|^3} \rho (\vec r')\, {d}^3 r'[/latex].

#### Key Terms

**Maxwell's equations**: a set of partial differential equations that, together with the Lorentz force law, form the foundation of classical electrodynamics, classical optics, and electric circuits**moment of inertia**: a measure of a body's resistance to a change in its angular rotation velocity

### Example 1

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: [latex-display]\displaystyle{I_z = \iiint_V \rho r^2\, dV}[/latex-display]### Example 2

The gravitational potential associated with a mass distribution given by a mass measure [latex]dm[/latex] on three-dimensional Euclidean space [latex]R^3[/latex] is: [latex-display]\displaystyle{V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{\left|\mathbf{x} - \mathbf{r}\right|}\,dm(\mathbf{r})}[/latex-display] If there is a continuous function [latex]\rho(x)[/latex] representing the density of the distribution at [latex]x[/latex], so that [latex]dm(x) = \rho (x)d^3x[/latex], where [latex]d^3x[/latex] is the Euclidean volume element, then the gravitational potential is: [latex-display]\displaystyle{V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{\left|\mathbf{x}-\mathbf{r}\right|}\,\rho(\mathbf{r})\,d^3\mathbf{r}}[/latex-display]### Example 3

In electromagnetism, Maxwell's equations can be written using multiple integrals to calculate the total magnetic and electric fields. In the following example, the electric field produced by a distribution of charges given by the volume charge density [latex]\rho (\vec r)[/latex] is obtained by a triple integral of a vector function: [latex-display]\displaystyle{\vec E = \frac {1}{4 \pi \epsilon_0} \iiint \frac {\vec r - \vec r'}{\| \vec r - \vec r' \|^3} \rho (\vec r')\, {d}^3 r'}[/latex-display] This can also be written as an integral with respect to a signed measure representing the charge distribution.## Center of Mass and Inertia

The center of mass for a rigid body can be expressed as a triple integral.### Learning Objectives

Use multiple integrals to find the center of mass of a distribution of mass### Key Takeaways

#### Key Points

- In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid.
- In the case of a system of particles [latex]P_i, i = 1, \cdots, n[/latex], each with mass [latex]m_i[/latex] that are located in space with coordinates [latex]\mathbf{r}_i, i = 1, \cdots, n[/latex], the coordinates [latex]\mathbf{R}[/latex] of the center of mass is given as [latex]\mathbf{R} = \frac{1}{M} \sum_{i=1}^n m_i \mathbf{r}_i[/latex].
- If the mass distribution is continuous with the density [latex]\rho (r)[/latex] within a volume [latex]V[/latex], the center of mass is expressed as [latex]\mathbf R = \frac 1M \int_V\rho(\mathbf{r}) \mathbf{r} dV[/latex].

#### Key Terms

**rigid body**: an idealized solid whose size and shape are fixed and remain unaltered when forces are applied; used in Newtonian mechanics to model real objects**centroid**: the point at the center of any shape, sometimes called the center of area or the center of volume

### A System of Particles

In the case of a system of particles [latex]P_i, i = 1, \cdots, n[/latex], each with mass [latex]m_i[/latex] that are located in space with coordinates [latex]\mathbf{r}_i, i = 1, \cdots, n[/latex], the coordinates [latex]\mathbf{R}[/latex] of the center of mass satisfy the condition: [latex-display]\displaystyle{\sum_{i=1}^n m_i(\mathbf{r}_i - \mathbf{R}) = 0}[/latex-display] Solve this equation for [latex]\mathbf{R}[/latex] to obtain the formula: [latex-display]\displaystyle{\mathbf{R} = \frac{1}{M} \sum_{i=1}^n m_i \mathbf{r}_i}[/latex-display] where [latex]M[/latex] is the sum of the masses of all of the particles.### A Continuous Volume

If the mass distribution is continuous with the density [latex]\rho (r)[/latex] within a volume [latex]V[/latex], then the integral of the weighted position coordinates of the points in this volume relative to the center of mass [latex]\mathbf{R}[/latex] is zero; that is: [latex-display]\displaystyle{\int_V \rho(\mathbf{r})(\mathbf{r}-\mathbf{R})dV = 0}[/latex-display] Solve this equation for the coordinates [latex]\mathbf{R}[/latex] to obtain: [latex-display]\displaystyle{\mathbf R = \frac 1M \int_V\rho(\mathbf{r}) \mathbf{r} dV}[/latex-display] where [latex]M[/latex] is the total mass in the volume. The integral is over the three dimensional volume, so it is a triple integral.## Licenses & Attributions

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